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10. Connectedness

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Last week, we talked about a topological property of metric spaces called compactness. This loosely describes the “size” of a metric space: compact metric have limited space, and things can’t be “infinitely spread out”, whatever that means. This is very vague, but the topological data of a metric space really does encode some notion of “size” — we’ll see some more of this when we talk about the Baire category theorem.

Today’s topic is connectedness, which is our first foray into the ways in which we can understand the shape of a metric space from its topology.

We should note here that if \(X=U\cup V\) for two disjoint open sets \(U\) and \(V\), then \(X\setminus U=V\) and \(X\setminus V=U\). Thus both \(U\) and \(V\) are clopen; reinterpreting the definition above, if \(X\) is connected, then its only clopen subsets are \(\varnothing\) and \(X\) itself.

This definition of connectedness may be funny at first, so it may be helpful to think about what it means to be disconnected. Consider the metric space \(X=\left[ 0,1 \right]\cup \left[ 2,3 \right]\). Both \(\left[ 0,1 \right]\) and \(\left[ 2,3 \right]\) are open in \(X\), and so they form a “disconnection” of \(X\); visually, we see that these two subsets form the “connected pieces” of \(X\).

More broadly speaking, there is a way to find connected pieces of a metric space:

Relationships to Continuity, Compactness, and Completeness

First, we should point out that connectedness has nothing to do with compactness or completeness whatsoever. Morally speaking, this makes a lot of sense — connectedness deals with the shape of a space while compactness deals with its size; completeness is not a topological property at all. In fact, consider the following table:

	Connected?	Compact?	Complete?
\((0,1)\)	Yes	No	No
\([0, 1]\)	Yes	Yes	Yes
\(\mathbb{R}\)	Yes	No	Yes
\((0, 1)\cup \left\lbrace 2 \right\rbrace\)	No	No	No
\([0, 1]\cup \left\lbrace 2 \right\rbrace\)	No	Yes	Yes
\((-\infty, 0]\cup [1, \infty)\)	No	No	Yes

Each of these examples are viewed as subsets of \(\mathbb{R}\) with the usual metric. I’ll leave it as an exercise to verify this table!

What’s more interesting is that connectedness does interact with continuous functions.

Proof

Suppose such an \(f\) did exist. Let \(U, V\) be disjoint nonempty open sets in \(Y\) such that \(Y=U\cup V\). Then \(f ^{-1}(U)\cup f ^{-1}(V)=X\) — if \(x\in X\), then either \(f(x)\in U\) (i.e. \(x\in f ^{-1}(U)\)) or \(f(x)\in V\) (i.e. \(x\in f ^{-1}(V)\)). But by continuity \(f ^{-1}(U)\) and \(f ^{-1}(V)\) are open; they must be disjoint since \(U\) and \(V\) are disjoint; and they are empty because \(f\) is surjective. This contradicts the connectedness of \(X\). \(\square\)

Here, we need to note two common mistakes.

1. It’s possible for \(f:X\to Y\) to be a surjective continuous function from a disconnected space onto a connected space. You can think of this as “folding” the connected components of \(X\) onto \(Y\).
2. It’s possible for \(f:X\to Y\) to be a continuous function from a connected space into a disconnected space; it just won’t be surjective. In fact, any constant function will be continuous!

This fact is very powerful. If you are tasked with proving that a metric space is connected, it’s not a bad idea to look for a surjection from a connected space. For instance, you can prove the circle \[S=\left\lbrace \left( x,y \right)\in \mathbb{R}^2 : x^2+y^2=1 \right\rbrace\] is connected because there is a surjective continuous map \(f:\mathbb{R}\to S\) via \(t\mapsto \left( \cos t, \sin t \right)\). The connectedness of \(\mathbb{R}\) is left as an exercise!

Our other option to show that metric spaces are connected is path connectedness, which intuitively means that you can draw a continuous line between any two points in your metric space.

You showed on your homework that every path connected metric space is necessarily connected. This makes perfect sense. However, not every connected metric space is path connected!

There are about two standard examples of this: the topologist’s sine curve, and the topologist’s hair brush. (There are probably more if you look hard enough.) In anticipation of the former being covered in lecture, let’s consider the topologist’s hair brush.

Define the following subset of \(\mathbb{R}^2\): \[S=\left\lbrace \left( x,0 \right) : x\in[0, 1] \right\rbrace \cup \left( \bigcup _{n=1} ^{\infty} \left\lbrace \left( \frac{1}{n} , y \right) : y\in[0, 1] \right\rbrace \right)\cup \left\lbrace (0, 1) \right\rbrace .\] In words, this is a horizontal “base” lying on the \(x\)-axis, then a bristle sticking straight up at \(x=\frac{1}{n}\) for each \(n\in \mathbb{N}\). Finally, there’s a single point added at the upper-left corner of the “brush”.

First, we’ll sketch the proof of connectedness. Let \(U\subseteq S\) be a clopen subset containing \(\left( 0,1 \right)\) (the funny point at the upper-left). Since \(U\) is open, there is some \(\epsilon>0\) such that \(B\left( \left( 0,1 \right),\epsilon \right)\subseteq U\); looking at the “tips” of the bristles, we get that \(\left( \frac{1}{n} , 1 \right)\in U\) for some \(n\).

Now consider the bristle containing \(\left( \frac{1}{n} ,1 \right)\); this bristle is connected, and the intersection of \(U\) with this bristle is nonempty and clopen (in the bristle). So, \(U\) must contain the whole bristle, and in particular, contains \(\left( \frac{1}{n} ,0 \right)\). Then by the same argument, \(U\) must contain the entire base of the brush. But the base of the brush intersects all of the bristles, and by repeating this argument yet again, \(U\) must contain everything. This (roughly) shows that the only nonempty clopen subset of \(S\) is \(S\) itself!

One way to think about this is that everything besides \(\left( 0,1 \right)\) is connected in a very intuitive way; thus the way only for \(S\) to be disconnected is for \(\left( 0,1 \right)\) to be its own little connected component. But, as the above argument shows, \(\left( 0,1 \right)\) is way too close to the bristles, so it can’t be disconnected from the rest of the brush!

Now let’s see why \(S\) is not path connected. Let \(p=\left( 0,1 \right)\), and let \(q=\left( 1,0 \right)\). We claim there is no path \(\gamma:[0, 1]\to S\) such that \(\gamma\left( 0 \right)=p\) and \(\gamma(1)=q\).

Intuitively, this should make sense: for a path to connect these two corners, it would have to “jump” over some of the bristles at the very start of the path before traveling down to the base and going to \(q\). But of course such a jump wouldn’t be continuous, so somehow the path needs to slide across the tips of the bristles at the very start of the path to be continuous! Visually, the only way for a path to go from the tip of one bristle to another is to go all the way down to the base and then back up. Together, this should create a contradiction.

Intuition is a double-edged sword, so let’s properly demonstrate why no such \(\gamma\) can exist. We’ll make good use of the intermediate value theorem! Begin by defining \[t_0 = \sup \left\lbrace t\in [0, 1] : \gamma(t) = p \right\rbrace.\] This set is nonempty and bounded above, so this supremum exists. By continuity, we necessarily have \(\gamma\left( t_0 \right)=p\). This is the last moment before \(\gamma\) leaves the upper-left corner! Of course, we must have \(t_0<1\), since \(\gamma\left( 1 \right)=q\).

Pick any \(\epsilon<1\); this makes it so that \(B\left( p, \epsilon \right)\) does not intersect the base of the brush. By continuity at \(t_0\), there exists some \(\delta>0\) such that \(\left\lvert t-t_0 \right\rvert<\delta\) implies \(d\left( p, \gamma(t) \right)<\epsilon\). In particular, taking \(t_1=t_0+\frac{\delta}{2}\) tells us that \(\gamma\left( t_1 \right)\) must land on a bristle — \(t_1>t_0\) means that \(\gamma\left( t_1 \right)\neq p\) by construction, and the continuity condition tells us that \(\gamma\left( t_1 \right)\) is not on the base of the brush. In particular, \(\gamma\left( t_1 \right)\) has an \(x\)-coordinate of \(\frac{1}{n}\) for some \(n\).

Now consider the function \(\pi:S\to [0, 1]\) that sends a point \(\left( x,y \right)\in S\) to its \(x\)-coordinate. That is, \(\pi\left( x,y \right)=x\). This \(\pi\) is continuous, and if you believe that the composition of two continuous functions is continuous, we get that \(\pi\circ \gamma:[0, 1]\to [0, 1]\) is continuous. This composite is the \(x\)-coordinate of our path.

Well, \(\pi\circ \gamma\left( t_0 \right)=0\) and \(\pi\circ \gamma\left( t_1 \right)=\frac{1}{n} \); by the intermediate value theorem, there exists some \(t’\) between \(t_0\) and \(t_1\) such that \(\pi\circ \gamma\left( t’ \right)=\frac{1}{n\sqrt 2}\). This means that \(\gamma\left( t’ \right)\) has an \(x\)-coordinate of \(\frac{1}{n\sqrt 2}\), which is an irrational number. Thus \(\gamma\left( t’ \right)\) cannot land on a bristle, since the bristles all have rational \(x\)-coordinates, and in fact \(\gamma\left( t’ \right)\) is on the base of the brush.

But this is a contradiction: since \(t_0< t’< t_1=t_0+\frac{\delta}{2}\), we have \(\left\lvert t_0 -t’\right\rvert< \delta\)! So \(\gamma\left( t’ \right)\) should not land on the base by continuity, and we conclude that no such \(\gamma\) can exist.