4.4 Connectedness and path-connectedness

A space is connected when it cannot be split into two separate open pieces. The connected subsets of the line are exactly the intervals, a continuous map sends connected sets to connected sets, and that one fact is the intermediate value theorem. Path-connected is stronger, and the topologist's sine curve shows the gap.

In 1.3 the intermediate value theorem said that a continuous function cannot jump over a value. Its proof used the supremum of the points where the function was still negative. That proof was about the interval $[a, b]$ and about the real numbers, and it left a question open. Is it a fact about functions, or about the interval? Bolzano’s 1817 proof already used no geometry, but it did use the order of the real line, and a function on a circle, a sphere or a disc has no order on its inputs to use. This chapter finds the property of $[a, b]$ that the proof used: it is in one piece. We make “one piece” exact with open sets, prove that the connected subsets of the line are the intervals, and show that continuous maps preserve it. The intermediate value theorem then becomes a corollary that holds for functions on any connected metric space.

Before you start

4.1: open sets and closed sets in a metric space, and that a union of open sets is open. 1.3: the intermediate value theorem and continuity with $\varepsilon$ and $\delta$ . 4.2: convergent sequences in a metric space.

Quick checkIs

[0, 1] \cup [2, 3]

a closed subset of

\mathbb R

Quick check

f

is continuous on

[0, 1]

with

f(0) = -2

and

f(1) = 5

. Which values must

f

take?

Quick checkTake the metric space

A = [0, 1] \cup [2, 3]

with the distance

|x - y|

. Is

[0, 1]

an open set of this space

A

The picture: can you cut it in two?

Take a subset $A$ of the line. A cut at the number $z$ splits $A$ into the part to the left of $z$ and the part to the right. Three things to do in the picture:

Pick $[0, 1]$ and $[2, 3]$ and drag the cut into the gap. The cut hits no point of $A$ , and both pieces are non-empty. Each piece is $A$ meets an open ray, so each is open in $A$ . The set has been separated.
Pick $[0, 3]$ and drag the cut anywhere. Either the cut sits on a point of $A$ (nothing is cut), or one side is empty. No cut ever works. This is the property that will be called connected.
Pick $[0, 1)$ and $(1, 2]$ . The point $1$ is missing from the set, and that missing point is a place where the cut lands. A missing point is enough to separate.

Where this came from· From 'a continuum' to 'no separation'

In 1883 Georg Cantor wanted to say what it means for a set of points to be a continuum, one unbroken piece. His definition used distances: any two points of the set can be joined by a finite chain of points of the set whose consecutive links are shorter than any tolerance you name. Arthur Schoenflies (1904) saw that distance is not needed and defined the notion purely with point sets. The form used today, that a set is connected when it is not the union of two separated non-empty sets, appeared independently with Nels Lennes, Frigyes Riesz and Felix Hausdorff in the first years of the twentieth century, and is often called the Lennes–Hausdorff definition. The problem they solved was to say “in one piece” without drawing a picture. Bernard Bolzano’s 1817 proof of the intermediate value theorem is the earliest result of this kind. It was proved long before anyone had a definition of the property it depends on.

The idea, in words

Why· One piece means: no way to build a fence

Suppose a set $A$ is in two pieces, like $[0, 1] \cup [2, 3]$ . Then you can build a fence in the gap. Each side of the fence is an open set of $A$ : no point of the left side touches the right side, so every point of the left side has room around it that stays on the left. A set in one piece has no fence. Any attempt to put the points into a “left side” and a “right side” leaves a point on the fence line with room on neither side.

A continuous function cannot tear a set. If $f$ sent an interval onto two separate pieces of the line, the fence between the pieces would pull back to a fence in the interval, and there is none. That is the whole proof of the intermediate value theorem, and it never mentions the order or the sup of the inputs. It only uses that the domain has no fence.

Definitions and theorems

Separations and connected sets

First we need open sets inside a subset. Take a subset $A$ of a metric space $X$ . With the same distance $d$ , $A$ is itself a metric space, and 4.1 applies to it: $B_A(x, r) = \{y \in A : d(x, y) < r\}$ , and a set $U \subseteq A$ is open in $A$ if each of its points has a ball $B_A(x, r) \subseteq U$ . The example in the “Before you start” box shows why this matters: $[0, 1]$ is open in $[0, 1] \cup [2, 3]$ .

Lemma B standard; Rudin Thm 2.30

Open sets inside a subset

Let $A \subseteq X$ and $U \subseteq A$ . Then $U$ is open in $A$ if and only if there is an open set $W$ of $X$ with $U = A \cap W$ .

In one line $U$ is open in $A$ if and only if $U = A \cap W$ for some open set $W$ of $X$ .

Why it is trueA ball of $A$ is a ball of $X$ with the points outside $A$ removed.

Proofstandard

The idea: A ball of the subspace A is the intersection of A with a ball of X. Forward: gather the balls of X into one open set W. Backward: shrink the ball of X around a point of A ∩ W so that it sits inside W.

(

\Rightarrow

) Let

U

be open in

A

. For each

x \in U

choose

r_x > 0

with

A \cap B(x, r_x) \subseteq U

By definition,

B_A(x, r) = A \cap B(x, r)

. The ball of

X

with its points outside

A

removed.

Put

W = \bigcup_{x \in U} B(x, r_x)

. It is open in

X

A union of open balls is open (4.1: balls are open, unions of open sets are open).

A \cap W = U

Each

x \in U

lies in its own ball, so

U \subseteq A \cap W

. Conversely a point of

A \cap W

is in

A \cap B(x, r_x) \subseteq U

for some

x

(

\Leftarrow

) Let

U = A \cap W

with

W

open, and

x \in U

. Choose

r > 0

with

B(x, r) \subseteq W

W

is open, so the point

x \in A \cap W

has a ball

B(x, r) \subseteq W

Then

B_A(x, r) = A \cap B(x, r) \subseteq A \cap W = U

, so

U

is open in

A

\blacksquare

Intersect both sides of

B(x, r) \subseteq W

with

A

Definition B

Separation

A separation of a metric space $A$ is a pair of sets $U, V$ with $A = U \cup V$ , $U \cap V = \varnothing$ , $U \ne \varnothing$ , $V \ne \varnothing$ , and both $U$ and $V$ open in $A$ . For a subset $A$ of a bigger space $X$ , “open in $A$ ” is as in the previous lemma.

In one lineA separation of a set $A$ is a way of writing $A = U \cup V$ with $U$ and $V$ disjoint, non-empty, and both open in $A$ .

Why it existsIt is the exact form of 'a fence through the gap': two pieces, each with room to move inside itself, with nothing shared.

Definition B

Connected set

A metric space $A$ is connected if it has no separation. A subset of a metric space is connected if it is connected as a metric space with the inherited distance. A set that is not connected is disconnected.

In one lineA set (a metric space) is connected if it has no separation: it cannot be written as two disjoint non-empty open pieces.

Why it existsIt is the property of 'being in one piece' that continuity preserves and that the intermediate value theorem needs.

Two quick consequences. The empty set and any one-point set are connected, because a separation needs two non-empty disjoint pieces. And because $V = A \setminus U$ is the complement of $U$ , a separation $A = U \cup V$ makes $U$ open and also closed in $A$ (its complement $V$ is open). That gives a second description.

Proposition B standard

Connected means no proper clopen subset

A metric space $A$ is connected if and only if its only subsets that are both open and closed (called clopen) are $\varnothing$ and $A$ .

In one lineA space $A$ is connected if and only if the only subsets that are both open and closed in $A$ are $\varnothing$ and $A$ .

Why it is trueA subset that is both open and closed, and neither empty nor everything, together with its complement is a separation; conversely, a separation gives such a subset.

Proofstandard

The idea: Both directions are the observation that a pair of complementary open sets is the same data as one clopen set.

(Not connected

\Leftarrow

a proper clopen set)

U

and

V = A \setminus U

are disjoint, non-empty and open in

A

, with union

A

: a separation.

Suppose

A

has a clopen subset

U

with

U \ne \varnothing

and

U \ne A

. Its complement

V = A \setminus U

is open (since

U

is closed) and non-empty (since

U \ne A

(Not connected

\Rightarrow

a proper clopen set) Given a separation

A = U \cup V

U

is clopen and proper.

\blacksquare

V

is open, and

U = A \setminus V

is its complement, so

U

is closed.

U

is open by definition of a separation.

U

is non-empty, and

U \ne A

because

V \ne \varnothing

and

V

is disjoint from

U

In 4.1 we noticed that $\varnothing$ and $\mathbb R$ are the only clopen subsets of $\mathbb R$ and left it unproved. By this proposition that statement is the same as ” $\mathbb R$ is connected”, which the next section proves.

Quick checkIs a set with exactly one point connected?

The next lemma is used twice below. It says a connected set cannot straddle a separation of a bigger set.

Lemma B standard

A connected subset lies on one side of a separation

Let $Z$ be a metric space with a separation $Z = P \cup Q$ , and let $C \subseteq Z$ be connected. Then $C \subseteq P$ or $C \subseteq Q$ .

In one lineIf $Z = P \cup Q$ is a separation and $C \subseteq Z$ is connected, then $C \subseteq P$ or $C \subseteq Q$ .

Why it is trueOtherwise $C \cap P$ and $C \cap Q$ would be a separation of $C$ .

Proofstandard

The idea: If C meets both P and Q, the pieces C ∩ P and C ∩ Q separate C.

Suppose

C

meets both

P

and

Q

. Then

C \cap P

and

C \cap Q

are open in

C

P

and

Q

are open in

Z

. By the lemma on open sets inside a subset (with

Z

as the big space and

C

as the subset),

C \cap P

and

C \cap Q

are open in

C

C = (C \cap P) \cup (C \cap Q)

is a separation of

C

, contradicting that

C

is connected. So

C

misses one of them, and lies in the other.

\blacksquare

They are non-empty by assumption, disjoint because

P \cap Q = \varnothing

, and their union is

C \cap (P \cup Q) = C

The connected subsets of the line

Definition B

Interval

A set $I \subseteq \mathbb R$ is an interval if for all $x, y \in I$ and every $z$ with $x < z < y$ , we have $z \in I$ .

In one lineA set $I \subseteq \mathbb R$ such that whenever $x < z < y$ and $x, y \in I$ , also $z \in I$ : it has no holes.

Why it existsIt is the exact statement of 'no gaps', and it covers $[a, b]$ , $(a, b)$ , half-open ones, rays, $\mathbb R$ , single points and the empty set in one definition.

This definition includes $[0, 1]$ , $(0, 1)$ , $[0, 1)$ , $[2, \infty)$ , $(-\infty, 3)$ , $\mathbb R$ , a single point, and $\varnothing$ . It excludes $[0, 1] \cup [2, 3]$ (take $x = 1$ , $y = 2$ , $z = 1.5$ ) and $\mathbb Q$ (take $x = 1$ , $y = 2$ , $z = \sqrt2$ ).

Theorem B Rudin Thm 2.47; Abbott Sec. 3.4

Connected subsets of the line are the intervals

A subset $A \subseteq \mathbb R$ is connected if and only if it is an interval.

In one lineA subset of $\mathbb R$ is connected if and only if it is an interval.

Why it is trueA hole at $z$ gives a cut through $z$ . In the other direction, a separation of an interval would give a last point of the left piece, which contradicts the openness of both pieces.

ProofRudin 2.47; Abbott Sec. 3.4

The idea: Not an interval: cut at the missing point. Interval: if it were separated, take the supremum of the left piece between a left point and a right point, and show the supremum can belong to neither piece. This is the proof of the intermediate value theorem in 1.3.

(

\Rightarrow

) Suppose

A

is not an interval. Pick

x < z < y

with

x, y \in A

and

z \notin A

We prove the contrapositive of (

\Rightarrow

): not an interval means not connected. Use the definition of interval, negated: there are

x, y \in A

and a

z

with

x < z < y

and

z \notin A

Let

U = A \cap (-\infty, z)

and

V = A \cap (z, \infty)

Open rays are open sets of

\mathbb R

, so these two are open in

A

by the lemma on open sets inside a subset. This is the cut in the picture.

They are non-empty, disjoint, and

U \cup V = A

. So

A

has a separation and is not connected.

x \in U

and

y \in V

. A point of

A

is below

z

, above

z

, or equal to

z

, and

z \notin A

Pick

a \in U

and

b \in V

. By swapping the names

U

and

V

we may assume

a < b

(they are different, since

U \cap V = \varnothing

(

\Leftarrow

) Suppose

I

is an interval with a separation

I = U \cup V

. We will reach a contradiction. Choose a point in each piece.

Then

[a, b] \subseteq I

, so each point of

[a, b]

is in

U

or in

V

I

is an interval and

a, b \in I

, so every point between them is in

I

Let

S = U \cap [a, b]

and

c = \sup S

. Then

a \le c \le b

, so

c \in I

S

is non-empty (

a \in S

) and bounded above by

b

, so the completeness axiom of 1.1 gives its supremum. This is the same set as in the proof of the intermediate value theorem of 1.3.Where the idea comes from: we want the last point of the left piece before the right piece begins. A last point is a supremum.

Case

c \in U

. Choose

r > 0

with

(c - r, c + r) \cap I \subseteq U

. Put

x = \min(c + \frac r2, b)

. Then

c < x \le b

x \in [a, b] \subseteq I

and

x

is in the ball, so

x \in U \cap [a, b] = S

. But

x > c = \sup S

, a contradiction.

c \in U

cannot be

b

since

b \in V

and

U \cap V = \varnothing

, so

c < b

U

is open in

I

, so some ball

(c - r, c + r) \cap I

lies in

U

Case

c \in V

. Choose

0 < r \le c - a

with

(c - r, c + r) \cap I \subseteq V

. Then

(c - r, c] \subseteq [a, c] \subseteq I

, so

(c - r, c] \subseteq V

, and

(c - r, c]

contains no point of

U

c \in V

cannot be

a

since

a \in U

, so

c > a

V

is open in

I

, so some ball

(c - r, c + r) \cap I

lies in

V

. Shrink

r

to at most

c - a

so the ball stays inside

[a, \infty)

c - r

is an upper bound of

S

smaller than

c

, contradicting that

c

is the least upper bound.

Every point of

S

is at most

c

(as

c

is an upper bound) and, by the previous line, not in

(c - r, c]

. So all points of

S

are at most

c - r

Both cases fail, so no separation exists.

I

is connected.

\blacksquare

c \in [a, b] \subseteq I = U \cup V

, so one of the two cases holds, and each is impossible.

In the picture: the first half is the cut in the visualizer, dragged into a gap. The second half is the visualizer for $[0, 3]$ , where no cut works: the supremum $c$ is the place the cut would have to go, and it has no side to belong to.

Proof workshopRebuild the proof that an interval is connected

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

Quick checkWhich of these subsets of

\mathbb R

is connected?

Worked example

A = \{x \in \mathbb R : x^2 > 5\} \cup \{x : x^2 < 3\}

connected? Into how many pieces does it break?

Solution.

Write $A$ as intervals. $x^2 < 3$ means $-\sqrt3 < x < \sqrt3$ , and $x^2 > 5$ means $x < -\sqrt5$ or $x > \sqrt5$ . So $A = (-\infty, -\sqrt5) \cup (-\sqrt3, \sqrt3) \cup (\sqrt5, \infty)$ . Solving each inequality gives intervals, and a connected subset of $\mathbb R$ is an interval.
Find a hole. Take $x = 0$ and $y = 3$ . Both lie in $A$ , and $z = 2$ lies between them. But $2^2 = 4$ is neither below $3$ nor above $5$ , so $2 \notin A$ . That is the failure of the interval property, so $A$ is not an interval.
Separate. $U = A \cap (-\infty, 2) = (-\infty, -\sqrt5) \cup (-\sqrt3, \sqrt3)$ and $V = A \cap (2, \infty) = (\sqrt5, \infty)$ . The cut at $z = 2$ , as in the theorem.
Count the pieces. Each of the three intervals is itself an interval, hence connected, and there are gaps between them, at $[-\sqrt5, -\sqrt3]$ and $[\sqrt3, \sqrt5]$ . So $A$ breaks into three connected pieces.

Answer. $A$ is not connected. It has three connected pieces. $\blacksquare$

Your answerHow many separate connected pieces (maximal intervals) does

\{x \in \mathbb R : x^2 > 5\} \cup \{x : x^2 < 3\}

have?

Worked example

Show that

\mathbb Q

is disconnected, and that its only connected subsets have at most one point.

Solution.

Suppose a connected subset $C \subseteq \mathbb Q$ has two points. Say $x < y$ in $C$ . We prove the contrapositive of “at most one point”: two points give a contradiction.
Use the theorem. $C$ is a connected subset of $\mathbb R$ , so it is an interval, and so every real number between $x$ and $y$ is in $C$ . The theorem applies to every connected subset of the line, rational or not.
Find an irrational number between them. Between any two rationals there is an irrational number (add $\frac{\sqrt2}{n}$ for large $n$ ). That number is in $C \subseteq \mathbb Q$ , a contradiction. Irrational numbers are not rational, so they cannot lie in $C$ .

Answer. Every connected subset of $\mathbb Q$ has at most one point; in particular $\mathbb Q$ is disconnected. $\blacksquare$ A space like this, with no connected piece bigger than a point, is called totally disconnected.

Continuous maps and connectedness

A continuous map $f : X \to Y$ between metric spaces is one for which, at every point $a$ of $X$ and for every $\varepsilon > 0$ , there is a $\delta > 0$ with $d_Y(f(x), f(a)) < \varepsilon$ whenever $d_X(x, a) < \delta$ . That is the definition of 1.3 with the two distances written $d_X$ and $d_Y$ . Chapter 4.5 gives it its name and studies it; here we need exactly one property, and we prove it from this definition.

Lemma B Rudin Thm 4.8

Continuous maps pull open sets back to open sets

Let $f : X \to Y$ be continuous (in the $\varepsilon$ - $\delta$ sense above) and let $W \subseteq Y$ be open. Then $f^{-1}(W)$ is open in $X$ .

In one lineIf $f : X \to Y$ is continuous and $W$ is open in $Y$ , then $f^{-1}(W) = \{x : f(x) \in W\}$ is open in $X$ .

Why it is trueAround $f(a)$ the open set $W$ contains a ball. Continuity gives a ball around $a$ that is sent into it.

ProofRudin 4.8

The idea: Take a point a in the preimage. Its image has a ball inside W; continuity turns that ball's radius into a δ-ball around a that maps inside it.

Let

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

, so

f(a) \in W

To show the preimage is open we give every point of it a ball. So start with an arbitrary point.

Choose

\varepsilon > 0

with

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

W

is open in

Y

, so the point

f(a)

W

has a ball inside

W

Choose

\delta > 0

with

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

whenever

d_X(x, a) < \delta

Continuity of

f

a

, with this

\varepsilon

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

. Since

a

was arbitrary,

f^{-1}(W)

is open.

\blacksquare

x \in B_X(a, \delta)

then

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

, which says

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

In the picture of 1.3 (ε-δ visualizer) $W$ is a horizontal band around $f(a)$ and $B(a, \delta)$ is the window of inputs whose graph stays in the band. The lemma says: the set of all inputs whose output lands in the band is a union of such windows.

Theorem B Rudin Thm 4.22

The continuous image of a connected set is connected

If $X$ is a connected metric space and $f : X \to Y$ is continuous, then the image $f(X)$ is connected.

In one lineIf $X$ is connected and $f : X \to Y$ is continuous, then $f(X)$ is connected.

Why it is trueA separation of the image would pull back, through the lemma, to a separation of $X$ .

ProofRudin 4.22

The idea: Contradiction: a separation of f(X) pulls back to a separation of X, because preimages of open sets are open, preimages of disjoint sets are disjoint, and the pieces are non-empty since every point of f(X) has a preimage.

Suppose

P, Q

are disjoint, non-empty, open in

f(X)

, with

P \cup Q = f(X)

Assume the opposite of what we want: the image has a separation

f(X) = P \cup Q

The sets

f^{-1}(P)

and

f^{-1}(Q)

are open in

X

View

f

as a map into the metric space

f(X)

; it is still continuous, since the distances are unchanged. Then apply the lemma to the open sets

P

and

Q

f(X)

They are disjoint, they cover

X

, and each is non-empty.

A point mapped into both

P

and

Q

would be in

P \cap Q = \varnothing

. Every point of

X

is sent into

f(X) = P \cup Q

. Every point of

P

is a value of

f

, so its preimage is non-empty.

This is a separation of

X

, a contradiction. So

f(X)

is connected.

\blacksquare

That is the definition of a separation of

X

, which is connected.

Proof workshopPick the reason for each line: the continuous image of a connected set

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

Suppose $f(X) = P \cup Q$ with $P, Q$ disjoint, non-empty and open in $f(X)$ .
Then $f^{-1}(P)$ and $f^{-1}(Q)$ are open in $X$ .
They are disjoint.
Their union is $X$ .
Each is non-empty.
So $X$ has a separation, contradicting that $X$ is connected.

Quick check

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

is continuous and every value

f(x)

is a rational number. What follows?

Corollary B Abbott Sec. 4.5; Rudin Thm 4.23

Intermediate value theorem from connectedness

Let $X$ be connected, $f : X \to \mathbb R$ continuous, $p, q \in X$ , and $y$ a real number between $f(p)$ and $f(q)$ . Then there is $c \in X$ with $f(c) = y$ . In particular, for $X = [a, b]$ this is the intermediate value theorem of 1.3.

In one lineIf $X$ is a connected metric space and $f : X \to \mathbb R$ is continuous, then $f$ takes every value between any two of its values.

Why it is trueThe image of a connected set is an interval, and an interval that contains two numbers contains everything between them.

ProofAbbott Sec. 4.5

The idea: The image f(X) is connected, so it is an interval, so it contains everything between f(p) and f(q).

f(X)

is a connected subset of

\mathbb R

The continuous image of a connected set is connected.

f(X)

is an interval.

A connected subset of

\mathbb R

is an interval (the theorem above).

f(p), f(q) \in f(X)

and

y

is between them, so

y \in f(X)

, which means

y = f(c)

for some

c \in X

\blacksquare

The definition of interval: it contains every number strictly between two of its members. If

y = f(p)

y = f(q)

it is a value already.

This is the same argument as 1.3’s, repackaged: the supremum is now inside the proof that intervals are connected, and the theorem about functions needs no more. The gain is that $X$ can be any connected metric space: a circle, a disc, a sphere.

Worked example

At every moment there are two opposite points on the equator with the same temperature.

Solution.

Set up. Let $T(\theta)$ be the temperature at the point of the equator at angle $\theta$ , so $T$ is continuous and $T(\theta + 2\pi) = T(\theta)$ . Temperature varies continuously along a path.
Compare opposite points. Let $g(\theta) = T(\theta) - T(\theta + \pi)$ for $0 \le \theta \le \pi$ . We want a zero of $g$ . $g(\theta) = 0$ says the point at angle $\theta$ and the opposite point (angle $\theta + \pi$ ) have equal temperature.
Look at the ends. $g(0) = T(0) - T(\pi)$ and $g(\pi) = T(\pi) - T(2\pi) = T(\pi) - T(0) = -g(0)$ . The end values are negatives of each other, so they have opposite signs (or both are zero).
Apply the theorem. $g$ is continuous on the connected set $[0, \pi]$ and takes a value $g(0)$ and its negative, so it takes the value $0$ between them: $g(\theta^*) = 0$ for some $\theta^* \in [0, \pi]$ . The corollary above, with the value $y = 0$ .

Answer. There is $\theta^*$ with $T(\theta^*) = T(\theta^* + \pi)$ . $\blacksquare$ The same argument works for any continuous quantity on a circle, such as pressure on a latitude line.

Your answerSuppose

T(\theta) = 2\cos\theta + \sin 2\theta

on the equator. For which

\theta \in [0, \pi]

T(\theta) = T(\theta + \pi)

? (Give

\theta

in radians.)

Worked example classic

A monk climbs a mountain path from 8:00 to 16:00 on Monday, and descends the same path from 8:00 to 16:00 on Tuesday. Show he is at the same spot at the same time of day on both days.

Solution.

Measure position along the path. Let $u(t)$ be his distance from the bottom, as a fraction $0$ to $1$ of the path, at time $t$ on Monday, and $w(t)$ the same on Tuesday. Both are continuous on $[8, 16]$ . He cannot teleport, so position is continuous in time.
Form the difference. $h(t) = u(t) - w(t)$ on $[8, 16]$ . We want $h(t^*) = 0$ . A zero is a time of day when both days found him at the same place on the path.
Check the signs. On Monday he starts at the bottom and on Tuesday at the top: $u(8) = 0$ , $w(8) = 1$ , so $h(8) = -1$ . At 16:00 he is at the top on Monday and at the bottom on Tuesday: $h(16) = 1 - 0 = 1$ . The two days have opposite directions, so the difference changes sign.
Apply the theorem. $h$ is continuous on the connected set $[8, 16]$ with $h(8) = -1 < 0 < 1 = h(16)$ , so $h(t^*) = 0$ for some $t^*$ . The corollary. Nothing about his speed, rests or direction changes is needed.

Answer. There is a time $t^*$ when he is at the same point on both days. $\blacksquare$ Imagine two monks, one climbing and one descending on the same day: they must meet.

Your answerSuppose on Monday he climbs at constant speed,

u(t) = (t - 8)/8

, and on Tuesday he descends so that

w(t) = 1 - s^2

where

s = (t - 8)/8

. At what time

t

(in hours, as a decimal) are the two positions equal?

Paths

Definition B

Path

A path in a metric space $X$ is a continuous map $\gamma : [0, 1] \to X$ (continuity as in the paragraph above, with $[0, 1]$ carrying the distance $|s - t|$ ). It goes from $\gamma(0)$ to $\gamma(1)$ .

In one lineA path in a space $X$ is a continuous map $\gamma$ from the interval $[0, 1]$ into $X$ : a journey that starts at $\gamma(0)$ and ends at $\gamma(1)$ without jumping.

Why it existsIt makes 'you can walk from $p$ to $q$ without leaving the set' into a mathematical object.

Definition B

Path-connected

A metric space $X$ is path-connected if for all $p, q \in X$ there is a path $\gamma$ in $X$ from $p$ to $q$ .

In one lineA space is path-connected if any two of its points are joined by a path inside it.

Why it existsIt is connectedness you can check by exhibiting an explicit journey, rather than by ruling out all possible fences.

The simplest path is a straight line. In a normed space (4.1), the segment from $p$ to $q$ is $\gamma(t) = (1 - t)p + tq$ . It is continuous because $\lVert \gamma(t) - \gamma(s) \rVert = |t - s|\,\lVert q - p \rVert$ , so $\delta = \varepsilon / \lVert q - p \rVert$ works (or any $\delta$ if $p = q$ ). A set $S$ in a normed space is convex if the segment between any two of its points stays in $S$ , and a convex set is path-connected, because the segment between two of its points is a path that stays in the set.

Paths can be reversed: if $\gamma$ goes from $p$ to $q$ , then $t \mapsto \gamma(1 - t)$ goes from $q$ to $p$ and is continuous. Paths can also be glued. If $\alpha$ is a path from $p$ to $m$ and $\beta$ a path from $m$ to $q$ , then $\gamma(t) = \alpha(2t)$ for $t \le \frac12$ and $\gamma(t) = \beta(2t - 1)$ for $t \ge \frac12$ is a path from $p$ to $q$ : both formulas give $m$ at $t = \frac12$ , so there is no jump, and continuity at $t = \frac12$ is checked by taking the smaller of the two $\delta$ values.

Theorem B Munkres Sec. 24

Path-connected implies connected

If $X$ is path-connected, then $X$ is connected.

In one lineA path-connected space is connected.

Why it is trueA path from one side of a separation to the other would be a connected set (the image of $[0, 1]$ ) that sits on both sides.

ProofRudin 4.22 with 2.47

The idea: Suppose X is separated. Join a point of one piece to a point of the other by a path. The path's image is connected, so by the lemma it lies on one side, but it touches both.

Pick

p \in U

and

q \in V

(both pieces are non-empty).

Contradiction: assume

X = U \cup V

is a separation.

Let

\gamma

be a path in

X

from

p

q

X

is path-connected.

C = \gamma([0, 1])

is a connected subset of

X

[0, 1]

is an interval, hence connected, and

\gamma

is continuous, so its image is connected.

C \subseteq U

C \subseteq V

A connected subset lies on one side of a separation.

But

C

contains

p \in U

and

q \in V

, a contradiction. So

X

is connected.

\blacksquare

p = \gamma(0) \in C

and

p \in U

but

p \notin V

;

q = \gamma(1) \in C

and

q \in V

but

q \notin U

Worked example

Show that

\mathbb R^2 \setminus \{(0, 0)\}

is path-connected, hence connected, while

\mathbb R \setminus \{0\}

is not connected.

Solution.

Two points of the punctured plane. Let $p, q \ne (0, 0)$ . If the segment from $p$ to $q$ misses the origin, it is the path. A segment is a path in a normed space.
The bad case. If the segment passes through the origin, then $q = -cp$ for some $c > 0$ (the points are on opposite sides of the origin on one line). Pick any $m$ not on that line, for example $m = (-p_2, p_1)$ (the vector $p$ turned by $90^\circ$ ). The origin is on the line through $p$ and $q$ , and $m$ is chosen off that line.
Go around. The segment from $p$ to $m$ misses the origin, since the origin would be on the line through $p$ and $m$ only if $m$ were a multiple of $p$ , and it is not. Likewise the segment from $m$ to $q$ . Glue the two. Two segments joined at $m$ give a path avoiding the origin.
Conclude. So $\mathbb R^2 \setminus \{(0,0)\}$ is path-connected, so connected. Path-connected implies connected.
The line is different. $\mathbb R \setminus \{0\}$ is not an interval ( $-1 < 0 < 1$ and $0$ is missing), so it is not connected: $U = (-\infty, 0)$ , $V = (0, \infty)$ . In one dimension there is no way around the missing point.

Answer. The punctured plane is connected; the punctured line is not. $\blacksquare$ This is the observation that $\mathbb R$ and $\mathbb R^2$ are different shapes, made precise in 4.5.

Your answerA path from

(1, 0)

(-1, 0)

in the punctured plane goes by way of the point

(0, 1)

along two straight segments. What is its total length?

Worked example

Show that the circle

S^1 = \{(x, y) : x^2 + y^2 = 1\}

is path-connected.

Solution.

Write the points by angles. Any two points are $p = (\cos\alpha, \sin\alpha)$ and $q = (\cos\beta, \sin\beta)$ . Every point of the unit circle has an angle.
Define a path. $\gamma(t) = (\cos((1 - t)\alpha + t\beta), \sin((1 - t)\alpha + t\beta))$ . Then $\gamma(0) = p$ , $\gamma(1) = q$ , and $\gamma(t)$ is on the circle. We move the angle at constant speed from $\alpha$ to $\beta$ .
Continuity. Writing $\phi(t)$ for the angle, $\lVert \gamma(t) - \gamma(s) \rVert_2^2 = 2 - 2\cos(\phi(t) - \phi(s)) = 4\sin^2\frac{\phi(t) - \phi(s)}{2}$ , so the distance is $2|\sin\frac{\phi(t) - \phi(s)}2| \le |\phi(t) - \phi(s)| = |\beta - \alpha|\,|t - s|$ . The inequality $|\sin u| \le |u|$ turns this into a bound by $|t - s|$ , so $\delta = \varepsilon / (|\beta - \alpha| + 1)$ works.

Answer. The circle is path-connected, hence connected. $\blacksquare$

Quick checkWhich statement is a theorem?

Open connected sets of $\mathbb R^n$ are path-connected

Connected and path-connected differ for strange sets. For the open sets we use (balls, regions where a function is defined), they agree.

Theorem B standard; Apostol Sec. 10.10

Open connected sets in R^n are path-connected

Let $U \subseteq \mathbb R^n$ be open and connected (with the usual distance). Then $U$ is path-connected.

In one lineIf $U \subseteq \mathbb R^n$ is open and connected, then any two points of $U$ can be joined by a path in $U$ .

Why it is trueThe set of points reachable from a base point is open (a small ball is all reachable by segments) and so is the set of unreachable points, so connectedness forces everything to be reachable.

Proofstandard

The idea: Fix p. Let P be the set of points of U that can be reached from p by a path in U. Show that P and its complement in U are both open. Connectedness leaves only P = U.

Fix

p \in U

and let

P = \{x \in U : \text{some path in } U \text{ goes from } p \text{ to } x\}

. Then

p \in P

, so

P \ne \varnothing

Constant path at

p

For each

q \in U

choose

r > 0

with

B(q, r) \subseteq U

. Then every

x \in B(q, r)

is joined to

q

by the segment from

q

x

, which lies in

B(q, r) \subseteq U

U

is open: every point

q \in U

has

B(q, r) \subseteq U

. A ball is convex: for

x \in B(q, r)

and

0 \le t \le 1

\lVert q + t(x - q) - q \rVert = t\lVert x - q \rVert < r

q \in P

, every

x \in B(q, r)

is also in

P

. So

P

is open.

Glue a path from

p

q

with the segment from

q

x

q \in U \setminus P

, no

x \in B(q, r)

is in

P

. So

B(q, r) \subseteq U \setminus P

and

U \setminus P

is open.

x \in B(q, r)

were in

P

, glue the path from

p

x

with the segment from

x

back to

q

(a segment run backwards is a segment). Then

q \in P

U \setminus P = \varnothing

and

P = U

. Every point of

U

is joined to

p

by a path. Any two points

x, y

are joined by going through

p

: reverse the path from

p

x

, then follow the path from

p

y

\blacksquare

U

is connected and

P \ne \varnothing

is open and has an open complement in

U

: if

U \setminus P

were non-empty,

P

and

U \setminus P

would be a separation of

U

Connected but not path-connected

Beyond the course: for understanding· The topologist's sine curve (B)

Let $G = \{(x, \sin\frac1x) : 0 < x \le 1\}$ be the graph of $\sin\frac1x$ , and $L = \{(0, y) : -1 \le y \le 1\}$ the vertical segment it oscillates toward. The topologist’s sine curve is $S = G \cup L$ with the usual distance of the plane. It is connected but not path-connected: you cannot walk from the segment onto the wiggles.

Lemma B Munkres Thm 23.4

Closure of a connected set is connected

If $A$ is connected and $A \subseteq B \subseteq \overline A$ , then $B$ is connected.

In one lineIf $A$ is connected and $A \subseteq B \subseteq \overline A$ , then $B$ is connected.

Why it is trueA separation of $B$ would put all of $A$ on one side, but every point of the other side is a limit of points of $A$ .

ProofMunkres 23.4

The idea: Suppose B has a separation P, Q. The connected set A lies in one of them, say P. A point of Q is in the closure of A, so a ball around it meets A, but the ball can be taken to lie in Q.

Suppose

B = P \cup Q

is a separation. Then

A \subseteq P

A \subseteq Q

; say

A \subseteq P

The lemma above: a connected subset of

B

lies on one side of a separation of

B

Then

b \in B \subseteq \overline A

. Choose

r

with

B(b, r) \cap B \subseteq Q

Q

is non-empty. Take

b \in Q

Q

is open in

B

, so there is

r > 0

with

B_B(b, r) \subseteq Q

, a ball of the subspace.

a \in B(b, r) \cap B \subseteq Q

and

a \in A \subseteq P

. Then

a \in P \cap Q = \varnothing

, a contradiction.

\blacksquare

b \in \overline A

, so the ball

B(b, r)

contains a point

a \in A

. Also

a \in A \subseteq B

Proposition B standard; Munkres Sec. 24

The topologist's sine curve

(a) $S$ is connected. (b) $S$ is not path-connected.

In one line $S = G \cup L$ is connected, but there is no path in $S$ from $(0, 0)$ to $(1, \sin 1)$ .

Why it is true $G$ is connected and every point of $L$ is a limit of points of $G$ . A path that leaves $L$ would have to oscillate between $+1$ and $-1$ infinitely often right after leaving, which no continuous second coordinate can do.

ProofMunkres Sec. 24

The idea: (a) G is the continuous image of the interval (0, 1], so connected; L lies in the closure of G; apply the lemma. (b) Take the last time a path is on L and look at what the path does right after it.

(a)

G

is the image of

(0, 1]

under this continuous map, so

G

is connected.

t \mapsto (t, \sin\frac1t)

is continuous on

(0, 1]

: the distance in the plane is at most

|t - s| + |\sin\frac1t - \sin\frac1s|

, since

\sqrt{a^2 + b^2} \le |a| + |b|

, and

\sin\frac1t

is continuous for

t > 0

(0, 1]

is an interval, so connected.

Every point

(0, y)

L

is the limit of the points

(x_n, y) \in G

, so

L \subseteq \overline G

For

-1 \le y \le 1

choose

\theta \in [-\frac\pi2, \frac\pi2]

with

\sin\theta = y

and put

x_n = 1/(\theta + 2\pi n)

for

n

large enough that

\theta + 2\pi n > 1

. Then

x_n \to 0

and

\sin\frac1{x_n} = y

By the lemma on closures,

S

is connected.

G \subseteq S

by definition, and

S = G \cup L \subseteq \overline G

because

G \subseteq \overline G

and

L \subseteq \overline G

. So the lemma applies with

A = G

and

B = S

Let

Z = \{t : \gamma_1(t) = 0\}

. It contains

0

, not

1

, and it is closed (a limit of times in

Z

is in

Z

by continuity of

\gamma_1

). Let

t^* = \sup Z

. Then

t^* \in Z

and

t^* < 1

(b) By contradiction. Suppose

\gamma = (\gamma_1, \gamma_2) : [0, 1] \to S

is a path from

(0, 0)

(1, \sin 1)

. Each coordinate is continuous, since

|\gamma_i(t) - \gamma_i(s)|

is at most the distance between

\gamma(t)

and

\gamma(s)

For

t \in (t^*, 1]

\gamma_1(t) > 0

and

\gamma_2(t) = \sin\frac{1}{\gamma_1(t)}

After

t^*

the path is off

L

, so on the graph

G

\gamma_2(t) = \sin(1/\gamma_1(t))

with

\gamma_1(t) > 0

Choose

\eta > 0

with

t^* + \eta \le 1

so that

|\gamma_2(t) - \gamma_2(t^*)| < 1

for all

t \in [t^*, t^* + \eta]

. Let

m = \gamma_1(t^* + \eta) > 0

Continuity of

\gamma_2

t^*

with

\varepsilon = 1

Choose

k

so large that

v = 1/(\frac\pi2 + 2\pi k)

and

v' = 1/(\frac{3\pi}2 + 2\pi k)

are both at most

m

. There are times

t, t' \in (t^*, t^* + \eta]

with

\gamma_1(t) = v

and

\gamma_1(t') = v'

The intermediate value theorem for

\gamma_1

[t^*, t^* + \eta]

, where it goes from

0

m

: it takes every value in

(0, m]

Then

\gamma_2(t) = 1

and

\gamma_2(t') = -1

. Both are within

1

\gamma_2(t^*)

, but

1

and

-1

are

2

apart, so no number is within

1

of both. Contradiction.

\blacksquare

\sin(\frac\pi2 + 2\pi k) = 1

and

\sin(\frac{3\pi}2 + 2\pi k) = -1

In words: right after a path leaves the vertical segment it must keep oscillating between the top and the bottom, and a continuous function cannot do that on a shrinking window. The set is in one piece (no fence) but there is no road.

Examples and non-examples

Worked example

A = \{(x, y) : xy \ne 0\}

connected?

Solution.

Describe $A$ . $A$ is the plane without the two axes: the four open quadrants. $xy \ne 0$ means $x \ne 0$ and $y \ne 0$ .
Try the cut. Let $U = \{(x, y) \in A : x > 0\}$ and $V = \{(x, y) \in A : x < 0\}$ . The line $x = 0$ is removed from $A$ , so it acts as the gap.
Check the pieces. $U = A \cap \{x > 0\}$ and $V = A \cap \{x < 0\}$ are intersections of $A$ with open half-planes, so they are open in $A$ . Both are non-empty ( $(1, 1) \in U$ , $(-1, 1) \in V$ ), disjoint, and their union is $A$ . The lemma on open sets inside a subset: the half-plane $x > 0$ is open in $\mathbb R^2$ , because the ball of radius $x$ around $(x, y)$ stays in it.

Answer. $A$ is not connected: it has a separation, and in fact four connected pieces. $\blacksquare$

Worked example

Show that a continuous function

f : \mathbb R \to \mathbb R

that takes only integer values is constant.

Solution.

Image is connected. $\mathbb R$ is an interval, so connected; $f$ is continuous, so $f(\mathbb R)$ is connected. Continuous image of a connected set.
Image is an interval inside $\mathbb Z$ . A connected subset of $\mathbb R$ is an interval, and $f(\mathbb R) \subseteq \mathbb Z$ . If it contained two integers $m < n$ , it would contain $m + \frac12$ , which is not an integer.
So one point. $f(\mathbb R)$ has at most one point, and $f$ is constant. The image is non-empty and has at most one element.

Answer. $f$ is constant. $\blacksquare$ The same argument shows a continuous function from a connected space to $\{0, 1\}$ is constant, which is how topologists phrase “connected” as a property of functions.

Worked example

Show that the union of the two coordinate axes,

X = \{(x, y) : xy = 0\}

, is path-connected, hence connected.

Solution.

Each axis is a line. The $x$ -axis $\{(x, 0)\}$ and the $y$ -axis $\{(0, y)\}$ are convex, so any two points on the same axis are joined by the segment, which stays on that axis. A segment between two points of a line lies in the line.
Points on different axes. Take $p = (a, 0)$ and $q = (0, b)$ . Go from $p$ along the $x$ -axis to the origin, then from the origin along the $y$ -axis to $q$ . The origin lies on both axes, so it is a meeting point of the two segments.
Glue. The two segments end and start at $O$ , so gluing gives a path from $p$ to $q$ inside $X$ . The gluing rule from the paragraph above.
Conclude. Any two points of $X$ are joined by a path in $X$ , so $X$ is path-connected, and path-connected sets are connected. The theorem above.

Answer. $X$ is path-connected, hence connected. The same idea proves a general fact: two connected sets that share a point have a connected union (a problem below). $\blacksquare$

Worked example classic

Show that every continuous

f : [0, 1] \to [0, 1]

has a fixed point: a number

c

with

f(c) = c

Solution.

Turn the goal into a zero. Put $g(x) = f(x) - x$ . We want $g(c) = 0$ . $f(c) = c$ exactly when $g(c) = 0$ , and the intermediate value theorem finds zeros.
Look at the ends. $g(0) = f(0) - 0 \ge 0$ because $f(0) \ge 0$ . $g(1) = f(1) - 1 \le 0$ because $f(1) \le 1$ . $f$ takes values in $[0, 1]$ , so it cannot be below $0$ at the left end or above $1$ at the right end.
Apply the theorem. $g$ is continuous on the connected set $[0, 1]$ , and $0$ lies between $g(1)$ and $g(0)$ . So $g(c) = 0$ for some $c$ . The corollary with $y = 0$ . If $g(0) = 0$ or $g(1) = 0$ we already have the fixed point.

Answer. $f(c) = c$ for some $c \in [0, 1]$ . $\blacksquare$ The recipe “subtract to make a function that must change sign” is the same one used for the equator and the monk.

Worked example

Let

X

be a connected metric space and

x \ne y

two points of it. Show that for every

r

with

0 < r < d(x, y)

there is a point

z \in X

with

d(x, z) = r

Solution.

Make a real function. Let $h(z) = d(x, z)$ , a map $X \to \mathbb R$ . We want to hit the value $r$ and the theorem is about real-valued functions.
It is continuous. By the reverse triangle inequality, $|h(z) - h(z')| \le d(z, z')$ , so $\delta = \varepsilon$ works. The reverse triangle inequality is from 4.1.
Look at two values. $h(x) = 0$ and $h(y) = d(x, y)$ , and $r$ lies strictly between them. That is the hypothesis $0 < r < d(x, y)$ .
Apply the theorem. $X$ is connected, so there is $z$ with $h(z) = r$ . The corollary, on the connected domain $X$ .

Answer. Such a $z$ exists. So $h$ takes every value in $[0, d(x, y)]$ , and a connected metric space with two points is uncountable (a problem below). $\blacksquare$

Where it is used

Used inside maths· Inside maths: zero gradient on a connected open set means constant (a preview of Unit 5)

In 1.4 a function with zero derivative on an interval is constant. In several variables the analogue needs the right kind of domain. Let $U \subseteq \mathbb R^n$ be open and connected, and $f : U \to \mathbb R$ differentiable with $\nabla f = 0$ at every point. Then $f$ is constant. The proof uses only two facts from Unit 5: the chain rule, so that $g(t) = f(q + t(x - q))$ has $g'(t) = \nabla f(q + t(x - q)) \cdot (x - q)$ , and the 1.4 result that a function with zero derivative is constant.

Theorem B standard; the chain rule is in Apostol Sec. 8.20

Zero gradient on a connected open set means constant

Let $U \subseteq \mathbb R^n$ be open and connected, and $f : U \to \mathbb R$ differentiable with $\nabla f = O$ on $U$ . Then $f$ is constant.

In one lineIf $U \subseteq \mathbb R^n$ is open and connected and $f$ is differentiable on $U$ with $\nabla f = O$ everywhere, then $f$ is constant on $U$ .

Why it is trueThe set where $f$ equals its value at one point is closed (continuity) and open (a segment inside a small ball, with zero derivative along it), so it is all of $U$ by connectedness.

Proofstandard

The idea: Let S be the set of points where f takes the value c = f(p). Show S is open using the segments in a ball (derivative zero along a segment), and closed because f is continuous. A nonempty clopen subset of a connected set is everything.

Let

c = f(p)

and

S = \{x \in U : f(x) = c\}

Fix a point

p \in U

S

contains

p

, so

S \ne \varnothing

U \setminus S = f^{-1}(\mathbb R \setminus \{c\})

is open in

U

, so

S

is closed in

U

f

is differentiable, so continuous (1.4 in one variable; the same in several variables, Unit 5).

\mathbb R \setminus \{c\}

is open, so the preimage under

f

is open in

U

by the lemma on preimages.

Let

q \in S

and

x \in B(q, r) \subseteq U

. Put

g(t) = f(q + t(x - q))

for

t \in [0, 1]

U

is open: choose

r

with

B(q, r) \subseteq U

. For

x \in B(q, r)

the segment from

q

x

stays in the ball (balls are convex).

g'(t) = \nabla f(q + t(x - q)) \cdot (x - q) = 0

for all

t

Chain rule, then

\nabla f = O

g(1) = g(0)

, that is,

f(x) = f(q) = c

. Hence

B(q, r) \subseteq S

and

S

is open.

A function with zero derivative on an interval is constant (1.4).

By the proposition on clopen sets,

S = U

\blacksquare

S

is non-empty, open and closed in the connected set

U

The connectedness cannot be dropped: on $U = (0, 1) \cup (2, 3)$ let $f = 0$ on $(0, 1)$ and $f = 1$ on $(2, 3)$ . Then $f' = 0$ everywhere on $U$ but $f$ is not constant.

Used in the real world· Image segmentation: counting the pieces of a picture

A scanned page is stored as a grid of black and white pixels. To read it, an OCR program first cuts it into connected components, one for each blob of touching black pixels: the letter “i” has two components (the stem and the dot), “O” has one. Two pixels count as connected when you can walk from one to the other through black pixels that touch, which is path-connectedness in a space with only finitely many points. The count depends on what “touching” means. For the grid below, where $1$ is black, 4-neighbour touching (sharing an edge) gives five components, and 8-neighbour touching (sharing an edge or a corner) gives three, because the pixel in the middle touches two blobs diagonally.

Your answerIn the grid above, how many connected components do the black pixels form when two pixels count as touching if they share an edge or a corner (8 neighbours)?

Used in the real world· Neural networks: a decision boundary lies between any two images

Let $s(x)$ be the difference between the score a classifier gives to “cat” and to “dog” for an image $x$ with pixel values in $[0, 1]$ . For a network built from continuous layers (matrix multiplications, ReLU, and so on) $s$ is continuous. The set of images is a cube in $\mathbb R^n$ , which is convex, so path-connected. Take a cat image $a$ with $s(a) > 0$ and a dog image $b$ with $s(b) < 0$ . The function $t \mapsto s((1 - t)a + tb)$ is continuous on $[0, 1]$ and changes sign, so by the intermediate value theorem there is an image on the straight segment with $s = 0$ : a decision boundary point exists. Binary search on $t$ , which is bisection, finds it. Attacks on classifiers that see only the predicted label use this: they repeatedly halve the segment between a correct and a wrong image to land next to the boundary. The existence of the boundary point is the theorem of this chapter. The search works because each halving keeps a sign change.

Python labFind the equal-temperature pair by bisection

import numpy as np

rng = np.random.default_rng(1)
K = 5
a, b = rng.normal(size=K), rng.normal(size=K)

def T(theta):
    t = sum(a[k]*np.cos((k+1)*theta) + b[k]*np.sin((k+1)*theta) for k in range(K))
    # t = 1.0 if (theta % (2*np.pi)) < np.pi else 0.0   # a temperature with a jump
    return t

def g(theta):
    return T(theta) - T(theta + np.pi)

lo, hi = 0.0, np.pi
print("g(0) =", round(g(lo), 4), "  g(pi) =", round(g(hi), 4), "  (opposite signs)")
for _ in range(40):
    mid = (lo + hi) / 2
    if g(lo) * g(mid) <= 0:
        hi = mid
    else:
        lo = mid
th = (lo + hi) / 2
print("theta* =", th)
print("T(theta*)      =", T(th))
print("T(theta*+pi)   =", T(th + np.pi))

Try this. First, run it as is and check the last two printed numbers agree. Then change the seed from 1 to 2 and K from 5 to 1: for K = 1 the function T is one sine wave, g is twice T, and the pair is where T is 0. Finally, swap in the broken temperature on the last line of the function T (remove the comment mark in front of it) so that T jumps from 1 to 0. Bisection still converges to a number, but T(theta) and T(theta + pi) now differ by 1. Why does the argument fail? Because a function with a jump is not continuous, and the corollary needs continuity.

Traps

Watch out· Connected is not the same as path-connected

Path-connected implies connected; the converse fails for the topologist’s sine curve. For an open set of $\mathbb R^n$ the two agree. When a problem says “connected”, you may use the theorem about intervals or the clopen test; “path-connected” lets you exhibit a road, which is usually the easier proof of connectedness when it works.

Watch out· Open in the subset is not open in the whole space

$[0, 1]$ is open in the space $[0, 1] \cup [2, 3]$ and not open in $\mathbb R$ . In the definition of separation, “open in $A$ ” means open in the space $A$ itself, which is why the pieces of $[0, 1] \cup [2, 3]$ can be closed intervals. The lemma on open sets inside a subset translates: the pieces are $A \cap W$ for open $W$ of the bigger space.

Watch out· Connected needs one piece, not that the pieces touch

$[0, 1] \cup (1, 2]$ is the interval $[0, 2]$ , connected, though written in two parts: the pieces touch at $1$ . But $[0, 1) \cup (1, 2]$ is not connected, because the common endpoint $1$ is missing. Two sets that touch, with the touching point in one of them, join into a connected set (see Problem on unions). Two sets with a missing point between them do not.

Watch out· The intermediate value theorem needs connected inputs

A continuous function on $[0, 1] \cup [2, 3]$ with $f = -1$ on $[0, 1]$ and $f = 1$ on $[2, 3]$ never takes the value $0$ . The theorem holds for $f$ on a connected domain. The hypotheses “continuous” and “takes both signs” are not enough.

Watch out· Do not stretch the intermediate value theorem past connectedness

A connected set gives you every value between two values, and nothing more. It does not give you a road. The box below shows a problem (two climbers) where knowing a set is connected is not enough and a path is needed.

Beyond the course: for understanding· The two climbers (B): where connected is not enough

Two climbers start at the two feet of a mountain range, drawn as a continuous profile $f : [0, 1] \to \mathbb R$ with $f(0) = f(1) = 0$ and $f > 0$ in between. Can they move, one from each end, so that at every moment they are at the same height, and meet at the end? Whittaker (1966) gave the problem that name, and Homma (1952) had solved a version of it. For a range with finitely many peaks and valleys they always can, and Whittaker gave a range with infinitely many oscillations where they cannot. The set of pairs of positions $(s, t)$ with $f(s) = f(t)$ is a closed set of the square. A route for the climbers is a PATH in that set from $(0, 1)$ to a point of the diagonal. Knowing the set is connected, which an intermediate value argument can give, is not enough: you need a road, and that is the difference this chapter has drawn between connected and path-connected. For one peak (rising then falling), the climbers match heights using the inverse of the rising part and the inverse of the falling part, and meet at the summit.

Problems

check

Which of these subsets of $\mathbb R$ are connected? (a) $(0, 1) \cup \{2\}$ . (b) $[0, 1] \cup [1, 5]$ . (c) $\mathbb R \setminus \{\pi\}$ . (d) $\{x : x^2 \le 4\}$ .

Solution

(a) Not connected: $x = \frac12$ , $y = 2$ , $z = 1.5$ . (b) Connected: it is $[0, 5]$ . (c) Not connected: $\pi$ is missing. (d) Connected: it is $[-2, 2]$ .

check

If $f : [0, 1] \to \mathbb R$ is continuous with $f(0) = 3$ and $f(1) = -2$ , must $f$ take the value $0$ ? The value $17$ ?

Solution

Yes for $0$ , because $0$ is between $-2$ and $3$ . Not necessarily for $17$ : $f(x) = 3 - 5x$ is continuous with these end values and never reaches $17$ .

check

Write what ” $X$ is not connected” says, using the words “open” and “disjoint”.

Solution

$X$ is the union of two disjoint, non-empty, open subsets of $X$ .

check

Is the unit circle $\{(x, y) : x^2 + y^2 = 1\}$ connected? Is the circle with the point $(1, 0)$ removed connected?

Solution

The circle is path-connected (example above), hence connected. Removing $(1, 0)$ leaves the points $(\cos\theta, \sin\theta)$ with $0 < \theta < 2\pi$ , the image of the interval $(0, 2\pi)$ under a continuous map, so it is connected.

use

Show that $\cos x = x$ has a solution in $[0, 1]$ .

Solution

Let $g(x) = \cos x - x$ , continuous on the connected set $[0, 1]$ . $g(0) = 1 > 0$ and $g(1) = \cos 1 - 1 < 0$ (since $\cos 1 < 1$ ). By the corollary, $g(c) = 0$ for some $c \in (0, 1)$ .

use

Let $f : [0, 1] \to \mathbb R$ be continuous with $f(0) = f(1)$ . Show there is $c \in [0, \frac12]$ with $f(c) = f(c + \frac12)$ .

Solution

Let $g(c) = f(c + \frac12) - f(c)$ on $[0, \frac12]$ . Then $g(0) = f(\frac12) - f(0)$ and $g(\frac12) = f(1) - f(\frac12) = f(0) - f(\frac12) = -g(0)$ . So $g$ takes a value and its negative, hence $0$ in between.

use

Is $A = \{(x, y) \in \mathbb R^2 : x^2 + y^2 \ne 1\}$ connected? Is it path-connected?

Solution

Not connected. $U = \{x^2 + y^2 < 1\}$ and $V = \{x^2 + y^2 > 1\}$ are open in the plane (each point has a ball around it inside: radius $1 - \lVert p \rVert$ for $U$ and $\lVert p \rVert - 1$ for $V$ ), non-empty, disjoint, and $U \cup V = A$ . A set that is not connected is not path-connected.

use

A function $f$ on a connected open set $U \subseteq \mathbb R^2$ has $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$ and is differentiable. Is $f$ constant if $U$ is the open annulus $1 < \lVert p \rVert < 2$ ? If $U$ is the union of two disjoint discs?

Solution

The annulus is open. It is connected because it is the image of the rectangle $(1, 2) \times [0, 2\pi]$ , which is convex and so connected, under the continuous map $(r, \theta) \mapsto (r\cos\theta, r\sin\theta)$ . So $f$ is constant by the theorem on zero gradient. For two disjoint discs, no: $f$ can take a different constant value on each disc.

prove

Prove: if $A$ and $B$ are connected subsets of a metric space and $A \cap B \ne \varnothing$ , then $A \cup B$ is connected.

Solution

Plan. Suppose $A \cup B = P \cup Q$ is a separation. Both $A$ and $B$ lie on one side, and they share a point, so they lie on the same side.

By the lemma on connected sets and separations, $A \subseteq P$ or $A \subseteq Q$ , and $B \subseteq P$ or $B \subseteq Q$ . Let $z \in A \cap B$ . If $z \in P$ , then $A \subseteq P$ (as $A$ lies on one side and contains a point of $P$ ) and $B \subseteq P$ . Then $A \cup B \subseteq P$ , so $Q = \varnothing$ , a contradiction. If $z \in Q$ the same argument shows $P = \varnothing$ . So there is no separation. $\blacksquare$

prove

Prove: if $f : X \to Y$ is continuous and $X$ is path-connected, then $f(X)$ is path-connected.

Solution

Plan. Compose paths with $f$ .

Let $y_1 = f(x_1)$ and $y_2 = f(x_2)$ be points of $f(X)$ . Take a path $\gamma$ in $X$ from $x_1$ to $x_2$ . Then $f \circ \gamma : [0, 1] \to f(X)$ is continuous: given $\varepsilon$ and $t$ , continuity of $f$ at $\gamma(t)$ gives $\eta$ with $d(f(x), f(\gamma(t))) < \varepsilon$ whenever $d(x, \gamma(t)) < \eta$ , and continuity of $\gamma$ at $t$ gives $\delta$ with $d(\gamma(s), \gamma(t)) < \eta$ whenever $|s - t| < \delta$ . It goes from $y_1$ to $y_2$ . $\blacksquare$

prove

Prove that a convex subset of $\mathbb R^n$ is path-connected, and so connected. Deduce that every open ball is connected, and that the punctured line $\mathbb R \setminus \{0\}$ is not convex.

Solution

Plan. A convex set contains the segment between any two of its points, and the segment is a path.

If $S$ is convex and $p, q \in S$ , then $\gamma(t) = (1 - t)p + tq \in S$ for all $t$ . It is continuous because $\lVert \gamma(t) - \gamma(s) \rVert = |t - s|\,\lVert q - p \rVert$ . So $S$ is path-connected, hence connected. An open ball is convex: for $x, y \in B(c, r)$ , $\lVert (1 - t)x + ty - c \rVert \le (1 - t)\lVert x - c \rVert + t\lVert y - c \rVert < r$ . For $\mathbb R \setminus \{0\}$ : the segment from $-1$ to $1$ contains $0$ . $\blacksquare$

prove

Prove: every connected subset of $\mathbb Q$ has at most one point, using only the theorem on connected subsets of the line.

Solution

Plan. Two points of the subset would force an interval inside $\mathbb Q$ .

Suppose $C \subseteq \mathbb Q$ is connected with $x < y$ in $C$ . Then $C$ is a connected subset of $\mathbb R$ , so an interval, so $(x, y) \subseteq C \subseteq \mathbb Q$ . But $(x, y)$ contains the irrational number $x + \frac{\sqrt2}{n}$ for $n$ so large that $\frac{\sqrt2}n < y - x$ . That number is not in $\mathbb Q$ , a contradiction. $\blacksquare$

proveMATH 375 style (Apostol 8.3)

Show that $\{(x, y) : y > x^2\}$ is open and connected. Is it path-connected?

Solution

Plan. It is a convex open set.

Open: the function $F(x, y) = y - x^2$ is continuous, and the set is $F^{-1}((0, \infty))$ , open by the lemma on preimages (4.5 shows how to see this for functions of two variables). Convex: if $y_1 > x_1^2$ and $y_2 > x_2^2$ then for $0 \le t \le 1$ , $(1-t)y_1 + ty_2 > (1-t)x_1^2 + tx_2^2 \ge ((1-t)x_1 + tx_2)^2$ , because $x \mapsto x^2$ is convex: $(1-t)x_1^2 + tx_2^2 - ((1-t)x_1 + tx_2)^2 = t(1-t)(x_1 - x_2)^2 \ge 0$ . So the segment stays in the set, which is path-connected, hence connected. $\blacksquare$

stretch

Prove that a continuous injective function $f : [0, 1] \to \mathbb R$ is strictly monotone.

Solution

Plan. Move two pairs of points toward each other and watch that the difference of values never reaches $0$ .

Since $f$ is injective, $f(0) \ne f(1)$ . Say $f(0) < f(1)$ (otherwise replace $f$ by $-f$ ). Let $x < y$ in $[0, 1]$ . For $t \in [0, 1]$ put $p(t) = tx$ and $q(t) = (1 - t) + ty$ . Then $q(t) - p(t) = (1 - t) + t(y - x) > 0$ , so $p(t) \ne q(t)$ , and $\varphi(t) = f(q(t)) - f(p(t))$ is never $0$ by injectivity. It is continuous (compositions of continuous maps), on the connected set $[0, 1]$ . $\varphi(0) = f(1) - f(0) > 0$ . If $\varphi(1) = f(y) - f(x)$ were negative, the intermediate value theorem would give a zero of $\varphi$ . So $f(y) > f(x)$ . Hence $f$ is strictly increasing. $\blacksquare$

stretch

Prove that a connected metric space with at least two points is uncountable.

Solution

Plan. The distance from one fixed point is a continuous function with an interval for its image.

Let $x \ne y$ in $X$ , and $h(z) = d(x, z)$ . By the reverse triangle inequality (4.1 problems), $|h(z) - h(z')| \le d(z, z')$ , so $h$ is continuous ( $\delta = \varepsilon$ ). The image $h(X)$ is connected, so an interval, and contains $h(x) = 0$ and $h(y) = d(x, y) > 0$ , hence contains $[0, d(x, y)]$ . This interval is uncountable, and $h$ maps $X$ onto a set containing it, so $X$ is uncountable (a countable set has a countable image). $\blacksquare$

Summary

A separation of a set is a splitting into two disjoint, non-empty pieces, each open in the set itself. A set is connected if it has none, equivalently if its only clopen subsets are the empty set and the whole set.
The connected subsets of the line are exactly the intervals. The proof is the supremum argument of 1.3: a separation of an interval would need a last point of the left piece, which belongs to neither piece.
A continuous map sends connected sets to connected sets, because it pulls open sets back to open sets. The intermediate value theorem is the special case where the image lies in $\mathbb R$ .
Path-connected (any two points joined by a continuous path) implies connected. Convex sets and balls are path-connected, the punctured plane is path-connected, and the punctured line is not connected.
For open sets of $\mathbb R^n$ , connected and path-connected are the same. In general they differ: the topologist’s sine curve is connected, with no path onto its wiggles (B).
Zero gradient on a connected open set means constant; connectedness is what carries the information between different pieces, and it fails for two separate discs.
Counting connected components is how images are segmented, and the existence of a decision boundary between two classes follows from the intermediate value theorem.

4.4 Connectedness and path-connectedness

The picture: can you cut it in two?

The idea, in words

Definitions and theorems

Separations and connected sets

Open sets inside a subset

Separation

Connected set

Connected means no proper clopen subset

A connected subset lies on one side of a separation

The connected subsets of the line

Interval

Connected subsets of the line are the intervals

Continuous maps and connectedness

Continuous maps pull open sets back to open sets

The continuous image of a connected set is connected

Intermediate value theorem from connectedness

Paths

Path

Path-connected

Path-connected implies connected

Open connected sets of $\mathbb R^n$ are path-connected

Open connected sets in R^n are path-connected

Connected but not path-connected

Closure of a connected set is connected

The topologist's sine curve

Examples and non-examples

Where it is used

Zero gradient on a connected open set means constant

Traps

Problems

Summary

Where this sits

Comes from

Leads to

Same idea elsewhere

4.4 Connectedness and path-connectedness

The picture: can you cut it in two?

The idea, in words

Definitions and theorems

Separations and connected sets

Open sets inside a subset

Separation

Connected set

Connected means no proper clopen subset

A connected subset lies on one side of a separation

The connected subsets of the line

Interval

Connected subsets of the line are the intervals

Continuous maps and connectedness

Continuous maps pull open sets back to open sets

The continuous image of a connected set is connected

Intermediate value theorem from connectedness

Paths

Path

Path-connected

Path-connected implies connected

Open connected sets of Rn\mathbb R^nRn are path-connected

Open connected sets in R^n are path-connected

Connected but not path-connected

Closure of a connected set is connected

The topologist's sine curve

Examples and non-examples

Where it is used

Zero gradient on a connected open set means constant

Traps

Problems

Summary

Where this sits

Comes from

Leads to

Same idea elsewhere

Open connected sets of $\mathbb R^n$ are path-connected