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Section 1.3 Connectedness

Definition 1.3.1.

Let \(X \subseteq \RR^n\) be a subspace. A subset \(S \subseteq X\) that is both open and closed (in \(X\)) is said to be clopen.

We shall say that \(X\) is connected if there are exactly two clopen subsets \(S \subseteq X\text{.}\)

If \(X \subseteq \RR^n\) is a nonempty subspace, then there are always at least two clospen subsets of \(X\text{:}\) the empty set \(\varnothing\text{,}\) and \(X\) itself. Thus if \(X\) is nonempty, then it is connected if and only if the only nonempty clopen subset is \(X\) itself.

The empty set \(\varnothing\text{,}\) however, is not connected: it has only one subset, itself, which is clopen.

The space \(\RR\) itself is connected. Let's prove this. There are at least two clopen subsets: \(\varnothing\) and \(\RR\) itself. Now we have to prove that there are no more. So let \(S \subseteq \RR\) be a clopen subset. If \(S \neq \varnothing\text{,}\) then there exists a point \(x \in S\text{.}\) Since \(S\) is open, there exists \(\varepsilon >0\) such that \(\left]x-\varepsilon, x+\varepsilon\right[ \subseteq S\text{.}\) Now let us consider the set

\begin{equation*} E \coloneq \left\{ \varepsilon >0 : \left]x-\varepsilon, x+\varepsilon\right[ \subseteq S \right\} \subseteq \RR\text{.} \end{equation*}

The set \(E\) is unbounded if and only if \(S = \RR\text{,}\) so assume that \(E\) in bounded. We aim to generate a contradiction. Since \(E\) is bounded, it admits a supremum \(\varepsilon_0\text{.}\) In particular, for every \(\varepsilon \lt \varepsilon_0\text{,}\) one has \(\left]x-\varepsilon, x+\varepsilon\right[ \subseteq S\text{.}\) Note that \(\left]x-\varepsilon_0, x+\varepsilon_0\right[ = \bigcup_{\varepsilon \lt \varepsilon_0} \left]x-\varepsilon, x+\varepsilon\right[\text{,}\) so it follows that \(\left]x-\varepsilon_0, x+\varepsilon_0\right[ \subseteq S\) as well. Thus \(\varepsilon_0 \in E\text{.}\)

Now consider the point \(x+\varepsilon_0\text{.}\) This is close to \(S\text{,}\) because the closure of \(S\) contains the closure of \(\left]x-\varepsilon_0, x+\varepsilon_0\right[\text{,}\) which is \([x-\varepsilon_0,x+\varepsilon_0]\text{.}\) Since \(S\) is closed, it follows that \(x +\varepsilon_0 \in S\text{.}\) The same analysis shows that \(x-\varepsilon_0 \in S\text{.}\)

Again since \(S\) is open, we may choose a \(\delta>0\) such that \(\left]x+\varepsilon_0-\delta, x+\varepsilon_0+\delta\right[ \subset S\) and \(\left]x-\varepsilon_0-\delta, x-\varepsilon_0+\delta\right[ \subset S\text{.}\) But now we have that \(\left]x-\varepsilon_0-\delta, x+\varepsilon_0+\delta\right[ \subset S\text{,}\) so \(\varepsilon_0+\delta \in E\text{,}\) which contradicts the maximality of \(\varepsilon_0\text{.}\)

This contradiction shows that \(E\) is unbounded, and so the only nonempty clopen subset of \(\RR\) is \(\RR\) itself.

The subspace

\begin{equation*} Y \coloneq \{(x,y) \in \RR^2 : |x|=1 \} \end{equation*}

we introduced above is not connected. Indeed, in addition to \(\varnothing\) and \(Y\) itself, the subset

\begin{equation*} Y_+ \coloneq \{(x,y) \in \RR^2 : x=1 \} \end{equation*}

is clopen. To see this, let us show that both \(Y_+\) and its complement

\begin{equation*} Y_- \coloneq Y \smallsetminus Y_+ = \{ (x,y) \in \RR^2 : x = -1 \} \end{equation*}

are open. The key observation here is that if \(u \in Y_+\) and \(v \in Y_-\text{,}\) then \(d(u,v) \geq 2\text{.}\) Consequently, if \(u \in Y_+\text{,}\) then \(B^2(u,1) \cap Y \subset Y_+\text{,}\) and, similarly, if \(v \in Y_-\text{,}\) then \(B^2(v,1) \cap Y \subset Y_-\text{.}\)

Since \(X_a \cap X_b\) is nonempty, the union \(X\) is nonempty too. Hence it suffices to show that if \(V \subseteq X\) is a nonempty clopen, then \(V = X\text{.}\)

Note that for any \(a\in A\text{,}\) the intersection \(V \cap X_a\) is clopen in \(X_a\text{.}\) For each \(a\in A\text{,}\) the subspace \(X_a\) is connected, so \(V \cap X_a\) is either \(\varnothing\) or \(X_a\text{.}\) In other words, for each \(a \in A\text{,}\) either \(V\) is disjoint from \(X_a\) or else \(X_a \subseteq V\text{.}\) Since \(V\neq\varnothing\text{,}\) there is at least one \(a_0 \in A\) such that \(X_{a_0} \subseteq V\text{.}\)

But now for any other \(a\in A\text{,}\) the nonempty intersection \(X_a \cap X_{a_0}\) is contained in \(X_a \cap V\text{;}\) thus \(X_a \subseteq V\) as well. We thus conclude that \(X \subseteq V\text{.}\)

We are now in a position to classify all the connected subspaces of \(\RR\text{.}\)

Let \(X \subseteq\RR\) be a nonempty subset. We'll say that \(X\) is an interval if and only if, for every \(a, b \in X\) and every \(x \in \RR\) such that \(a \leq x \leq b\text{,}\) we have \(x\in X\text{.}\) Assume that \(X\) is an interval. If \(X\) is bounded above, then there exists a supremum \(b \in \RR\text{.}\) If \(X\) is bounded below, then there exists an infimum \(a \in \RR\text{.}\) Now, depending upon whether \(a\) and \(b\) exist, there are three options for an interval \(X\text{:}\)

  • a interval of finite length such as \([a,b]\text{,}\) \(\left]a,b\right]\text{,}\) \(\left[a,b\right[\text{,}\) or \(\left]a,b\right[\text{;}\)

  • a ray \(\left]a,+\infty\right[\text{,}\) \(\left[a,+\infty\right[\text{,}\) \(\left]-\infty, b\right[\text{,}\) or \(\left]-\infty, b\right]\text{;}\) or

  • the line \(\RR\) itself.

Here now is our claim: a subspace \(X \subseteq \RR\) is connected if and only if it is an interval — hence if and only if it is of one of the three forms above. To prove this, we must prove two things:

  • first, that any interval is connected, and

  • second, that any subspace that is not an interval is not connected.

To prove the first statement, assume first that \(X\) is a closed interval \([a,b]\text{.}\) Assume that \(V \subseteq [a,b]\) is a clopen subset. Suppose that \(x \in [a,b]\) is a point such that \(x \notin V\text{;}\) we aim to show that \(V\) is empty.

If there are points \(s\in V\) such that \(s \lt x\text{,}\) let

\begin{equation*} s_0 \coloneq \sup \{s\in V : s\lt x\}\text{.} \end{equation*}

Thus \(s_0 \leq x\text{.}\) Observe that \(s_0\) is close

to \(V\text{.}\) Since \(V\) is closed, it follows that \(s_0 \in V\text{.}\) Since \(x \notin V\text{,}\) it follows that \(s_0\lt x\text{.}\) Since \(V\) is open, there is a \(\varepsilon >0\) such that \(\left]s_0-\varepsilon,s_0+\varepsilon\right[ \cap [a,b]\) is contained in \(V\text{.}\) In particular, there exist points \(s \in V\) such that \(s_0 \lt s \lt x\text{,}\) which contradicts the definition of \(s_0\text{.}\) Consequently, there are no points \(s\in V\) such that \(s\lt x\text{.}\)

On the other hand, if there are points \(s \in V\) such that \(s>x\text{,}\) then let

\begin{equation*} s_1 \coloneq \inf \{s\in V : s>x\}\text{.} \end{equation*}

We can replay the same argument as above to generate a contradiction. Since \(V\) is closed, \(s_1 \in V\text{.}\) Since \(V\) is open, there are points \(s\in V\) such that \(x\lt s\lt s_1\text{,}\) which is a contradiction. It therefore follows that \(V=\varnothing\text{,}\) as desired.

So far, we have shown that a closed interval \([a,b]\) is always connected. Now for a general interval \(X\text{,}\) note that if \(x\in X\text{,}\) then we may write

\begin{equation*} X = \bigcup_{\substack{a,b\in X, \\ a\leq x\leq b} } [a,b]\text{.} \end{equation*}

This is a union of connected subspaces of \(\RR\) such that any two intersect (because they all contain \(x\)). Hence \(X\) is connected as well, thanks to the previous proposition.

Now let us prove the converse. Assume that \(X \subset \RR\) is not an interval. We aim to prove that \(X\) is not connected; it will suffice to construct a nonempty proper clopen \(V \subseteq X\text{.}\) For this, choose real numbers \(a\lt x\lt b\) such that \(a,b \in X\) but \(x \notin X\text{.}\) Now consider the subset

\begin{equation*} V = \left[x,+\infty\right[ \cap X = \left]x,+\infty\right[ \cap X\text{.} \end{equation*}

The first expression proves that \(V\) is closed; the second proves that \(V\) is open. Since \(b \in V\) but \(a\notin V\text{,}\) it follows that \(V\) is nonempty and proper.

Here's our first truly topological theorem. We will think of this as a generalization of the intermediate value theorem.

Assume that \(X\) is connected. In particular, \(X\) is nonempty, hence so is \(Y\text{.}\)

Let \(V \subseteq Y\) be a nonempty clopen. Then \(f^{-1}(V) \subseteq X\) is clopen since \(f\) is continuous, and it is nonempty since \(f\) is a surjection. Since \(X\) is connected, it follows that \(f^{-1}(V) = X\) itself. This implies that the image \(f(X)\) is contained in \(V\text{.}\) Since \(f\) is a surjection, it follows that \(V=Y\text{.}\) Thus \(Y\) has exactly two clopen subsets: \(\varnothing\) and \(Y\text{.}\)

Let \(X \subseteq \RR^n\) be a connected subspace, and let \(f \colon X \to \RR\) be a continuous map. Then the image \(f(X)\) is an interval. This is our friend, the intermediate value theorem.

Consider the following assertion: there are two antipodal

points on the Earth's equator that, at ground level, have exactly the same temperature.

In other words, there are two coordinates of the form

\begin{equation*} 0^{\circ}\text{ S, } \; x^{\circ}\text{E} \andeq 0^{\circ}\text{ N, } \; (180-x)^{\circ}\text{W} \end{equation*}

where at ground level the temperature is exactly the same.

Here's the proof. The Earth's equator is homeomorphic to the circle \(S^1\text{,}\) which is a connected subspace of \(\RR^2\text{.}\) Temperature at the equator is thus a continuous function \(T \colon S^1 \to \RR\text{.}\) Now define a related continuous map \(g \colon S^1 \to \RR\) by the formula

\begin{equation*} g(x) = T(x)-T(-x)\text{.} \end{equation*}

Now since \(x\) and \(-x\) are antipodal points, our claim will be proved if we can show that there is a point \(x\in S^1\) such that \(g(x)=0\text{.}\) If \(g(1,0)=0\text{,}\) then we are done. Otherwise, our formula ensures that \(g(-1,0)=g(1,0)\text{,}\) so \(g(1,0)\) and \(g(-1,0)\) have different signs.

But since the image \(g(S^1) \subseteq\RR\) is an interval, it follows that \(0 \in g(S^1)\text{.}\) Hence there exists \(x \in S^1\) such that \(g(x)=0\text{.}\)

Since \(\RR \smallsetminus\{0\}\) and the subspace

\begin{equation*} Y = \{(x,y) \in \RR^2 : |x|=1\} \end{equation*}

are homeomorphic, it follows that \(\RR\smallsetminus \{0\}\) is not connected.

Since \(\RR\) is connected, it follows that it is not homeomorphic to \(\RR \smallsetminus \{0\}\text{.}\)

A good way to confirm that a nonempty subspace \(X \subseteq \RR^n\) is connected is to confirm that you can “walk” from every point to every other point without leaving \(X\text{.}\) For any two points \(x,y \in X\text{,}\) a path from \(x\) to \(y\) in \(X\) is a continuous map

\begin{equation*} \gamma \colon [0,1] \to X \end{equation*}

such that \(\gamma(0)=x\) and \(\gamma(1) = y\text{.}\) If \(X\) has the property that for every pair of points from \(x\) to \(y\text{,}\) there exists a path from \(x\) to \(y\) in \(X\text{,}\) then \(X\) is connected.

Why? Well, choose \(x\in X\text{.}\) For every path \(\gamma\) from \(x\) to another point in \(X\text{,}\) consider the image \(\gamma([0,1])\text{;}\) this is connected, and our assumption on \(X\) ensures that \(X\) is the union of all these images, all of which contain \(x\text{.}\) Hence \(X\) is connected.

The line \(\RR\) is not homeomorphic to \(S^1\text{.}\) Indeed, suppose it were; choose a homeomorphism \(f \colon \RR \to S^1\text{.}\) This will restrict to a homeomorphism \(\RR \smallsetminus \{0\} \to S^1\smallsetminus\{f(0)\}\text{.}\) However, we claim that \(S^1 \smallsetminus \{f(0)\}\) is connected. In fact, we'll do better: we shall prove that \(S^1 \smallsetminus \{f(0)\}\) is homeomorphic to \(\RR\text{.}\)

To make our formulas simpler, let's use complex numbers.

We have:

\begin{equation*} S^1 = \{z \in \CC : |z| = 1 \}\text{.} \end{equation*}

Let \(w = f(0)\text{.}\) We may first construct a homeomorphism \(S^1\smallsetminus\{w\} \to S^1 \smallsetminus \{1\}\) by the rule \(z \mapsto z/w\text{.}\) Since the inverse map is \(z \mapsto wz\text{,}\) which is also continuous, we are done.

With this done, we now construct a homeomorphism \(\left]0,1\right[ \to S^1 \smallsetminus \{1\}\) by the rule \(t \mapsto \exp(2\pi i t)\text{.}\) This map is a continuous bijection, but what about its inverse \(r\text{?}\) To make it easy to deduce the continuity of \(r\) from well-known bits of analysis, we write it as:

\begin{equation*} r(x,y) = \frac{1}{2}-\frac{1}{\pi}\arctan\left(\frac{y}{1-x}\right)\text{.} \end{equation*}

It now follows that \(S^1\smallsetminus\{1\}\) is homeomorphic to an open interval and thus to \(\RR\) itself.