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<y) 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<y. 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<z<y 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


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)<c<f(y), 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)<c<f(y)), 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


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<t<1. 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<c: (t,\sin\frac{1}{t})\in E\right\}

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


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.

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