4.5 Continuity between metric spaces

Replace the two absolute values in the epsilon-delta definition by two distances and you have continuity for maps between any metric spaces. It is the same as continuity along sequences, and the same as 'preimages of open sets are open', which needs no distance at all and is where topology begins. Several variables bring new failures (limits that depend on the path), and homeomorphisms say when two spaces have the same shape.

The definition of continuity in 1.3 used the real line twice: once for the input, as the distance $|x - a|$ , and once for the output, as $|f(x) - f(a)|$ . Both are distances. The proofs used only the four properties of distance listed in 4.1, so the definition makes sense for a map from any metric space to any other: a map from the plane to the line, from the space of functions to the line (the integral), from bit strings to numbers. This chapter writes down that definition, proves it is the same as continuity along sequences, and then proves the surprising third form: a map is continuous exactly when preimages of open sets are open. The third form mentions no number $\varepsilon$ or $\delta$ and no distance, only open sets. That is the form topology (MATH 551) takes as the definition, and the reason 4.1 spent time on unions and finite intersections of open sets.

Before you start

1.3: continuity with $\varepsilon$ and $\delta$ and the order of the quantifiers. 4.1: open sets, balls, and that balls are open. 4.2: convergent sequences in a metric space. 4.3 and 4.4: compact and connected sets, which continuity preserves.

Quick checkIn the definition of continuity at

a

, who chooses

\delta

and when?

Quick checkIs

(1, 4) \cup (6, 7)

an open subset of

\mathbb R

Quick checkIn a metric space, what does

x_n \to x

mean?

The picture: preimages of open intervals

A map $f$ takes inputs to outputs. An open interval $V$ on the output axis is a region with room around each of its points. The set of all inputs that land in $V$ is the preimage of $V$ , drawn in orange on the input axis. Three things to do:

Start with $f(x) = x^2$ and slide the centre and radius of $V$ . The preimage is always one or two open intervals, never including an endpoint. An endpoint would be sent to the edge of $V$ , and $V$ has no edge points.
Switch to the jump function and put $V$ around $1$ only. The preimage is $[0, \infty)$ , which contains $0$ . But $0$ has no room on its left. The jump has torn the line, and the preimage fails to be open.
Put $V$ around $-1$ only. Now the preimage is $(-\infty, 0)$ , open. A discontinuous function can have some good preimages. Continuity means ALL open sets have open preimages, and one bad $V$ is enough to fail.

Where this came from· Lipschitz, and a definition without numbers

In the 1800s Augustin-Louis Cauchy proved that the equation $y' = f(x, y)$ with a starting value has a solution, and Rudolf Lipschitz later isolated the condition on $f$ that makes the solution unique: the output changes by at most a fixed multiple of the change in $y$ . That condition, now called a Lipschitz condition, is a quantitative form of continuity, and the multiplier is the contraction constant that 4.2 used to prove the existence theorem. Separately, Felix Hausdorff’s 1914 book defined spaces by open sets alone, with no distance. In a space like that the $\varepsilon$ - $\delta$ definition is unavailable, and the only way to say that a map does not tear the space is that preimages of open sets are open. For metric spaces the two definitions agree, which is the first theorem below. The problem both solved was to say what continuous means without relying on the real line.

The idea, in words

Why· Three ways to say 'the map does not tear'

With numbers ( $\varepsilon$ - $\delta$ ). Points that are close go to points that are close, with a recipe for how close. You can compute with this.
With sequences. If a sequence of inputs converges to $a$ , the outputs converge to $f(a)$ . You can test with this: one bad sequence disproves continuity.
With open sets. If you fence off an open region in the output, the inputs that land in it form an open region. You can prove with this: it works cleanly with unions, intersections and composition, and needs no distance.

The theorem of this chapter says the three are the same property. The first is the one you reach for to find a $\delta$ . The second is the one that disproves continuity quickly. The third is the one that will carry over to spaces that have open sets and no distance.

Definitions and theorems

Definition B

Limit of a map between metric spaces

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, $A \subseteq X$ , $f : A \to Y$ , and let $a$ be a limit point of $A$ (a point of $X$ , which may or may not be in $A$ ). Then $L \in Y$ is the limit of $f$ as $x \to a$ , written $\lim_{x \to a} f(x) = L$ , if

\forall \varepsilon > 0\ \ \exists \delta > 0\ \ \forall x \in A:\quad 0 < d_X(x, a) < \delta \implies d_Y(f(x), L) < \varepsilon.

The condition $0 < d_X(x, a)$ excludes $x = a$ , as in 1.3. Taking $X = Y = \mathbb R$ recovers that definition.

In one line $\lim_{x \to a} f(x) = L$ if for every $\varepsilon > 0$ there is $\delta > 0$ such that $d_Y(f(x), L) < \varepsilon$ whenever $0 < d_X(x, a) < \delta$ .

Why it existsIt is the limit of 1.3 with $|x - a|$ and $|f(x) - L|$ replaced by the distances of the two spaces, so it makes sense for maps from vectors, functions or bit strings.

Definition B

Continuous map between metric spaces

A map $f : X \to Y$ between metric spaces is continuous at $a \in X$ if

\forall \varepsilon > 0\ \ \exists \delta > 0\ \ \forall x \in X:\quad d_X(x, a) < \delta \implies d_Y(f(x), f(a)) < \varepsilon.

It is continuous if it is continuous at every point of $X$ . In words of balls: for every ball $B_Y(f(a), \varepsilon)$ there is a ball $B_X(a, \delta)$ that $f$ sends inside it.

In one line $f : X \to Y$ is continuous at $a$ if for every $\varepsilon > 0$ there is $\delta > 0$ such that $d_Y(f(x), f(a)) < \varepsilon$ whenever $d_X(x, a) < \delta$ ; it is continuous if it is continuous at every point.

Why it existsIt says nearby inputs have nearby outputs, in whatever way each space measures nearness.

If $a$ is a limit point of $X$ , then $f$ is continuous at $a$ exactly when $\lim_{x \to a} f(x) = f(a)$ , as in 1.3. If $a$ is an isolated point (a ball $B(a, \delta)$ holds no other point of $X$ ), continuity at $a$ is automatic: with that $\delta$ the only $x$ allowed is $a$ itself. In particular every map out of a discrete metric space is continuous.

Theorem B Rudin Thm 4.2 and 4.8; Abbott Thm 4.3.2; Munkres Thm 18.1

Three faces of continuity

For a map $f : X \to Y$ between metric spaces, the following are equivalent.

(a) $f$ is continuous ( $\varepsilon$ - $\delta$ ) at every point.

(b) For every $a \in X$ and every sequence $x_n \to a$ in $X$ , $f(x_n) \to f(a)$ in $Y$ .

(d) For every closed set $F \subseteq Y$ , the preimage $f^{-1}(F)$ is closed in $X$ .

In one lineFor $f : X \to Y$ between metric spaces: $f$ is continuous (with $\varepsilon$ and $\delta$ ) iff $x_n \to a$ implies $f(x_n) \to f(a)$ iff the preimage of every open set is open iff the preimage of every closed set is closed.

Why it is trueSequences and numbers are linked by taking $\delta = 1/n$ ; numbers and open sets by replacing a ball in the target by a ball in the source; open and closed by taking complements.

ProofRudin 4.8; Munkres 18.1

The idea: (a)⇒(b) and (b)⇒(a): the sequential criterion of 1.3, with distances. (a)⇒(c): a ball around f(a) inside W pulls back to a ball around a. (c)⇒(a): apply (c) to the ball B(f(a), ε), which is open. (c)⇔(d): complements.

Assume (a). Let

x_n \to a

and

\varepsilon > 0

. Choose

\delta > 0

with

d_Y(f(x), f(a)) < \varepsilon

whenever

d_X(x, a) < \delta

(a) ⇒ (b). Continuity at

a

gives a

\delta

for each

\varepsilon

Choose

N

with

d_X(x_n, a) < \delta

for

n > N

. Then

d_Y(f(x_n), f(a)) < \varepsilon

for

n > N

, so

f(x_n) \to f(a)

x_n \to a

with tolerance

\delta

Suppose

f

is not continuous at

a

. There is

\varepsilon > 0

such that for every

\delta > 0

some

x

has

d_X(x, a) < \delta

and

d_Y(f(x), f(a)) \ge \varepsilon

(b) ⇒ (a), by contrapositive. Negate the definition of continuity at

a

: there is some

\varepsilon > 0

that no

\delta

answers.

Choose

x_n

with

d_X(x_n, a) < \frac1n

and

d_Y(f(x_n), f(a)) \ge \varepsilon

. Then

x_n \to a

but

f(x_n) \not\to f(a)

, so (b) fails.

Use

\delta = \frac1n

for each

n

and pick the bad point

x_n

Assume (a). Choose

\varepsilon > 0

with

B_Y(f(a), \varepsilon) \subseteq W

, then

\delta > 0

from continuity at

a

(a) ⇒ (c). Take

W

open in

Y

and a point

a \in f^{-1}(W)

. Then

f(a) \in W

and

W

is open, so a ball around

f(a)

lies in

W

B_X(a, \delta) \subseteq f^{-1}(W)

. Every point of

f^{-1}(W)

has a ball inside it:

f^{-1}(W)

is open.

d_X(x, a) < \delta

then

d_Y(f(x), f(a)) < \varepsilon

, so

f(x) \in W

Assume (c). The preimage

f^{-1}(B_Y(f(a), \varepsilon))

is open in

X

and contains

a

a

and

\varepsilon > 0

. The ball

B_Y(f(a), \varepsilon)

is open (4.1: open balls are open).

So there is

\delta > 0

with

B_X(a, \delta) \subseteq f^{-1}(B_Y(f(a), \varepsilon))

. That says

d_X(x, a) < \delta

implies

d_Y(f(x), f(a)) < \varepsilon

: (a) holds.

Definition of open: some ball around

a

lies inside it.

f^{-1}(Y \setminus F) = X \setminus f^{-1}(F)

F \subseteq Y

, a point

x

is in

f^{-1}(Y \setminus F)

iff

f(x) \notin F

iff

x \notin f^{-1}(F)

Hence (c) and (d) are equivalent.

\blacksquare

A set is closed iff its complement is open. If (c) holds and

F

is closed,

Y \setminus F

is open, so

f^{-1}(Y \setminus F)

is open, so its complement

f^{-1}(F)

is closed. Conversely, the same with open and closed swapped.

In the picture: (a) is the window of inputs in the first visualizer, (c) is the orange set with its open ends. For the jump function with $V$ around $1$ , the orange set is $[0, \infty)$ , so (c) fails, and (b) fails too: the sequence $x_n = -\frac1n \to 0$ has $f(x_n) = -1$ , not tending to $f(0) = 1$ .

Proof workshopRebuild (c) implies (a): open preimages give epsilon-delta

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Worked example

Show that if

f : \mathbb R \to \mathbb R

is continuous and

f(q) = 0

for every rational

q

, then

f(x) = 0

for every real

x

Solution.

Take any real $x$ . We must show $f(x) = 0$ . We know $f$ only on the rationals, so we reach $x$ through them.
Approach $x$ by rationals. Choose rationals $q_n$ with $|q_n - x| < \frac1n$ . Then $q_n \to x$ . Every interval contains a rational number (density of $\mathbb Q$ ).
Use the sequence form of continuity. By form (b) of the three-faces theorem, $f(q_n) \to f(x)$ . Continuity at $x$ sends convergent sequences to convergent sequences.
Read off the limit. Every $f(q_n) = 0$ , so the sequence $f(q_n)$ is constantly $0$ , and its limit is $0$ . Limits are unique, so $f(x) = 0$ . A constant sequence converges to that constant, and a sequence has only one limit (4.2).

Answer. $f$ is zero everywhere. The same argument shows that two continuous functions that agree on a dense set agree everywhere. $\blacksquare$

Quick checkLet

f(x) = x^2

as a map

\mathbb R \to \mathbb R

. Which of these sets is NOT open?

Theorem B Rudin Thm 4.7 and 4.9, 4.10; Apostol Sec. 8.4 to 8.5; MATH 375 Lecture 19

Building continuous maps

(a) If $f : X \to Y$ and $g : Y \to Z$ are continuous, so is $g \circ f$ .

(b) If $f, g : X \to \mathbb R$ are continuous and $c \in \mathbb R$ , then $f + g$ , $cf$ and $fg$ are continuous, and $f/g$ is continuous on the set where $g \ne 0$ .

(c) A map $F = (F_1, \dots, F_m) : X \to \mathbb R^m$ is continuous iff each $F_i : X \to \mathbb R$ is. The projection $\pi_i : \mathbb R^n \to \mathbb R$ , $\pi_i(x) = x_i$ , is continuous.

In one lineCompositions of continuous maps are continuous; sums, multiples, products and quotients (with non-zero denominator) of continuous real functions are continuous; a map into $\mathbb R^m$ is continuous iff each of its $m$ coordinate functions is; coordinate projections are continuous.

Why it is trueComposition: preimages of open sets pull back twice. Algebra: pass through sequences and use the limit laws of 1.2. Coordinates: convergence in $\mathbb R^m$ is coordinatewise (4.2).

ProofRudin 4.7 to 4.10

The idea: (a) with open sets. (b) with sequences, since 1.2 proved the limit laws. (c) with sequences, since convergence in R^m is coordinatewise.

(a)

(g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W))

Use form (c) of the previous theorem. For

W

open in

Z

g^{-1}(W)

is open in

Y

because

g

is continuous.

Hence

g \circ f

is continuous. Compare the

\varepsilon

\delta

proof, which needs two nested choices of

\delta

Then

f^{-1}

of that open set is open in

X

because

f

is continuous. So the preimage of every open

W

is open.

By the algebraic limit laws of 1.2,

f(x_n) + g(x_n) \to f(a) + g(a)

c f(x_n) \to c f(a)

and

f(x_n)g(x_n) \to f(a)g(a)

(b) Let

x_n \to a

X

. By form (b) of the previous theorem,

f(x_n) \to f(a)

and

g(x_n) \to g(a)

The same for

f(x_n)/g(x_n) \to f(a)/g(a)

when

g(a) \ne 0

. By form (b) again, the sums, products and quotients are continuous.

The quotient law of 1.2 requires

g(a) \ne 0

and then

g(x_n) \ne 0

eventually; on the set

\{g \ne 0\}

we use that set as the space.

F

satisfies (b) of the previous theorem iff each

F_i

does.

F(x_n) \to F(a)

\mathbb R^m

iff

F_i(x_n) \to F_i(a)

for every

i

The projection

\pi_i

is continuous.

\blacksquare

|\pi_i(x) - \pi_i(y)| = |x_i - y_i| \le \lVert x - y \rVert_\infty \le \lVert x - y \rVert_2

(the comparison theorem of 4.1), so

\delta = \varepsilon

works.

Together these say: every polynomial in the coordinates is continuous on $\mathbb R^n$ (built from projections, sums and products), every rational function is continuous where its denominator is non-zero, and every composition of such with $\sin$ , $\exp$ , $\log$ and the other functions of 1.3 is continuous on the set where it is defined. In JEE you “knew” this. Here it is a theorem with a proof.

Worked example MATH 375 HW 11 (Apostol 8.5 #1 b, f, j)

Find the set where each function is continuous: (a)

\log(x^2 + y^2)

. (b)

\arcsin\dfrac{x}{\sqrt{x^2 + y^2}}

. (c)

\arccos\sqrt{x/y}

Solution.

(a) Find the domain, then compose. $x^2 + y^2$ is a polynomial, continuous everywhere, and positive except at the origin. $\log$ is continuous on $(0, \infty)$ . By the composition rule, $\log(x^2 + y^2)$ is continuous at every point where it is defined, the set $\mathbb R^2 \setminus \{(0, 0)\}$ .
(b) The inner function. $u(x, y) = x/\sqrt{x^2 + y^2}$ is a quotient of continuous functions with non-zero denominator, so it is continuous on $\mathbb R^2 \setminus \{(0, 0)\}$ . Its values satisfy $|u| \le 1$ because $|x| \le \sqrt{x^2 + y^2}$ . The domain of the whole function is where the inner function is defined, which is everywhere except the origin; $\arcsin$ needs its input in $[-1, 1]$ , and that always holds.
(b) The outer function. $\arcsin$ is continuous on the CLOSED interval $[-1, 1]$ , including the endpoints $\pm 1$ . So the composite is continuous at every point except the origin, including the points $(x, 0)$ with $x \ne 0$ , where $u = \pm 1$ and the value is $\pm\frac\pi2$ . A tempting answer, “discontinuous along $y = 0$ ”, is wrong: $\arcsin$ has no jump at $\pm 1$ .
(c) The domain. $\sqrt{x/y}$ needs $y \ne 0$ and $x/y \ge 0$ , and $\arccos$ needs $\sqrt{x/y} \le 1$ , that is $x/y \le 1$ . So the domain is $\{y \ne 0,\ 0 \le x/y \le 1\}$ . We find where the function is defined before asking where it is continuous.
(c) Continuity on the domain. $x/y$ is continuous for $y \ne 0$ ; $\sqrt{\ }$ is continuous on $[0, \infty)$ ; $\arccos$ is continuous on $[-1, 1]$ . So the function is continuous at every point of its domain, including the boundary cases $x = 0$ and $x = y$ . Again the outer functions are continuous on closed intervals, so no extra points are lost.

Answer. (a) Continuous on $\mathbb R^2 \setminus \{0\}$ , its whole domain. (b) Continuous on $\mathbb R^2 \setminus \{0\}$ , its whole domain. (c) Continuous on its whole domain $\{y \ne 0,\ 0 \le x/y \le 1\}$ . $\blacksquare$

Worked example MATH 375 HW 11 (Apostol 8.3 #1c, #2m)

Show that

S = \{(x, y) : (x^2 + y^2 - 1)(4 - x^2 - y^2) > 0\}

is open, and describe it. Show that each level set

\{(x, y) : \cos(x + y) = c\}

is closed.

Solution.

Write $S$ as a preimage. Let $F(x, y) = (x^2 + y^2 - 1)(4 - x^2 - y^2)$ , a polynomial, so continuous on $\mathbb R^2$ . Then $S = F^{-1}((0, \infty))$ . $S$ is the set of inputs whose output lies in the open interval $(0, \infty)$ .
Apply the theorem. $(0, \infty)$ is open in $\mathbb R$ , so $S$ is open in $\mathbb R^2$ by form (c). This replaces a ball-by-ball argument at every point of $S$ .
Describe $S$ . A product is positive when both factors are positive or both negative. Both positive: $x^2 + y^2 > 1$ and $x^2 + y^2 < 4$ , that is $1 < r < 2$ with $r$ the distance from the origin. Both negative: $x^2 + y^2 < 1$ and $x^2 + y^2 > 4$ , impossible. So $S$ is the open annulus between the circles of radius $1$ and $2$ about the origin. (The posted key sketched two overlapping discs of unequal size, which does not match this $F$ .)
The boundary. $\partial S$ is the two circles $r = 1$ and $r = 2$ , where $F = 0$ . Points where $F = 0$ are in the closure of $S$ but not in $S$ .
Level sets. $g(x, y) = \cos(x + y)$ is continuous, and $\{c\}$ is closed in $\mathbb R$ , so $g^{-1}(\{c\})$ is closed by form (d). Since $\cos(x + y)$ depends only on $x + y$ , the level set is a family of parallel lines $x + y = \text{constant}$ of slope $-1$ , for example $x + y = \frac\pi2 + m\pi$ for $c = 0$ .

Answer. $S$ is the open annulus $1 < x^2 + y^2 < 4$ , which is open, with boundary the two circles. Every level set of $\cos(x + y)$ is closed (a family of parallel lines). $\blacksquare$ The rule: a set defined by strict inequalities between continuous functions is open, and by non-strict inequalities or equations is closed.

Worked example MATH 375 Lecture 19; Apostol 8.3

Show that a product of open sets is open: if

A_1, A_2 \subseteq \mathbb R

are open, so is

A_1 \times A_2 \subseteq \mathbb R^2

Solution.

Write the product with projections. $A_1 \times A_2 = \pi_1^{-1}(A_1) \cap \pi_2^{-1}(A_2)$ . A point $(x, y)$ is in the product iff $x \in A_1$ and $y \in A_2$ , which is the same as $\pi_1(x, y) \in A_1$ and $\pi_2(x, y) \in A_2$ .
Each piece is open. The projections are continuous, and $A_1$ , $A_2$ are open, so $\pi_1^{-1}(A_1)$ and $\pi_2^{-1}(A_2)$ are open (form (c)). These are strips: $A_1 \times \mathbb R$ and $\mathbb R \times A_2$ .
A finite intersection of open sets is open. So $A_1 \times A_2$ is open. The theorem of 4.1.

Answer. $A_1 \times A_2$ is open. $\blacksquare$ The proof in Spagnolie’s lecture gets the same result by hand: take $r_1, r_2$ with $B_{r_1}(a_1) \subseteq A_1$ and $B_{r_2}(a_2) \subseteq A_2$ , let $r = \min(r_1, r_2)$ ; a point of the disc $B_r(a)$ has $|x - a_1| < r \le r_1$ and $|y - a_2| < r \le r_2$ , so it is in $A_1 \times A_2$ . The minimum is the finite-intersection step of 4.1 again.

Lipschitz maps

Definition B

Lipschitz map

A map $f : X \to Y$ between metric spaces is Lipschitz if there is a constant $L \ge 0$ with $d_Y(f(x), f(y)) \le L\,d_X(x, y)$ for all $x, y \in X$ . The smallest such $L$ is the Lipschitz constant of $f$ . A Lipschitz map with $L < 1$ is a contraction in the sense of 4.2.

In one line $f : X \to Y$ is Lipschitz with constant $L$ if $d_Y(f(x), f(y)) \le L\,d_X(x, y)$ for all $x, y$ : the output distance is at most $L$ times the input distance.

Why it existsIt is continuity with an explicit recipe, $\delta = \varepsilon / L$ , valid at every point at once.

Theorem B standard

Lipschitz implies uniformly continuous

If $f : X \to Y$ is Lipschitz with constant $L$ , then $f$ is uniformly continuous: for every $\varepsilon > 0$ , $\delta = \varepsilon / (L + 1)$ satisfies $d_X(x, y) < \delta \implies d_Y(f(x), f(y)) < \varepsilon$ for all $x, y$ .

In one lineA Lipschitz map is uniformly continuous, hence continuous.

Why it is trueIf distances are multiplied by at most $L$ , the tolerance $\delta = \varepsilon / L$ works at every point of the space.

Proofstandard

The idea: Multiply the input distance by L; choose δ so the result is below ε. The +1 only avoids dividing by zero when L = 0.

Let

\varepsilon > 0

and

\delta = \varepsilon / (L + 1)

Arbitrary tolerance;

\delta

depends on

\varepsilon

and

L

but not on

x

y

d_X(x, y) < \delta

then

d_Y(f(x), f(y)) \le L\,d_X(x, y) \le L\delta < \varepsilon

\blacksquare

The Lipschitz inequality, then

\delta (L) < \varepsilon

since

L/(L+1) < 1

Worked example links to 3.5

Norms, projections and distance-from-a-point are 1-Lipschitz; a matrix is Lipschitz with constant its operator norm. For

A = \begin{bmatrix} 3 & 0 \\ 4 & 5 \end{bmatrix}

find the constant and a

\delta

for

\varepsilon = 0.1

Solution.

A norm. By the triangle inequality $\lVert x \rVert \le \lVert x - y \rVert + \lVert y \rVert$ , so $\lVert x \rVert - \lVert y \rVert \le \lVert x - y \rVert$ ; swapping $x$ and $y$ gives $|\lVert x \rVert - \lVert y \rVert| \le \lVert x - y \rVert$ . This reverse triangle inequality says the norm, as a map $V \to \mathbb R$ , is Lipschitz with $L = 1$ .
The same for distance from a point. For fixed $p$ , $h(x) = d(x, p)$ has $|h(x) - h(y)| \le d(x, y)$ . The reverse triangle inequality of 4.1: $d(x, p) \le d(x, y) + d(y, p)$ , and symmetric.
A projection. $|\pi_i(x) - \pi_i(y)| = |x_i - y_i| \le \lVert x - y \rVert_2$ . One coordinate is at most the 2-norm (4.1).
A matrix. For $T(x) = Ax$ on $\mathbb R^2$ with the usual length, $\lVert Ax - Ay \rVert = \lVert A(x - y) \rVert \le \lVert A \rVert\,\lVert x - y \rVert$ . The defining property of the operator norm of 3.5, applied to the vector $x - y$ and the linearity of $A$ .
Compute $\lVert A \rVert$ . $A^{\mathsf T}A = \begin{bmatrix} 25 & 20 \\ 20 & 25 \end{bmatrix}$ has eigenvalues $45$ and $5$ , so $\sigma_1 = \sqrt{45} = 3\sqrt5 \approx 6.708$ . The operator norm is the largest singular value, the square root of the largest eigenvalue of $A^{\mathsf T}A$ .
Find $\delta$ . $\delta = \varepsilon / \lVert A \rVert = 0.1 / 6.708 \approx 0.0149$ . Then $\lVert x - y \rVert < \delta$ gives $\lVert Ax - Ay \rVert < \lVert A \rVert \delta = 0.1$ , at every pair of points at once.

Answer. $L = 3\sqrt5$ and $\delta \approx 0.0149$ work for the whole plane. $\blacksquare$

Your answerWhat is the Lipschitz constant (operator norm) of

T(x) = Ax

for

A = \begin{bmatrix} 3 & 0 \\ 4 & 5 \end{bmatrix}

with the usual length? (Give a decimal.)

Quick checkWhich of these maps

\mathbb R \to \mathbb R

is NOT Lipschitz?

Uniform continuity between metric spaces

Uniform continuity was defined in 1.3 for functions on the line. The same definition, with distances, makes sense for maps between metric spaces. We do not define a new term; we state the general version and its main facts.

Theorem B Rudin Thm 4.19 and Ex. 4.13; Abbott Sec. 4.4; standard

Uniform continuity between metric spaces

Let $f : X \to Y$ be a map between metric spaces. It is uniformly continuous if

\forall \varepsilon > 0\ \ \exists \delta > 0\ \ \forall x, y \in X:\quad d_X(x, y) < \delta \implies d_Y(f(x), f(y)) < \varepsilon.

Then:

(a) Lipschitz implies uniformly continuous, and uniformly continuous implies continuous.

(b) If $(x_n)$ is Cauchy in $X$ and $f$ is uniformly continuous, then $(f(x_n))$ is Cauchy in $Y$ .

In one lineA map $f : X \to Y$ is uniformly continuous when for every $\varepsilon > 0$ one $\delta > 0$ works for all pairs of points. Lipschitz maps are uniformly continuous; uniformly continuous maps are continuous and send Cauchy sequences to Cauchy sequences; a continuous map on a compact space is uniformly continuous.

Why it is trueA single $\delta$ lets two points that are close in $X$ have close images, and 'close to each other' is exactly what a Cauchy sequence says; continuity at one point gives no such statement about two moving points.

ProofRudin 4.19

The idea: (a) The first part is the previous theorem; the second takes y = a in the definition. (b) Feed the uniform δ to the Cauchy condition. (c) is proved in 4.3.

(a) Uniformly continuous implies continuous at every

a

, with the same

\delta

. Lipschitz implies uniformly continuous by the previous theorem.

Put

y = a

in the definition: the

\delta

that works for all pairs works in particular for the pairs

(x, a)

There is

N

with

d_X(x_m, x_n) < \delta

for all

m, n > N

, because

(x_n)

is Cauchy.

(b) Let

\varepsilon > 0

and take the

\delta

from uniform continuity.

d_Y(f(x_m), f(x_n)) < \varepsilon

for

m, n > N

(f(x_n))

is Cauchy.

Uniform continuity applied to the pair

(x_m, x_n)

\blacksquare

Part (b) gives a quick way to see that a continuous map is not uniformly continuous. The function $f(x) = \frac1x$ on $(0, 1]$ is continuous. The sequence $x_n = \frac1n$ is Cauchy in $(0, 1]$ , but $f(x_n) = n$ is not Cauchy. So $f$ is not uniformly continuous, matching the example in 1.3 by a different route.

Worked example

Show that

f(x) = \sqrt x

is uniformly continuous on

[0, \infty)

but not Lipschitz there.

Solution.

A key inequality. For $x \ge y \ge 0$ , $(\sqrt x - \sqrt y)^2 \le (\sqrt x - \sqrt y)(\sqrt x + \sqrt y) = x - y$ . So $|\sqrt x - \sqrt y| \le \sqrt{|x - y|}$ . The first step holds because $0 \le \sqrt x - \sqrt y \le \sqrt x + \sqrt y$ . The product of the two brackets is the difference of squares.
Choose $\delta$ . Given $\varepsilon > 0$ , take $\delta = \varepsilon^2$ . If $|x - y| < \delta$ then $|\sqrt x - \sqrt y| < \sqrt\delta = \varepsilon$ . The same $\delta$ works for every pair, which is uniform continuity.
Suppose it were Lipschitz. Then $|\sqrt x - \sqrt 0| \le L\,|x - 0|$ for some $L > 0$ and all $x \ge 0$ , that is $\sqrt x \le Lx$ , so $1 \le L\sqrt x$ for $x > 0$ . Divide by $\sqrt x$ .
Contradiction. For $0 < x < \frac1{L^2}$ we get $L\sqrt x < 1$ . So no constant $L$ works near $0$ , where the graph is vertical.

Answer. $\sqrt x$ is uniformly continuous but not Lipschitz. So the arrow Lipschitz $\Rightarrow$ uniformly continuous cannot be reversed. $\blacksquare$

Beyond the course: for understanding· Extending a uniformly continuous map to the closure (B)

Suppose $D$ is a dense subset of a metric space $X$ (every ball of $X$ meets $D$ ), $Y$ is complete, and $f : D \to Y$ is uniformly continuous. Then $f$ extends to a unique continuous map $F : X \to Y$ , which is also uniformly continuous. Sketch: for $x \in X$ pick $d_n \in D$ with $d_n \to x$ ; then $(d_n)$ is Cauchy, so $(f(d_n))$ is Cauchy by (b), and converges in $Y$ by completeness; call the limit $F(x)$ . Interleaving two sequences tending to $x$ shows the limit does not depend on the choice. This is how $a^x$ for irrational $x$ is defined from rational exponents, and how an integral of a function on $[a, b]$ extends from step functions. If $f$ is only continuous, not uniformly, the extension can fail: $\sin\frac1x$ on $(0, 1]$ cannot be extended continuously to $0$ .

Quick checkA student says: 'The map

f : (0, 1] \to \mathbb R

f(x) = \frac1x

, is continuous, and

(0, 1]

is bounded, so

f

must be uniformly continuous.' Where is the error?

Several variables: where the new danger is

For $F : \mathbb R^n \to \mathbb R^m$ the definition is the same, and the new difficulty is that " $x \to a$ " can happen along infinitely many different paths. The definition of the limit requires the SAME value $L$ for every approach, so a function can have a value along every line and still have no limit.

Worked example MATH 375 Lecture 19

Show that

f(x, y) = \dfrac{x - y}{x + y}

has no limit at the origin.

Solution.

Approach along a line. Put $y = mx$ and let $x \to 0$ with $x \ne 0$ . Then $f(x, mx) = \dfrac{x(1 - m)}{x(1 + m)} = \dfrac{1 - m}{1 + m}$ (for $m \ne -1$ ). The factor $x$ cancels, so along each line the value is constant, and the limit along that line is that constant.
The limit depends on the line. $m = 0$ gives $1$ , $m = 1$ gives $0$ , $m = 3$ gives $-\frac12$ . Different lines give different limits.
Turn it into a sequence argument. The sequences $(\frac1n, 0)$ and $(\frac1n, \frac1n)$ both tend to the origin, but $f$ along them is constantly $1$ and constantly $0$ . By the sequential form of the limit, a limit $L$ would need $f(p_n) \to L$ for every sequence $p_n \to (0, 0)$ ; two sequences with different limits disprove it.

Answer. $\lim_{(x, y) \to (0, 0)} \frac{x - y}{x + y}$ does not exist. $\blacksquare$ (The function is not defined on the line $y = -x$ , and the surface is torn along it.)

Your answerFor

f(x, y) = \frac{x - y}{x + y}

, what value does

f

approach as

(x, y) \to (0, 0)

along the line

y = 3x

Python labApproach the origin along different paths

import numpy as np

fs = {
  "(x-y)/(x+y)":        lambda x, y: (x - y) / (x + y),
  "x*y^2/(x^2+y^4)":    lambda x, y: x * y**2 / (x**2 + y**4),
  "x*y/(x^2+y^2)":      lambda x, y: x * y / (x**2 + y**2),
}
paths = {
  "y = 0":    lambda t: (t, 0.0 * t),
  "y = x":    lambda t: (t, t),
  "y = 3x":   lambda t: (t, 3 * t),
  "x = y^2":  lambda t: (t**2, t),
  "x = -y^2": lambda t: (-t**2, t),
}
for name, f in fs.items():
    print(name)
    for pname, p in paths.items():
        vals = []
        for t in (1e-1, 1e-3, 1e-5):
            x, y = p(t)
            with np.errstate(all="ignore"):
                vals.append(float(f(x, y)))
        print("  along", pname.ljust(9), [round(v, 4) for v in vals])

Try this. Run it. The first function has a different limit on each line through the origin. The second, f(x, y) = x*y2/(x2 + y4), goes to 0 along every straight line but not along the parabola x = y2, where the value is always 1/2. Try the path (t3, t) and (-t2, t) as well. Then try the third function x*y/(x2 + y2) and find a line along which it approaches 0.5.

Worked example MATH 375 Piazza, Nov 21 (@105)

Show that

f(x, y) = \dfrac{x y^2}{x^2 + y^4}

(with

f(0, 0) = 0

) is not continuous at the origin, although its limit along every straight line through the origin is

0

Solution.

Along a line $y = mx$ . $f(x, mx) = \dfrac{m^2 x^3}{x^2 + m^4 x^4} = \dfrac{m^2 x}{1 + m^4 x^2} \to 0$ as $x \to 0$ . The numerator has one factor of $x$ left, the denominator tends to $1$ . The vertical line $x = 0$ gives $f = 0$ directly.
Along the parabola $x = y^2$ . $f(y^2, y) = \dfrac{y^2 \cdot y^2}{y^4 + y^4} = \dfrac12$ for every $y \ne 0$ . The numerator and denominator are both of order $y^4$ on this curve, so the ratio is constant.
Conclude. The sequence $p_n = (\frac1{n^2}, \frac1n) \to (0, 0)$ has $f(p_n) = \frac12 \not\to 0 = f(0, 0)$ . By form (b) of the three-faces theorem, $f$ is not continuous at $0$ . One sequence whose images stay away is enough. Checking all lines is not checking all paths.

Answer. $f$ is not continuous at the origin. $\blacksquare$ (Piazza’s remark: an $\varepsilon$ - $\delta$ argument is needed for rigour. The sequence argument above is one.)

Worked example MATH 375 Piazza @119 and HW 11 (Apostol 8.5)

A function can be continuous in each variable separately and still not be continuous:

f(x, y) = \dfrac{xy}{x^2 + y^2}

f(0, 0) = 0

Solution.

Separately continuous. Fix $y_0$ . If $y_0 \ne 0$ , $x \mapsto \frac{x y_0}{x^2 + y_0^2}$ is a quotient of polynomials with non-zero denominator, so continuous. If $y_0 = 0$ , $x \mapsto f(x, 0) = 0$ is constant. The same holds with the roles of $x$ and $y$ swapped. Each one-variable slice is a continuous function of one variable.
Not jointly continuous. Along $y = mx$ the value is $\dfrac{m}{1 + m^2}$ , which is $\frac12$ for $m = 1$ . So $f(\frac1n, \frac1n) = \frac12 \not\to 0 = f(0, 0)$ . The sequence $(\frac1n, \frac1n) \to (0, 0)$ , and the outputs stay at $\frac12$ .

Answer. $f$ is continuous in each variable separately and discontinuous at $(0, 0)$ . $\blacksquare$ Piazza @119 corrected a student’s “proof” that separate continuity implies joint continuity: it used the extreme value theorem on a function that was never shown to be continuous. Always check what the hypotheses of a theorem are before applying it.

Worked example MATH 375 HW 11 (Apostol 8.5 #5)

Let

f(x, y) = x \sin\frac1y

for

y \ne 0

f(x, 0) = 0

. Show

f \to 0

at the origin, but that one of the two iterated limits does not exist.

Solution.

Squeeze. $|\sin\frac1y| \le 1$ , so $|f(x, y)| \le |x|$ for $y \ne 0$ , and $f(x, 0) = 0 \le |x|$ . So $|f(x, y) - 0| \le |x| \le \lVert (x, y) \rVert_2$ . Given $\varepsilon$ , take $\delta = \varepsilon$ : if $\lVert (x, y) \rVert < \delta$ then $|f(x, y)| < \varepsilon$ . (The posted key wrote $-x \le f \le x$ , which is only correct for $x \ge 0$ ; the absolute value fixes it.)
First iterated limit. $\lim_{y \to 0}\bigl[\lim_{x \to 0} f(x, y)\bigr]$ : for each fixed $y \ne 0$ the inner limit is $0 \cdot \sin\frac1y = 0$ . So the outer limit is $0$ . Take the inner limit in $x$ first, with $y$ fixed.
Second iterated limit. For fixed $x \ne 0$ , $\lim_{y \to 0} x \sin\frac1y$ does not exist, because $\sin\frac1y$ oscillates between $-1$ and $1$ infinitely often near $0$ . The inner limit is already undefined, so the iterated limit $\lim_{x \to 0}\bigl[\lim_{y \to 0} f(x, y)\bigr]$ makes no sense.

Answer. The limit at the origin is $0$ , and one iterated limit exists while the other does not. A limit in two variables does not imply that an iterated limit exists, as this example shows. The converse also fails: for $\frac{xy}{x^2 + y^2}$ (the previous example) both iterated limits are $0$ , yet there is no limit at the origin. $\blacksquare$

Quick checkA function

f(x, y)

has limit

0

at the origin along every straight line through the origin. What can you conclude?

Homeomorphisms

Definition B

Homeomorphism

A homeomorphism between metric spaces $X$ and $Y$ is a bijection $f : X \to Y$ such that $f$ and its inverse $f^{-1} : Y \to X$ are both continuous. If one exists, $X$ and $Y$ are homeomorphic.

In one lineA homeomorphism is a bijection $f : X \to Y$ such that both $f$ and $f^{-1}$ are continuous: a one-to-one correspondence that tears and glues nothing, in either direction.

Why it existsIt is the equivalence of topology: two spaces related by one have the same open sets, the same convergent sequences, the same compact and connected subsets, though possibly different distances.

By form (c) of the three-faces theorem, a bijection $f$ is a homeomorphism exactly when $U$ is open in $X$ iff $f(U)$ is open in $Y$ (the preimage under $f^{-1}$ of $U$ is $f(U)$ ). So it matches up the open sets of the two spaces. An isometry (a bijection preserving distances) is a homeomorphism with $L = 1$ both ways. More generally a bijection with both $f$ and $f^{-1}$ Lipschitz is a homeomorphism; the identity map between two equivalent norms is one.

Worked example

Show that

(-1, 1)

and

\mathbb R

are homeomorphic, and that

[0, 2\pi) \to S^1

t \mapsto (\cos t, \sin t)

is a continuous bijection but not a homeomorphism.

Solution.

A candidate. $f(x) = \dfrac{x}{1 - |x|}$ from $(-1, 1)$ to $\mathbb R$ , with inverse $g(y) = \dfrac{y}{1 + |y|}$ . $f$ stretches the ends of the interval out to infinity.
They are inverse to each other. For $x \ge 0$ : $g(f(x)) = \dfrac{x/(1-x)}{1 + x/(1-x)} = \dfrac{x/(1-x)}{1/(1-x)} = x$ ; the case $x < 0$ is the same by symmetry. And $f(g(y)) = y$ similarly. So both are bijections.
Both are continuous. $f$ is a quotient of continuous functions with denominator $1 - |x| > 0$ on $(-1, 1)$ , and $g$ has denominator $1 + |y| \ge 1$ . The building theorem: quotients and the continuity of $|\cdot|$ .
So they are homeomorphic. An open interval and the whole line have the same topology, though one is bounded and the other not, and one is complete and the other not. Boundedness and completeness belong to the distance, not to the topology.
The circle map is a continuous bijection. $t \mapsto (\cos t, \sin t)$ is continuous on $[0, 2\pi)$ (shown in 4.4: Lipschitz with $L = 1$ ) and takes each point of the circle exactly once. One angle in $[0, 2\pi)$ for each point.
Its inverse is not continuous. Take $p_n = (\cos(-\frac1n), \sin(-\frac1n)) \to (1, 0)$ . The angle of $p_n$ in $[0, 2\pi)$ is $2\pi - \frac1n \to 2\pi$ , but the angle of $(1, 0)$ is $0$ . Sequential continuity of the inverse fails at $(1, 0)$ : nearby points on the circle have very different angles when the circle is cut open at $t = 0$ .

Answer. $(-1, 1) \cong \mathbb R$ . The map $[0, 2\pi) \to S^1$ is a continuous bijection whose inverse is not continuous, so not a homeomorphism. $\blacksquare$

Theorem B Rudin Thm 4.14 and 4.22; Munkres Sec. 18 and 26

Properties preserved by homeomorphisms

Let $h : X \to Y$ be a homeomorphism. Then $U$ is open in $X$ iff $h(U)$ is open in $Y$ ; $x_n \to x$ iff $h(x_n) \to h(x)$ ; $K$ is compact iff $h(K)$ is; $C$ is connected iff $h(C)$ is. Completeness and boundedness are NOT preserved.

In one lineA homeomorphism sends open sets to open sets, convergent sequences to convergent sequences, compact sets to compact sets and connected sets to connected sets, in both directions.

Why it is trueEach of these properties was shown to be preserved by continuous maps, and a homeomorphism is continuous both ways.

Proofstandard

The idea: Apply the three-faces theorem and the preservation theorems of 4.3 and 4.4 to h and to h inverse.

Open sets:

U

open implies

h(U)

open (continuity of

h^{-1}

), and

h(U)

open implies

U

open (continuity of

h

h(U) = (h^{-1})^{-1}(U)

is the preimage of

U

under the continuous map

h^{-1}

, and

U = h^{-1}(h(U))

is the preimage of

h(U)

under

h

Sequences:

x_n \to x

implies

h(x_n) \to h(x)

, and conversely using

h^{-1}

Form (b) of the three-faces theorem for

h

and for

h^{-1}

Compact and connected sets: likewise, by the compact image theorem and the connected image theorem.

The continuous image of a compact set is compact (4.3), and of a connected set is connected (4.4), applied to

h

and to

h^{-1}

So completeness and boundedness are properties of the distance, not of the shape.

\blacksquare

The example above:

(-1, 1)

is not complete and

\mathbb R

is;

(-1, 1)

is bounded and

\mathbb R

is not;

f

is a homeomorphism between them.

Proposition B Rudin Thm 4.17; Munkres Thm 26.6

A continuous bijection from a compact space is a homeomorphism

If $K$ is a compact metric space and $f : K \to Y$ is a continuous bijection onto a metric space $Y$ , then $f^{-1}$ is continuous.

In one lineIf $K$ is compact and $f : K \to Y$ is a continuous bijection onto a metric space, then $f^{-1}$ is continuous, so $f$ is a homeomorphism.

Why it is trueA closed subset of $K$ is compact, its image is compact hence closed, and 'sends closed sets to closed sets' is exactly 'the inverse pulls closed sets back to closed sets'.

ProofRudin 4.17

The idea: Show that f sends closed sets to closed sets; then (f inverse) pulls closed sets back to closed sets, which is form (d).

Let

F \subseteq K

be closed. Then

F

is compact.

A closed subset of a compact set is compact (4.3).

f(F)

is compact in

Y

The continuous image of a compact set is compact (4.3).

f(F)

is closed in

Y

Compact sets are closed (4.3).

So the preimage under

f^{-1}

of every closed set is closed:

f^{-1}

is continuous.

\blacksquare

Since

f

is a bijection,

(f^{-1})^{-1}(F) = f(F)

This explains the circle example: $[0, 2\pi)$ is not compact, so the proposition does not apply, and its conclusion indeed fails. (The map $[0, 2\pi] \to S^1$ has a compact domain, but it is not a bijection, because both ends go to $(1, 0)$ . Gluing the two ends of the interval together is the idea of a quotient space, which MATH 551 studies.)

Worked example

Show that

\mathbb R

and

\mathbb R^2

are not homeomorphic, and that

[0, 1]

and

(0, 1)

are not homeomorphic.

Solution.

Suppose $h : \mathbb R \to \mathbb R^2$ is a homeomorphism. Let $p = h(0)$ . We look for a property that distinguishes them after one point is removed.
Restrict. $h$ maps $\mathbb R \setminus \{0\}$ onto $\mathbb R^2 \setminus \{p\}$ , and this restriction is still continuous with a continuous inverse. Restricting a map to a subset, and its inverse to the image, keeps continuity with the same $\delta$ .
Compare connectedness. $\mathbb R \setminus \{0\}$ is not connected (it is not an interval). $\mathbb R^2 \setminus \{p\}$ is path-connected, hence connected (the punctured plane, 4.4; translating $p$ to the origin does not change this). A homeomorphism preserves connectedness in both directions, but one space is connected and the other is not.
Contradiction. So $\mathbb R \not\cong \mathbb R^2$ . This is a small case of the invariance of dimension; the general statement for $\mathbb R^n$ and $\mathbb R^m$ is much harder and is a MATH 551 theorem.
The interval. $[0, 1]$ is compact and $(0, 1)$ is not (4.3), and a homeomorphism preserves compactness. $(0, 1)$ is not closed, so not compact.

Answer. Neither pair is homeomorphic. $\blacksquare$

Quick checkIs the continuous bijection

f : [0, 2\pi) \to S^1

f(t) = (\cos t, \sin t)

, a homeomorphism?

Equivalent norms again, and all norms on R^n

4.1 defined equivalent norms and proved that equivalent norms have the same open sets. Now we can say more and settle the question 4.1 left open.

Proposition B standard

Equivalent norms give the same continuous maps

If $m\,\lVert x \rVert_a \le \lVert x \rVert_b \le M\,\lVert x \rVert_a$ for all $x \in V$ , then the identity map $(V, \lVert \cdot \rVert_a) \to (V, \lVert \cdot \rVert_b)$ is Lipschitz with constant $M$ , its inverse is Lipschitz with constant $1/m$ , and so both are continuous. Hence a map $f : X \to V$ is continuous into $(V, \lVert \cdot \rVert_a)$ iff it is continuous into $(V, \lVert \cdot \rVert_b)$ , and likewise for maps out of $V$ .

In one lineIf two norms on $V$ are equivalent, the identity map between the two normed spaces is a bi-Lipschitz homeomorphism, so a map into or out of $V$ is continuous for one norm iff it is for the other.

Why it is trueThe defining inequalities $m\lVert x \rVert_a \le \lVert x \rVert_b \le M\lVert x \rVert_a$ say that the identity is Lipschitz with constant $M$ one way and $1/m$ the other.

Proofstandard

The idea: The two inequalities are Lipschitz inequalities for the identity map. Composition of continuous maps does the rest.

\lVert x - y \rVert_b \le M\,\lVert x - y \rVert_a

: the identity is

M

-Lipschitz from

a

b

Apply the right inequality to the vector

x - y

\lVert x - y \rVert_a \le \frac1m \lVert x - y \rVert_b

: its inverse is

\frac1m

-Lipschitz.

Apply the left inequality to

x - y

f

continuous into

a

gives

\mathrm{id} \circ f

continuous into

b

, and conversely.

\blacksquare

Lipschitz maps are continuous. A map

f

into

(V, a)

followed by the continuous identity is a map into

(V, b)

; the composition of continuous maps is continuous. The reverse direction uses the inverse.

Theorem B standard

All norms on R^n are equivalent

Let $\lVert \cdot \rVert$ be any norm on $\mathbb R^n$ . Then it is equivalent to $\lVert \cdot \rVert_1$ , and so any two norms on $\mathbb R^n$ are equivalent. In particular they have the same open sets, convergent sequences, Cauchy sequences, and continuous maps.

In one lineFor any norm $\lVert \cdot \rVert$ on $\mathbb R^n$ there are constants $0 < m \le M$ with $m\lVert x \rVert_1 \le \lVert x \rVert \le M\lVert x \rVert_1$ for all $x$ .

Why it is trueThe norm is a continuous function on the compact 'unit sphere' of the 1-norm, so it has a positive minimum and a finite maximum there; scaling extends the bounds to every vector.

Proofstandard

The idea: Upper bound by the triangle inequality on the standard basis. That makes the norm Lipschitz for the 1-norm, so continuous. The unit sphere of the 1-norm is compact (Heine–Borel), so the norm attains a minimum m there, which is positive because the norm of a non-zero vector is positive. Scale.

Let

M = \max_i \lVert e_i \rVert

. Then

\lVert x \rVert \le \sum |x_i|\,\lVert e_i \rVert \le M\,\lVert x \rVert_1

Write

x = \sum x_i e_i

and apply the triangle inequality (N3) repeatedly and homogeneity (N2).

|\lVert x \rVert - \lVert y \rVert| \le \lVert x - y \rVert \le M\,\lVert x - y \rVert_1

: the function

x \mapsto \lVert x \rVert

is Lipschitz, hence continuous, on

(\mathbb R^n, \lVert \cdot \rVert_1)

Reverse triangle inequality for the norm, then the bound proved in the previous step applied to

x - y

The sphere

S = \{x : \lVert x \rVert_1 = 1\}

is compact.

It is closed (the 1-norm is continuous and the sphere is the preimage of the closed set

\{1\}

) and bounded (each

|x_i| \le 1

). Heine–Borel applies to

\mathbb R^n

with the 1-distance, by the equivalence of the 1-, 2- and max-norms in 4.1.

So there is

x_0 \in S

with

\lVert x_0 \rVert = m = \min_S \lVert x \rVert

The extreme value theorem on a compact set (4.3), applied to the continuous function

\lVert \cdot \rVert

S

;

S

is non-empty.

m > 0

x_0 \ne O

since

\lVert x_0 \rVert_1 = 1

, and a norm is positive on non-zero vectors (N1).

m\,\lVert x \rVert_1 \le \lVert x \rVert \le M\,\lVert x \rVert_1

for all

x

For

x \ne O

the vector

x/\lVert x \rVert_1

lies on

S

, so

\lVert x/\lVert x \rVert_1 \rVert \ge m

. Homogeneity (N2) gives

\lVert x \rVert \ge m\,\lVert x \rVert_1

. For

x = O

both sides are

0

Any two norms are equivalent to

\lVert \cdot \rVert_1

, hence to each other.

\blacksquare

Equivalence is symmetric and transitive: if

a \sim 1

and

b \sim 1

then

a \sim b

, multiplying constants.

In the picture of 4.1 (diamond, disc and square) this says every other norm’s unit ball is also squeezed between two scaled copies of the diamond. A consequence: every linear map $T : \mathbb R^n \to \mathbb R^m$ is continuous whichever norms are used, since it is Lipschitz for the 2-norms (the operator norm) and norms are equivalent.

Beyond the course: for understanding· In infinite dimensions linear maps can be discontinuous (B)

Let $C^1[0, 1]$ be the continuously differentiable functions with the sup norm $\lVert f \rVert_\infty$ , and $D f = f'$ the derivative, mapping into $C[0, 1]$ with the sup norm. $D$ is linear. Take $f_n(x) = x^n / n$ . Then $\lVert f_n \rVert_\infty = \frac1n \to 0$ , so $f_n \to 0$ , but $D f_n = x^{n-1}$ has $\lVert D f_n \rVert_\infty = 1$ , which does not tend to $\lVert D 0 \rVert = 0$ . So $D$ is not continuous at $0$ . In finite dimensions every linear map is continuous (the theorem above); in infinite dimensions the choice of norm decides, and differentiation is the standard example of an unbounded operator. This is part of why analysis on function spaces needs care.

Where it is used

Used inside maths· Inside maths: the doorway to topology, and to derivatives in several variables

Form (c) of the three-faces theorem is the definition of continuity in topology: a topological space is a set with a collection of “open” sets satisfying the rules proved in 4.1 (the empty set and the whole space, arbitrary unions, finite intersections), and a map is continuous when preimages of open sets are open. Everything in this chapter that was proved from form (c) alone, compositions, preservation of compact and connected sets, homeomorphisms, carries over. What does not: Cauchy sequences, completeness, boundedness, uniform continuity, which use the distance. That is the dividing line between analysis (MATH 521) and topology (MATH 551). In Unit 5 the derivative of $F : \mathbb R^n \to \mathbb R^m$ at a point $a$ is defined as the linear map $T$ with $\lVert F(a + h) - F(a) - T h \rVert / \lVert h \rVert \to 0$ ; that needs the limit of a map between metric spaces, defined here, and an interior point, from 4.1, so that $a + h$ makes sense in every direction.

Used in the real world· Robustness of neural networks: Lipschitz constants as certificates

An image classifier is a map $f : \mathbb R^n \to \mathbb R^k$ from pixels to class scores. If $f$ is Lipschitz with constant $L$ for the usual length, then changing the input by a vector of length $\varepsilon$ changes the score vector by length at most $L\varepsilon$ . The score of class $i$ minus the score of class $j$ is $(e_i - e_j) \cdot f(x)$ , and $\lVert e_i - e_j \rVert_2 = \sqrt2$ , so by Cauchy–Schwarz this difference changes by at most $\sqrt2\,L\varepsilon$ . Hence if the winning class beats every other by a margin $\mu$ , the prediction cannot change for any input within distance $\mu / (\sqrt2 L)$ . This is a certificate: no attack of that size can succeed.

A network $x \mapsto W_2\,\mathrm{ReLU}(W_1 x)$ is a composition. A matrix multiplication is Lipschitz with constant its operator norm (the largest singular value, 3.5). ReLU applied to each entry is 1-Lipschitz, since $|\max(a, 0) - \max(b, 0)| \le |a - b|$ . Compositions multiply Lipschitz constants, so $L \le \lVert W_2 \rVert\,\lVert W_1 \rVert$ . For the two-layer network in the lab below, $\lVert W_1 \rVert \approx 1.513$ and $\lVert W_2 \rVert \approx 1.208$ , so $L \le 1.828$ . At the input $(1, -0.5)$ the margin is $1.705$ , so the certified radius is $\frac{1.705}{\sqrt2 \cdot 1.828} \approx 0.659$ . The lab then searches for the true distance to the decision boundary and finds about $0.9$ . The certificate is correct and conservative: the product of norms overestimates how much a real network can stretch a particular input. Computing the exact Lipschitz constant of a ReLU network is NP-hard even for two layers (Virmaux and Scaman, 2018), which is why a product of layer norms is the standard cheap bound.

Your answerA network has layer operator norms

2.0

and

1.5

, so

L \le 3

. At an input, the winning class beats the runner-up by a margin of

3

. What is the certified radius

\mu / (\sqrt2\,L)

? (Give a decimal.)

Python labCertify a tiny network, then look for the true boundary

import numpy as np

W1 = np.array([[1.2, -0.7], [0.4, 0.9], [-0.5, 0.3]])
W2 = np.array([[0.8, -0.6, 0.2], [-0.3, 0.5, 0.9]])
f = lambda x: W2 @ np.maximum(W1 @ x, 0)

L = np.linalg.norm(W1, 2) * np.linalg.norm(W2, 2)     # product of operator norms
x0 = np.array([1.0, -0.5])
out = f(x0)
margin = abs(out[0] - out[1])
radius = margin / (np.sqrt(2) * L)
print("L <=", round(L, 3), " margin =", round(margin, 3), " certified radius =", round(radius, 3))

top = np.sign(out[0] - out[1])
best = np.inf
for a in np.linspace(0, 2 * np.pi, 720, endpoint=False):
    d = np.array([np.cos(a), np.sin(a)])
    for r in np.linspace(0.005, 4, 800):
        o = f(x0 + r * d)
        if np.sign(o[0] - o[1]) != top:
            best = min(best, r)
            break
print("true distance to the nearest class change is about", round(best, 3))

Try this. Run it. The certified radius is guaranteed and the true distance to a class change is larger. Change x0 to [0.5, 1.5] or [2.0, -1.0] and compare the gap between the certified and the true distance. Then scale W1 by 3 (W1 = 3 * W1): L multiplies by 3, the margin also grows, but the certified radius stays about the same because both scale together. Why does multiplying the weights by 3 not help robustness at all?

Used in the real world· Topology in data analysis (a light touch): counting pieces at every scale

Given a cloud of data points in $\mathbb R^n$ , one can ask how many separate clusters it has. The answer depends on the scale $\varepsilon$ at which you look. Replace each point by an open ball of radius $\varepsilon/2$ ; two balls overlap exactly when the points are closer than $\varepsilon$ to each other. The union of balls has some number of connected pieces (4.4). Plot that number against $\varepsilon$ : for the points $0, 1, 2.5, 10, 11$ on a line, at $\varepsilon = 0.5$ there are 5 pieces, at $\varepsilon = 1.4$ there are 3 ( $\{0, 1\}$ , $\{2.5\}$ , $\{10, 11\}$ ), at $\varepsilon = 1.6$ there are 2 (the point $2.5$ joins $1$ ), and at $\varepsilon = 8$ there is 1. The scales at which the count changes record the gaps in the data, and a long stretch of $\varepsilon$ with the same count signals a real cluster structure. This count of components over all scales is the zero-dimensional part of persistent homology, a method used on shape data, protein structures and neural activity. It is a stability statement too: moving every point by less than $\delta$ changes the scales at which components merge by at most $2\delta$ , which is a Lipschitz property of the whole procedure.

Your answerFor the points

0, 1, 2.5, 10, 11

on a line, two points are joined when they are closer than

\varepsilon = 1.4

to each other. How many connected pieces are there?

Traps

Watch out· Limits along every line do not give a limit

The function $xy^2/(x^2 + y^4)$ tends to $0$ along every straight line through the origin and has value $\frac12$ along the parabola $x = y^2$ . To show a two-variable limit exists, bound $|f(x, y) - L|$ by something that tends to $0$ as $\lVert (x, y) \rVert \to 0$ (the squeeze), or convert to polar form. To show it does not exist, exhibit two paths or sequences with different limits.

Watch out· Separately continuous is not jointly continuous

$xy/(x^2 + y^2)$ is continuous in $x$ for each fixed $y$ and in $y$ for each fixed $x$ , and is discontinuous at the origin. Continuity of $F : \mathbb R^2 \to \mathbb R$ is a statement about nearby points in the plane, not only nearby points on the two coordinate lines. A proof that uses only the two one-variable slices has proved nothing about $F$ .

Watch out· Continuous maps pull open sets back; they do not push them forward

$f(x) = x^2$ sends $(-1, 1)$ to $[0, 1)$ , and a constant map sends every open set to a point. The theorem is about preimages: $f^{-1}(\text{open})$ is open. A map for which images of open sets are open is called an open map, a stronger and different property.

Watch out· A continuous bijection need not have a continuous inverse

$[0, 2\pi) \to S^1$ is the standard example. To prove a map is a homeomorphism, show that both directions are continuous, or that the domain is compact and the target is a metric space (the compact-domain proposition above).

Watch out· Continuity depends only on the open sets; uniform continuity and completeness depend on the distance

The identity map from $((0, \infty), |x - y|)$ to $((0, \infty), |\frac1x - \frac1y|)$ is a homeomorphism (the open sets agree), but it is not uniformly continuous: the sequence $\frac1n$ is Cauchy in the first space, and its image $n$ is not Cauchy in the second ( $|\frac1m - \frac1n|$ becomes $|m - n|$ after the map). So a property that is invariant under homeomorphism, such as compactness or connectedness, is different from one that is not, such as Cauchy sequences, completeness and uniform continuity.

Watch out· Do not forget the domain: arcsin and arccos are continuous on closed intervals

When finding where $\arcsin(u(x, y))$ is continuous, first find where $u$ is defined and continuous, and note $|u| \le 1$ there. Then $\arcsin$ is continuous on $[-1, 1]$ , ends included. A frequent mistake (it appeared in a posted solution key) is to drop the boundary points where $|u| = 1$ , which are perfectly good points of continuity.

Problems

check

Is $f(x) = \lfloor x \rfloor$ (the greatest integer at most $x$ ) continuous as a map $\mathbb R \to \mathbb R$ ? As a map $\mathbb Z \to \mathbb R$ (with $|x - y|$ on $\mathbb Z$ )? Where does it fail on $\mathbb R$ ?

Solution

On $\mathbb R$ it is discontinuous at every integer: at $n$ , the sequence $n - \frac1k \to n$ has $f = n - 1 \not\to n$ . As a map out of $\mathbb Z$ it is continuous (it is the inclusion $\mathbb Z \to \mathbb R$ , since $\lfloor n \rfloor = n$ ): every point of $\mathbb Z$ is isolated, since $B(n, \frac12) = \{n\}$ , and continuity at an isolated point is automatic.

check

For $f(x) = x^2$ , describe $f^{-1}((1, 4))$ and $f^{-1}([1, 4])$ . Which is open, which closed?

Solution

$f^{-1}((1, 4)) = (-2, -1) \cup (1, 2)$ , open. $f^{-1}([1, 4]) = [-2, -1] \cup [1, 2]$ , closed. This is the pattern: open preimage of an open set, closed preimage of a closed set.

check

Use a sequence to show that $f(x, y) = \dfrac{x y}{x^2 + y^2}$ (with $f(0,0) = 0$ ) is not continuous at the origin.

Solution

The points $(\frac1n, \frac1n) \to (0, 0)$ , but $f(\frac1n, \frac1n) = \frac{1/n^2}{2/n^2} = \frac12 \not\to 0 = f(0, 0)$ . By the sequential form of continuity, $f$ is not continuous at the origin.

check

Which of $|x|$ , $x^2$ , $\sqrt{|x|}$ and $\sin x$ are Lipschitz on $\mathbb R$ ?

Solution

$|x|$ (constant $1$ ) and $\sin x$ (constant $1$ , since $|\cos| \le 1$ and the mean value theorem). Not $x^2$ (the slope $2x$ is unbounded) and not $\sqrt{|x|}$ (near $0$ , $|\sqrt h - \sqrt 0| = \sqrt h$ is not at most $L h$ for small $h$ ).

useMATH 375 HW 11 (Apostol 8.5 #1j) with correction

Find the set where $f(x, y) = \arccos\sqrt{x/y}$ is defined and show it is continuous on all of that set.

Solution

Defined iff $y \ne 0$ and $0 \le x/y \le 1$ . On this set, $x/y$ is continuous, $\sqrt{\ }$ is continuous on $[0, \infty)$ and $\arccos$ is continuous on $[-1, 1]$ , so the composition is continuous on the whole domain. This includes the points $x = 0$ (value $\frac\pi2$ ) and $x = y \ne 0$ (value $0$ ). The posted key kept only the strict inequalities.

useMATH 375 HW 11 (Apostol 8.3 #2 style)

Sketch $S = \{(x, y) : xy > 1\}$ , show that it is open, and find its boundary and its closure.

Solution

$S = F^{-1}((1, \infty))$ for the continuous $F(x, y) = xy$ , so $S$ is open. It is the region above the branch of the hyperbola $xy = 1$ in the first quadrant and below the branch in the third quadrant (two pieces: $x, y > 0$ with $y > 1/x$ , and $x, y < 0$ with $y < 1/x$ ). The boundary is the hyperbola $\{xy = 1\}$ (two branches), and the closure is $\{xy \ge 1\}$ .

use

Show that $\lim_{(x, y) \to (0, 0)} \dfrac{x^2 - y^2}{x^2 + y^2}$ does not exist.

Solution

Along $y = mx$ the value is $\frac{1 - m^2}{1 + m^2}$ , which is $1$ for $m = 0$ and $0$ for $m = 1$ . Two paths, two limits, so no limit.

use

Does $\dfrac{x^2 y}{x^2 + y^2}$ (value $0$ at the origin) have limit $0$ at the origin? Prove it.

Solution

Yes. $|x^2 y| \le (x^2 + y^2)\,|y|$ because $x^2 \le x^2 + y^2$ , so $\left| \frac{x^2 y}{x^2 + y^2} \right| \le |y| \le \lVert (x, y) \rVert$ . Given $\varepsilon$ , $\delta = \varepsilon$ works. So the function is continuous at the origin.

prove

Prove that a map $f : X \to Y$ is continuous iff $f(\overline A) \subseteq \overline{f(A)}$ for every $A \subseteq X$ .

Solution

Plan. Use sequences and the description of the closure by sequences (4.2).

( $\Rightarrow$ ) If $x \in \overline A$ there are $a_n \in A$ with $a_n \to x$ (4.2). By continuity $f(a_n) \to f(x)$ , so $f(x)$ is a limit of points of $f(A)$ , hence in $\overline{f(A)}$ .

( $\Leftarrow$ ) Let $F \subseteq Y$ be closed, and $A = f^{-1}(F)$ . Then $f(A) \subseteq F$ , so $\overline{f(A)} \subseteq F$ ( $F$ is closed). By hypothesis $f(\overline A) \subseteq \overline{f(A)} \subseteq F$ , so $\overline A \subseteq f^{-1}(F) = A$ . So $A$ is closed. By form (d), $f$ is continuous. $\blacksquare$

prove

Prove that if $f, g : X \to \mathbb R$ are continuous, then $\max(f, g)$ is continuous.

Solution

Plan. Write the maximum with absolute values.

$\max(f, g) = \frac12(f + g + |f - g|)$ . The map $f - g$ is continuous (building theorem), $|\cdot| : \mathbb R \to \mathbb R$ is 1-Lipschitz hence continuous, so $|f - g|$ is a composition of continuous maps, hence continuous; the sum and the multiple by $\frac12$ are continuous. $\blacksquare$

prove

Prove that if $f : X \to Y$ and $g : Y \to Z$ are uniformly continuous, then $g \circ f$ is uniformly continuous.

Solution

Plan. Work backwards from the tolerance in $Z$ : first the $\delta$ of $g$ , then the $\delta$ of $f$ for that tolerance.

Let $\varepsilon > 0$ . Uniform continuity of $g$ gives $\eta > 0$ with $d_Z(g(u), g(v)) < \varepsilon$ whenever $d_Y(u, v) < \eta$ , for all $u, v \in Y$ . Uniform continuity of $f$ , with tolerance $\eta$ , gives $\delta > 0$ with $d_Y(f(x), f(x')) < \eta$ whenever $d_X(x, x') < \delta$ , for all $x, x' \in X$ . If $d_X(x, x') < \delta$ , then $d_Y(f(x), f(x')) < \eta$ , so $d_Z(g(f(x)), g(f(x'))) < \varepsilon$ . The number $\delta$ depends only on $\varepsilon$ , not on the points. $\blacksquare$

prove

Prove that $[0, 1)$ and $(0, 1)$ are not homeomorphic.

Solution

Plan. Remove a point and compare connectedness.

Suppose $h : [0, 1) \to (0, 1)$ is a homeomorphism and put $p = h(0)$ . Then $h$ restricts to a homeomorphism from $[0, 1) \setminus \{0\} = (0, 1)$ onto $(0, 1) \setminus \{p\}$ . The first is an interval, hence connected. The second is $(0, p) \cup (p, 1)$ , which is not an interval, hence not connected. Connectedness is preserved by homeomorphisms, a contradiction. $\blacksquare$

prove

Prove that a linear map $T : V \to W$ between normed spaces is continuous iff it is bounded: $\lVert Tx \rVert \le C\lVert x \rVert$ for some constant $C$ and all $x$ .

Solution

Plan. Bounded gives Lipschitz. For the converse use continuity at $O$ with $\varepsilon = 1$ , and scale.

If $\lVert Tx \rVert \le C\lVert x \rVert$ , then $\lVert Tx - Ty \rVert = \lVert T(x - y) \rVert \le C\lVert x - y \rVert$ : Lipschitz, so continuous. Conversely, suppose $T$ is continuous at $O$ . Take $\varepsilon = 1$ : there is $\delta > 0$ with $\lVert Tz \rVert < 1$ whenever $\lVert z \rVert < \delta$ . For $x \ne O$ put $z = \frac{\delta}{2\lVert x \rVert}x$ , so $\lVert z \rVert = \frac\delta2 < \delta$ . Then $\lVert Tz \rVert = \frac{\delta}{2\lVert x \rVert}\lVert Tx \rVert < 1$ , so $\lVert Tx \rVert < \frac2\delta \lVert x \rVert$ . Take $C = \frac2\delta$ . $\blacksquare$

stretch

Prove the extension theorem in the box above: a uniformly continuous $f : D \to Y$ on a dense subset $D \subseteq X$ , with $Y$ complete, extends to a continuous $F : X \to Y$ .

Solution

Plan. Define $F(x)$ as the limit of $f(d_n)$ along any sequence $d_n \to x$ in $D$ , and check it is well defined.

Let $x \in X$ . By density, pick $d_n \in D$ with $d_n \to x$ . The sequence $(d_n)$ is Cauchy, so $(f(d_n))$ is Cauchy (part (b) of the uniform continuity theorem), and converges in $Y$ since $Y$ is complete. If $d_n' \to x$ is another sequence, the interleaved sequence $d_1, d_1', d_2, d_2', \dots$ also tends to $x$ , so its image is Cauchy and converges, so the limits of $f(d_n)$ and $f(d_n')$ agree. Define $F(x)$ to be this limit. For $x \in D$ , the constant sequence gives $F(x) = f(x)$ . For uniform continuity of $F$ : given $\varepsilon$ , take $\delta$ for $\frac\varepsilon3$ from $f$ . If $d_X(x, x') < \frac\delta3$ , choose $d, d' \in D$ with $d_X(d, x)$ , $d_X(d', x') < \frac\delta3$ and $d_Y(f(d), F(x))$ , $d_Y(f(d'), F(x')) < \frac\varepsilon3$ (possible by the definition of $F$ with sequences). Then $d_X(d, d') < \delta$ , so $d_Y(F(x), F(x')) < \frac\varepsilon3 + \frac\varepsilon3 + \frac\varepsilon3 = \varepsilon$ . Uniqueness: a continuous extension is determined on the dense set. $\blacksquare$

stretch

Prove that every norm on $\mathbb R^n$ gives a complete metric space.

Solution

Plan. Equivalence of norms moves Cauchy sequences and limits between the given norm and the 1-norm, where completeness is known (4.2).

Let $\lVert \cdot \rVert$ be a norm, with $m\lVert x \rVert_1 \le \lVert x \rVert \le M\lVert x \rVert_1$ . Let $(x_k)$ be Cauchy for $\lVert \cdot \rVert$ . Then $\lVert x_k - x_l \rVert_1 \le \frac1m \lVert x_k - x_l \rVert$ , so $(x_k)$ is Cauchy for the 1-norm. $\mathbb R^n$ is complete for the 1-distance (4.2), so $\lVert x_k - x \rVert_1 \to 0$ for some $x$ . Then $\lVert x_k - x \rVert \le M\lVert x_k - x \rVert_1 \to 0$ , so $x_k \to x$ in the given norm. $\blacksquare$

Summary

A map $f : X \to Y$ between metric spaces is continuous when the $\varepsilon$ - $\delta$ condition of 1.3 holds with the distances $d_X$ and $d_Y$ . The limit of a map between metric spaces is defined the same way with $0 < d_X(x, a)$ .
Three equivalent forms. $\varepsilon$ - $\delta$ at every point; $x_n \to a$ implies $f(x_n) \to f(a)$ ; the preimage of every open set is open (and equivalently of every closed set is closed). The last needs no distance, which is where topology begins.
Continuous maps are closed under composition, sums, products, quotients and combination into coordinates; projections and norms are continuous. This proves continuity for polynomials, rational functions and elementary compositions on their domains.
A Lipschitz map ( $d(f(x), f(y)) \le L\,d(x, y)$ ) is uniformly continuous; a matrix is Lipschitz with constant its operator norm. Uniformly continuous maps send Cauchy sequences to Cauchy sequences, and continuous maps on compact spaces are uniformly continuous.
In several variables a limit must be the same along every path. Different lines giving different limits, or a parabola giving a different limit from the lines, disproves it. Separate continuity does not give joint continuity.
A homeomorphism is a bijection continuous in both directions. It preserves open sets, sequences, compactness and connectedness, not completeness or boundedness. A continuous bijection from a compact space is automatically a homeomorphism. $\mathbb R$ and $\mathbb R^2$ are not homeomorphic.
All norms on $\mathbb R^n$ are equivalent, proved with compactness of the unit sphere, so they give the same open sets and the same continuous maps. In infinite dimensions this fails, and linear maps can be discontinuous.
A product of layer Lipschitz constants bounds a neural network’s sensitivity and gives a certified robustness radius.

The picture: preimages of open intervals

The idea, in words

Definitions and theorems

Limit of a map between metric spaces

Continuous map between metric spaces

Three faces of continuity

Building continuous maps

Lipschitz maps

Lipschitz map

Lipschitz implies uniformly continuous

Uniform continuity between metric spaces

Uniform continuity between metric spaces

Several variables: where the new danger is

Homeomorphisms

Homeomorphism

Properties preserved by homeomorphisms

A continuous bijection from a compact space is a homeomorphism

Equivalent norms again, and all norms on R^n

Equivalent norms give the same continuous maps

All norms on R^n are equivalent

Where it is used

Traps

Problems

Summary

Where this sits

Comes from

Leads to

Same idea elsewhere