## Connected Spaces and Path-Connected Spaces

This post isn’t related to anything I’ve been focusing on as of late, but it’s in response to a student’s question. Last week, a student asked me about proving that every path-connected space is connected–at least, I think that’s what he asked. I asked if he had tried to prove it, to which he responded “not yet.” Not wishing to deprive him of self-discovery and being a tad lazy, I told him to try on his own and get back to me if he ran into trouble.

As I have not heard back from him on this matter, I assume that he proved the result on his own. To make up for my laziness, I have decided to write a short exposition of the topological property of connectedness. We begin by reviewing some definitions.

We say a topological space $(X,\tau)$ is connected if the only nonempty open and closed subset of $X$ is $X$ itself. If $x_{1},x_{2}$ are two points in $X$, then a path between $x_{1}$ and $x_{2}$ is a continuous function $\gamma:[0,1]\rightarrow X$ such that $\gamma(0)=x_{1}$ and $\gamma(1)=x_{2}$. If every two points in $X$ are joined by a path, then we say that $X$ is path-connected.

There are a number of equivalent definitions of a connected space, and we briefly list some here:

1. The only subsets of $X$ both open and closed are $\emptyset, X$.
2. $X$ cannot be written as the union of two disjoint, nonempty open sets.
3. $X$ cannot be written as the union of two disjoint, nonempty closed sets.
4. $X$ cannot be written as the union of two separated sets.
5. If $f: X\rightarrow\left\{0,1\right\}$ is continuous, then $f$ is constant.

What are some examples of connected spaces? Your intuition might suggest to you that the closed unit interval $[0,1]$ is connected, and indeed, your intuition would be correct. To see this, recall that an equivalent definition of a connectedness space is that the only subsets, which are both open and closed, are $\emptyset$ and $X$. Let $E$ be a nonempty subset of $[0,1]$ which is both open and closed. Define real number $t_{0}$ and $t_{\max}$ by

$\displaystyle t_{0}=\inf E,\indent t_{\max}=\sup\left\{t\in E: (t_{0},t)\subset E\right\}$

I claim that $t_{\max}=1$. Note that since $E$ is closed, both $t_{0}$ and $t_{\max}$ belong to $E$. If $t_{\max}<1$, then since $E$ is open, there exists $\varepsilon>0$ such that $(t_{\max}-\varepsilon,t_{\max}+\varepsilon)\subset E$. Taking $\varepsilon$ smaller if necessary, we may assume that $t_{\max}+\varepsilon<1$. By definition of $t_{\max}$, there exists a point $t\in (t_{0},t_{\max}+\varepsilon)$ which does not belong to $E$ (in particular, $t\neq t_{\max}$). We see that $t\notin (t_{0},t_{\max})$ since this interval is a subset of $E$. Similarly, $t\notin (t_{\max},t_{\max}+\varepsilon)$ since this interval is also a subset of $E$, by choice of $\varepsilon$. We arrive at a contradiction.

A completely analogous argument shows that $t_{0}=0$, and we conclude that $[0,1]\subset E\Longleftrightarrow E=[0,1]$. $\Box$

We can prove a more general result concerning connected subsets of the real line. This result can be found in Rudin, Principles of Mathematical Analysis Theorem 2.47.

Lemma 1. $E\subset\mathbb{R}$ is connected if and only if

$\displaystyle x,y\in E\text{ and }z\in (x,y)\Longrightarrow z\in E$

Proof. Suppose there exists $x,y\in E$ (w.l.o.g. $x) and $z\in (x,y)\cap E^{c}$. We can then write $E$ as the union of two disjoint open subsets $A_{z}$ and $B_{z}$ defined by

$\displaystyle A_{z}:=E\cap (-\infty,z),\indent B_{z}:=E\cap(z,\infty)$

$x\in A_{z}$ and $y\in B_{z}$ so these two sets are nonempty. Hence, $E$ is the union of nonempty, disjoint open sets, which implies that $E$ is not connected.

Suppose $E$ is not connected. Then there exist two disjoint, open (with respect to the subspace topology) sets $A$ and $B$ such that $E=A\cup B$. Choose $x\in A$ and $y\in B$, where w.l.o.g. $x. Define a point $z\in \mathbb{R}$ by

$\displaystyle z:=\sup\left\{t\in \mathbb{R}:t\in A\cap [x,y]\right\}$

Denote the closures of $A$ and $B$ in $\mathbb{R}$ by $\overline{A}$ and $\overline{B}$, respectively. It is clear that $z\in\overline{A}$ and therefore $z\notin B$ by definition of the subspace topology. If $z\notin A$, then $x and $z\notin E$, since $A$ and $B$ partition $E$. If $z\in A$, then $z\notin\overline{B}$, so there exists $\varepsilon>0$ such that $(z-\varepsilon,z+\varepsilon)\cap \overline{B}=\emptyset$. Choosing $z'$ in this interval, we obtain a point which is in neither $A$ nor $\overline{B}$. Hence, $z'\notin E$. $\Box$

“Path-connected implies connected” follows readily from the lemma.

Proposition 2. If $(X,\tau)$ is a path-connected topological space, then it is also connected.

Proof. Suppose there exist two nonempty, disjoint open sets $U_{1}$ and $U_{2}$ such that

$\displaystyle X=U_{1}\cup U_{2}$

Let $x_{1},x_{2}$ be points in $U_{1}$ and $U_{2}$, respectively. By hypothesis that $X$ is path-connected, there exists a continuous function $\gamma:[0,1]\rightarrow X$ such that $\gamma(0)=x_{1}$ and $\gamma(1)=x_{2}$. By continuity, the pre-images $\gamma^{-1}(U_{1})$ and $\gamma^{-1}(U_{2})$ are open. But then $[0,1]$ is the union of two disjoint, nonempty open sets since

$\displaystyle[0,1]=\gamma^{-1}(X)=\gamma^{-1}(U_{1})\cup\gamma^{-1}(U_{2})$,

which contradicts the connectedness of $[0,1]$. $\Box$

The converse is not true, and there are two well-known examples of connected topological spaces which are not path-connected: the deleted comb space and the topologist’s sine curve. We will only cover the second example. Before doing so, I want to describe some properties of connected spaces.

First, continuity preserves connectedness and path-connectedness.

Proposition . If $(X,\tau_{X})$ and $(Y,\tau_{Y})$ are topological spaces, where $X$ is connected (path-connected), and $f: X\rightarrow Y$ is continuous, then $f(X)$ is a connected (path-connected) subspace of $Y$.

Proof. Let $E$ be a nonempty subset of $f(X)$ which is both open and closed in $f(X)$ equipped with the subspace topology. By definition of the subspace topology, we can write

$\displaystyle E=E'\cap f(X), \indent E=E''\cap f(X)$

where $E'$ is open in $Y$ and $E''$. Since $f$ is continuous, $f^{-1}(E')$ and $f^{-1}(Y)$ are open in $X$, so that $f^{-1}(E)=f^{-1}(E')\cap f^{-1}(Y)$ is open in $X$. Similarly, $f^{-1}(E'')$ and $f^{-1}(Y)$ are closed in $X$, so that $f^{-1}(E)$ is closed in $X$. Since $X$ is connected, we must have that $f^{-1}(E)=X$, which implies that $E=f(X)$.

Now suppose that $X$ is path-connected. Fix two points $f(x_{1})$ and $f(x_{2})$ in $f(X)$, and let $\gamma:[0,1]\rightarrow X$ be a path between $x_{1}$ and $x_{2}$. Then $f\circ\gamma:[0,1]\rightarrow f(X)$ is a continuous function between $f(x_{1})$ and $f(x_{2})$. $\Box$

The following proposition is a generalization of an elementary result in Calculus known as the intermediate value theorem.

Proposition . Suppose $X$ is a connected topological space, and $f: X\rightarrow\mathbb{R}$ is continuous. If $f(x), for two points $x,y\in X$, then there exists a point $z\in X$ such that $f(x)=c$.

Proof. Suppose there exist $x,y\in X$ and some $c\in\mathbb{R}$ between $f(x)$ and $f(y)$ (w.l.o.g. $f(x)), such that $f^{-1}(\left\{c\right\})=\emptyset$. Since $f$ is continuous, the two pre-images

$\displaystyle f^{-1}(-\infty,c),\indent f^{-1}(c,\infty)$

form a partition of $X$ into two nonempty, open sets. But this contradicts our hypothesis that $X$ is connected. $\Box$

Let’s go back to the topologist’s sine curve. Consider $\mathbb{R}^{2}$ endowed with the standard topology. We define the topologist’s sine curve by taking the set

$\displaystyle T=\left\{\left(x,\sin\frac{1}{x}\right):x\in(0,1]\right\}\cup\left\{(0,0)\right\}$

and endowing it with the subspace topology. What topological properties does $T$ have? You might be tempted to say that $T$ is closed in $\mathbb{R}^{2}$, but this is incorrect; $T$ is neither open nor closed. The first claim is obvious, so consider the second. For any $x\in (0,1]$, there exists some $t\in (0,\frac{\pi}{2}]$ such that $x=\sin t$. Since $\sin$ is $2\pi$-periodic, for any $n\in\mathbb{N}$,

$\displaystyle x=\sin t=\sin(t+2n\pi)=\sin\frac{1}{\left(t+2n\pi\right)^{-1}}, \indent \left(t+2n\pi\right)^{-1}\downarrow 0, n\rightarrow\infty$

Hence, $(0,x)$ is a limit point of $T$ that does not belong to $T$.

If we adjoin the set $\left\{0\right\}\times(0,1]$ to $T$, then we obtain the closed topologist’s sine curve. Indeed, let $(c_{1},c_{2})$ be a point $\mathbb{R}^{2}\setminus T$. If $c_{1}\notin[0,1]$, then it’s clear that there exists an open disk about $(c_{1},c_{2})$ not intersecting $T$, so assume $c_{1}\in (0,1]$ and therefore $\sin\frac{1}{c_{1}}\neq c_{2}$. Since $\sin\frac{1}{x}$ is continuous on $(0,\infty)$, there exists $\delta>0$ such that

$\displaystyle t\in[c_{1}-\delta,c_{1}+\delta]\Longrightarrow \left|\sin\frac{1}{t}-\sin\frac{1}{c_{1}}\right|<\frac{\left|\sin\frac{1}{c_{1}}-c_{2}\right|}{2}=\varepsilon$

For $t\in (c_{1}-\delta,c_{1}+\delta)$, $\sin\frac{1}{t}\in(c_{2}-\varepsilon,c_{2}+\varepsilon)$ implies that

$\displaystyle\left||\alpha-c_{2}\right|<\left|\sin\frac{1}{t}-\sin\frac{1}{c_{1}}\right|+\left|\sin\frac{1}{t}-c_{2}\right|<2\varepsilon=\left|\alpha-c_{2}\right|$,

which is a contradiction. Hence, the open square $(c_{1}-\delta,c_{1}+\delta)\times (c_{2}-r,c_{2}+r)$ does not intersect $T$. Since $T$ is also bounded, the Heine-Borel theorem implies that $T$ is compact.

To see that $T$ is connected, let $E\subset T$ be a nonempty subset, which is both open and closed. Suppose $E\neq T$. Since $E^{c}$ is also open, there exists $t\in E^{c}$ satisfying $0. Define

$\displaystyle a:=\inf\left\{t\in [0,1]: \left(t,\sin\frac{1}{t}\right)\in E\right\},\indent b=\sup\left\{t\in[0,1]:\left(t,\sin\frac{1}{t}\right)\in E\right\}$

Since $E$ is closed, $a,b\in E$. There exists $c\in(a,b)\cap E^{c}$. Otherwise $[a,b]=E$, which implies that $E=T$, since it is an open and closed set. Define

$\displaystyle x:=\sup\left\{t

Since $E$ is closed, $x\in E$. Since $E$ is open, there exists $\varepsilon>0$ such that

$\displaystyle\left|\left(u,\sin\frac{1}{u}\right)-\left(x,\sin\frac{1}{x}\right)\right|<\varepsilon\Longrightarrow\left(u,\sin\frac{1}{u}\right)\in E$

But by definition of $x$, for any $\varepsilon>0$, there exists $u\in (x,x+\varepsilon)$ such that $(u,\sin\frac{1}{u})\in E^{c}$, which is a contradiction.

To show that $T$ is not path-connected, we will prove that $T$ does not contain a path from $(0,0)$ to $(\frac{2}{\pi},1)$. Assume the contrary, and let $\gamma:[0,1]\rightarrow X$ denote such a path. Since $\gamma$ is continuous, there exists $\delta>0$ such that

$\displaystyle t\in(0,\delta)\Longrightarrow\left|\gamma(t)\right|<\frac{1}{2}$

Let $\gamma_{1}$ denote the projection of $\gamma$ onto the first cooridnate. Being the composition of continuous functions, $\gamma_{1}$ is a continuous function $[0,1]\rightarrow[0,1]$. Choose $n\in\mathbb{N}$ sufficiently large so that

$\displaystyle\gamma_{1}(0)=0<\frac{2}{(2n+1)\pi}<\gamma_{1}(\frac{1}{2}\delta)$

By the intermediate value theorem, there exists a point $c\in(0,\frac{\delta}{2})$ such that

$\displaystyle\gamma_{1}(c)=\frac{2}{(2n+1)\pi}\Longrightarrow \gamma(c)=\left(\frac{2}{(2n+1)\pi},\sin\left(\frac{2}{(2n+1)\pi}\right)^{-1}\right)=\left(\frac{2}{(2n+1)\pi},(-1)^{n}\right)$

But, $\left|\gamma(c)\right|\geq\frac{1}{2}$, which is a contradiction.