Connectedness
Contents
Connected Spaces
I decided to postpone the second half of my construction of the rational numbers because I needed to break up the monotony a bit.
I'm finally about to show you some really cool stuff. In this post, I'm going to prove the Intermediate Value Theorem and the One-Dimensional Brouwer Fixed Point Theorem, which are two results that are undeniably and unreasonably useful. In order to prove them, however, we will need to study the notion of connectedness. I'll probably save path-connectedness for a later time, since this post will be long enough as is.
Connectedness is pretty much exactly what you'd expect it to be - if something is connected then that means it is all one coherent piece. This is a very hand-wavy statement though, so let's try to come up with a real definition that makes sense in the context of topological spaces. It might be better to start by considering what it is that makes a space disconnected. Let's look at this picture of a space $X$ which we should certainly call disconnected (as a subspace of the plane $\R^2$ with the standard topology):
The space $X$ is very clearly comprised of two distinct components that are separated from each other. Now here is the key insight: each of these components is both open and closed in $X$. For this particular example, we can see that each component is open because it is the intersection of $X$ with an open set in $\R^2$ (remember that this is the definition of an open set in the subspace topology).
Similarly, each component is closed because it is the intersection of $X$ with a closed set in $\R^2$.
So in disconnected spaces, there can be numerous sets that are both open and closed. This is something that is not true of connected spaces!
Definition. A space $X$ is connected if its only subsets which are both open and closed are $\varnothing$ and $X$ itself.
Definition. A space $X$ is disconnected if it is not connected, i.e., there exists a nonempty proper subset $A\subset X$ which is both open and closed in $X$.
We may occasionally also wish to ask whether subsets of a topological space are connected, and there is an easy way to look at this by using definitions we already have:
Definition. A subset $A$ of a topological space $X$ is connected if $A$ is connected when viewed as a subspace of $X$.
This is just one of several equivalent and common ways that we may define connectedness. Let's take a look at two more, since they will be convenient for proofs later in this post. For the first equivalent definition, it is somewhat easier to talk about disconnectedness.
Theorem. A topological space $X$ is disconnected if and only if there exist nonempty open sets $U,V\subset X$ for which $U\cap V=\varnothing$ and $U\cup V=X$.
Proof. Suppose first that $X$ is disconnected. That is, there exists a nonempty proper subset $A\subset X$ which is both open and closed in $X$. Then by definition, $X-A$ is also both open and closed, and $X-A$ is nonempty because $A\ne X$. We observe that $A\cap(X-A)=\varnothing$ and $A\cup(X-A)=X$. Thus, we have demonstrated nonempty open sets $U=A$ and $V=X-A$ for which $U\cap V=\varnothing$ and $U\cup V=X$.
Suppose next that there exist nonempty open sets $U,V\subset X$ for which $U\cap V=\varnothing$ and $U\cup V=X$. Then $V=X-U$ is also closed since it is the complement of the open set $U$. Note also that $V$ is nonempty by hypothesis, and $V\ne X$ because $U$ is nonempty. It follows that $X$ is disconnected.
Just looking at this new phrasing of disconnectedness, we see that it is immediately applicable to my above example. Clearly that space $X$ is the union of two disjoint, nonempty open sets. Before I move on, let me formally state this new definition in the context of connectedness.
Definition. If a space $X$ is disconnected, then a separation of $X$ is a pair of nonempty open sets $U,V\subset X$ for which $U\cap V=\varnothing$ and $U\cup V=X$.
Corollary. A topological space $X$ is connected if it does not have a separation.
The next equivalent definition of connectedness may look a little bit strange to you right now, but it will make our lives a lot easier in two of the proofs that lie ahead. Recall that a constant function $f$ has $f(x_1)=f(x_2)$ for all $x_1,x_2$ in its domain.
Theorem. A topological space $X$ is connected if and only if every continuous function $f:X\to\{0,1\}$ is constant, where $\{0,1\}$ is equipped with the discrete topology.
Proof. We prove each direction by contraposition. We argue first that if a continuous function $f:X\to\{0,1\}$ is not constant, then $X$ is not connected.
Suppose there exists a nonconstant continuous function $f:X\to\{0,1\}$. Since $f$ is not constant, it takes on at least two values in its codomain and thus it must be the case that $f[X]=\{0,1\}$, so clearly $f$ is surjective. Define $U=f^{-1}\big[{0}\big]$ and $V=f^{-1}\big[{1}\big]$. Certainly $U\cap V=\varnothing$ and $U\cup V=X$. Also, $U$ and $V$ are nonempty because $f$ is surjective. It follows that $U$ and $V$ form a separation of $X$, and so $X$ is not connected.
We argue next that if $X$ is not connected, then there exists a continuous function $f:X\to\{0,1\}$ which is not constant.
Suppose $X$ is not connected. Then there exists a separation $U,V$ of $X$. Define $f:X\to\{0,1\}$ by
$$f(n) =
\begin{cases}
0 & \text{if } x\in U, \\
1 & \text{if } x\in V.
\end{cases}$$This function is nonconstant because both $U$ and $V$ are nonempty. It is obvious that $f^{-1}\big[\{1\}\big]=U$ and $f^{-1}\big[\{0\}\big]=V$, which are both open in $X$. Furthermore, $f^{-1}\big[\{0,1\}\big]=X$ and $f^{-1}[\varnothing]=\varnothing$ are both open in $X$. We have demonstrated that the preimage of every open set in $\{0,1\}$ is open in $X$, and so $f$ is continuous by definition.
Let's rewrite this as yet another definition of connectedness.
Definition. A topological space $X$ is connected if every continuous function $f:X\to\{0,1\}$ is constant.
This definition may seem a little bit stranger than the other two, but it is perfectly natural if we recall that continuous functions map points that are close together to points that are close together. In a connected space, all points are close together and thus a continuous function cannot bridge the gap between $0$ and $1$ - it must stay constant for all values of the domain. Of course, there is nothing special about $\{0,1\}$. We could just has easily used any other discrete two-point space. I'm about to prove a useful theorem whose proof is made much easier by this new definition of connectedness, but first we will need the following lemma.
Lemma. Let $X,Y$ denote topological spaces, let $A$ denote a subspace of $X$ and let $f:X\to Y$ be a continuous map. The restriction $f\mid_A:A\to Y$, defined by $f\mid_A(a)=f(a)$ for each $a\in A$, is continuous.
Proof. Choose any open set $U\subseteq Y$. Since $f$ is continuous, $f^{-1}[U]$ is open in $X$. Thus, $f\mid_A^{-1}[U]=A\cap f^{-1}[U]$ is open in $A$ by the definition of the subspace topology, and thus $f\mid_A$ is continuous.
Now we have the machinery to prove the following intuitive theorem:
Theorem. Let $X$ denote a topological space and suppose $\{A_i\}_{i=1}^n$ is a collection of $n$ connected subsets of $X$. If $A_i\cap A_{i+1}$ is nonempty for each $1\leq i< n$, then the set $\bigcup\limits_{i=1}^n A_i$ is connected.
Proof. We will proceed by induction on $n$, with our base case $n=1$ being too obvious to require proof.
Suppose $\{A_i\}_{i=1}^{n+1}$ are connected with $A_i\cap A_{i+1}\ne\varnothing$ for every $1\le i< n+1$, and that $\bigcup\limits_{i=1}^n A_i$ is connected. Suppose also that $f:\bigcup\limits_{i=1}^{n+1}A_i\to\{0,1\}$ is continuous. Then $f\mid_{\bigcup\limits_{i=1}^n A_i}:\bigcup\limits_{i=1}^n A_i\to\{0,1\}$ and $f\mid_{A_{n+1}}:A_{n+1}\to\{0,1\}$ are also continuous, and they are thus constant because their domains are connected. Since $A_n\cap A_{n+1}$ is nonempty, certainly $\left(\bigcup\limits_{i=1}^n A_i\right)\cap A_{n+1}$ is nonempty, and so there exists a point $x\in\left(\bigcup\limits_{i=1}^n A_i\right)\cap A_{n+1}$. Thus, because $f$ must be well defined, $f\mid_{\bigcup\limits_{i=1}^n A_i}(x)=f\mid_{A_{n+1}}(x)$. Since $f\mid_{\bigcup\limits_{i=1}^n A_i}$ and $f\mid_{A_{n+1}}$ are both constant and they agree at one point, they must agree at every point. Thus, $f$ is constant and so $\bigcup\limits_{i=1}^{n+1} A_i$ is connected.
The proof is simple but looks a bit ugly rendered this way, for which I apologize. The theorem itself should be fairly obvious, and is in fact quite easy to think of visually. Take the following diagram, for instance:
This is a collection of six connected subsets of $\R^2$, each of which intersects the next in precisely one point. (They may or may not correspond to the six Bagel Bites I am about to enjoy.) From our theorem, it follows that their union is connected. Of course, we could have any number of connected subsets that intersect each other in different ways, as long as the successive pairwise intersections are nonempty. I will not prove it here, but this result also applies for any collection of connected subsets (even uncountably infinite).
I mentioned a few posts back that much of the study of topology is focused on distinguishing between non-homeomorphic spaces. Connectedness offers a way of doing exactly that, because it is a topological property. What I mean by that is this: if two spaces are homeomorphic and one is connected, then so is the other.
Theorem. Let $X,Y$ denote topological spaces, let $f:X\to Y$ be a homeomorphism and suppose $X$ is connected. Then $Y$ is connected.
Proof. We will prove the contrapositive. Suppose $Y$ is not connected, so that there exist open sets $U,V$ that separate $Y$. We argue that $f^{-1}[U]$ and $f^{-1}[V]$ form a separation of $X$.
Clearly $f^{-1}[U]$ and $f^{-1}[V]$ are open because $f$ is continuous, and they are each nonempty because $f$ is surjective. In addition, because $U\cap V=\varnothing$, we have that
$$\begin{align}
f^{-1}[U]\cap f^{-1}[V]&=f^{-1}[U\cap V] \\
&=f^{-1}[\varnothing] \\
&=\varnothing.
\end{align}$$Similarly, because $U\cup V=Y$, we have that
$$\begin{align}
f^{-1}[U]\cup f^{-1}[V]&=f^{-1}[U\cup V] \\
&=f^{-1}[Y] \\
&=X.
\end{align}$$It follows that $f^{-1}[U]$ and $f^{-1}[V]$ form a separation of $X$, so $X$ is not connected.
Really all we used was continuity and surjectivity of $f$, so saying that $f$ was a homeomorphism actually weakened the result a little bit. It's actually true that all continuous functions preserve connectedness! I won't prove it though, because it's really just a teensy modification of the above proof.
Clearly we can distinguish between topological spaces as follows: if one space is connected and another isn't, then the spaces are not homeomorphic. However, we can also use connectedness in a somewhat trickier way to distinguish between certain connected spaces!
Definition. A cutset $A$ of a connected topological space $X$ is a subset of $X$ for which $X-A$ is disconnected.
Definition. A cutpoint $x$ of a connected topological space $X$ is a point of $X$ for which $\{x\}$ is a cutset, i.e., $X-\{x\}$ is disconnected.
Example. Let's look again the connected set $A=\bigcup\limits_{i=1}^6 A_i$ depicted below, and consider the points $x$ and $y$.
Is $x$ a cutpoint of $A$? No, it is not because $$A-\{x\}=(A_1-\{x\})\cup\bigcup\limits_{i=2}^5 A_i\cup(A_6-\{x\}).$$ Each of the sets being unioned are connected and the intersection of each set with the next is nonempty, so $A-\{x\}$ is connected. If we permute the indices of the $A_i$, the same logic tells us that $y$ is not a cutpoint of $A$.
However, $\{x,y\}$ is a cutset of $A$ (or, equivalently, $y$ is a cutpoint of $A-\{x\}$). That's because
$$\begin{align}
U &= (A_1-\{x\})\cup(A_2-\{y\}) \\
V &= (A_3-\{y\})\cup\bigcup\limits_{i=4}^5 A_i\cup (A_6-\{x\})
\end{align}$$is a separation of $A-\{x,y\}$.
How can we use cutsets to distinguish between connected spaces? I'm glad you ask.
Theorem. Let $f:X\to Y$ be a homeomorphism between connected spaces $X$ and $Y$. If $A$ is a cutset of $X$, then $f[A]$ is a cutset of $Y$.
Proof. Since $X-A$ is disconnected, there exists a separation $U,V$ of $X-A$. We argue that $f[U]$ and $f[V]$ form a separation of $Y-f[A]$. Because $f$ is a homeomorphism, $f^{-1}$ is continuous and thus $f[U]=(f^{-1})^{-1}[U]$ and $f[V]=(f^{-1})^{-1}[V]$ are open in $Y$. Since $U$ and $V$ are nonempty, certainly $f[U]$ and $f[V]$ are nonempty. Observe that, because $U\cap V=\varnothing$,
$$\begin{align}
f[U]\cap f[V]&=(f^{-1})^{-1}[U]\cap(f^{-1})^{-1}[V]\\
&=(f^{-1})^{-1}[U\cap V]\\
&=f[U\cap V]\\
&=f[\varnothing]\\
&=\varnothing.
\end{align}$$Furthermore, because $U\cup V=X-A$ and $f$ is bijective,
$$\begin{align}f[U]\cup f[V]&=f[U\cup V]\\
&=f[X-A]\\
&=f[X]-f[A]\\
&=Y-f[A].
\end{align}$$Thus, $f[U]$ and $f[V]$ separate $Y-f[A]$. It follows that $f[A]$ is a cutset of $Y$.
To see how this can be applied, let's take a look at another example.
Example. The subspace $[0,2\pi)$ of $\R$ and the unit circle $S^1=\{(x,y)\in\R^2\mid x^2+y^2=1\}$ as a subspace of $\R^2$ are not homeomorphic. I have not yet shown that either of these sets is connected, but for now we will take it for granted that they both are. Any point $x\in [0,2\pi)$ with $x\ne 0$ is a cutpoint of $[0,2\pi)$ because $$[0,2\pi)-\{x\}=[0,x)\cup (x,2\pi),$$ which is already expressed as the union of its separation. On the other hand, $S^1$ has no cutpoints. For any $x\in S^1$, the space $S^1-\{x\}$ is homeomorphic to an open interval (which is connected). Any cutset of $S^1$ would therefore consist of at least two points, and so cannot be the homeomorphic image of a cutpoint for $[0,2\pi)$.
I was just very hand-wavy by claiming that certain sets were connected without proving it. Let me at least prove that $\R$ is connected, from which it will immediately follow that open intervals are connected (because they are homeomorphic to $\R$). The proof will require the following fact about the real numbers, which itself needs to be prefaced with the following definitions.
Definition. Given a subset $A$ of $\R$, a number $x\in\R$ is an upper bound for $A$ if $x\ge a$ for every $a\in A$.
Definition. A subset $A$ of $\R$ is bounded above if an upper bound for $A$ exists.
Definition. Given a subset $A$ of $\R$, a number $x\in\R$ is the supremum (or least upper bound) of $A$ if $x$ is an upper bound for $A$ and $x\le y$ for every upper bound $y$ of $A$.
Least Upper Bound Property. Any nonempty subset of $\R$ which is bounded above has a supremum.
This is one of the defining properties of the real numbers, and it is in fact equivalent to the fact that every Cauchy sequence in $\R$ converges (if you've taken an analysis class, this should seem familiar). We will take the least upper bound property for granted, although it can be proven from the construction of the real numbers as either Dedekind cuts or equivalence classes of rational Cauchy sequences. You don't need to understand any of this to move forward. Just smile and nod while you wait for me to shut up.
Here's the proof that $\R$ is connected. It's pretty gross.
Theorem. The set $\R$ of real numbers, equipped with the standard topology, is connected.
Proof. We proceed by contradiction, supposing that $\R$ is not connected. This means that there exist disjoint open sets $U$ and $V$ which separate $\R$. Choose $u\in U$ and $v\in V$, and assume without loss of generality that $u<v$. Define $U'=[u,v]-V$ and $V'=[u,v]-U$. Using De Morgan's law and the fact that $V\cap U=\varnothing$, it follows that
$$\begin{align}
U'\cup V'&=\big([u,v]-V\big)\cup\big([u,v]-U\big)\\
&=[u,v]-(V\cap U)\\
&=[u,v]-\varnothing\\
&=[u,v].
\end{align}$$The set $U'$ is nonempty because $u\in U'$ and it is bounded above by $v$, and so $U'$ has a supremum, $s$. Clearly $s\in [u,v]$, so either $s\in U'$ or $s\in V'$.
Suppose first that $s\in U'$. Then because $U'$ is open in $[u,v]$ and $v\notin U'$, there exists $d\in[u,v]$ for which $[s,d)\subseteq U'$. Then for every $x\in (s,d)$, we have that $x\in U'$ and $x>s$, which is a contradiction because $s$ is the supremum of $U'$.
Suppose then that $s\in V'$. Then because $V'$ is open in $[u,v]$ and $u\notin V'$, there exists $d\in (u,s)$ for which $(d,s]\subseteq V'$. Then for every $x\in (d,s)$, we have that $x<s$ and $x>y$ for every $y\in U'$, which is again a contradiction because $s$ is the supremum of $U'$.
Since $s\notin U'$ and $s\notin V'$, it must be true that $s\notin [u,v]$, which contradicts its definition.
Some Cool Results
It's time for some fun stuff now. Let's start with the intermediate value theorem, which you may have experienced and ignored in a calculus class.
Intermediate Value Theorem. Let $X$ denote a connected space and let $f:X\to\R$ be continuous. If $x,y\in f[X]$ and $c\in [x,y]$, then $c\in f[X]$.
Proof. It suffices to consider $c\in (x,y)$, because if $c\in\{x,y\}$ there is nothing to prove. Since $X$ is connected and $f$ is continuous, $f[X]$ is connected in $\R$. We proceed by contradiction.
Suppose $c\notin f[X]$ and define $U'=(-\infty,c)$ and $V'=(c,\infty)$. Clearly $U'\cap V'=\varnothing$ and $f[X]\subseteq U'\cup V'$. We argue that $U=U'\cap f[x]$ and $V=V'\cap f[X]$ form a separation of $f[X]$. Since $x\in U'$ and $y\in V'$, we have that $U$ and $V$ are nonempty. Furthermore,
$$\begin{align}
U\cap V &= (U'\cap f[X])\cap(V'\cap f[X])\\
&= (U'\cap V')\cap f[X]\\
&= \varnothing\cap f[X]\\
&= \varnothing.
\end{align}$$In addition, because $f[X]\subseteq U'\cup V'$, we have that
$$\begin{align}
U\cup V &= (U'\cap f[X])\cup(V'\cap f[X])\\
&= (U'\cup V')\cap f[X]\\
&= f[X].
\end{align}$$Thus, $U$ and $V$ separate $f[X]$, which is a contradiction.
This theorem has many important consequences, but I won't spend too long talking about them right now because I want to get to the Brouwer fixed point theorem. We'll see the intermediate value theorem again though, I promise.
Of course, before I can talk about a fixed point theorem, it would help if I told you what fixed points were.
Definition. Given a set $X$, a function $f:X\to X$ has a fixed point $x\in X$ if $f(x)=x$.
So fixed points are points which stay fixed after a function acts on them.
One-Dimensional Brouwer Fixed Point Theorem. Every continuous map $f:[-1,1]\to[-1,1]$ has a fixed point.
Proof. We proceed by contradiction. Suppose $f:[-1,1]\to[-1,1]$ is continuous but has no fixed point. That is, $f(x)\ne x$ for every $x\in[-1,1]$. Define $g:[-1,1]\to\{-1,1\}$ by
$$g(x) =
\begin{cases}
-1 & \text{if } f(x) > x, \\
\phantom{-}1 & \text{if } f(x) < x.
\end{cases}$$There is an alternative way of writing this: $$g(x)=\frac{f(x)-x}{\abs{f(x)-x}}.$$ Because $f(x)\ne x$ for every $x\in[-1,1]$, clearly $f(x)-x\ne 0$ and $g$ is thus a continuous function. Because $g$ is continuous and its codomain is a discrete two-point set, it must be that $g$ is constant. However, we also have that $g(-1)=1$ and $g(1)=-1$, and so $g$ is not constant - a contradiction.
This might all seem a bit abstract right now, so maybe a diagram will shed some light on why this theorem makes sense.
Here's a graph of some continuous function $f:[-1,1]\to[-1,1]$. Clearly any such function must intersect the diagonal of the square in at least one point. Any such intersection is a fixed point of $f$.
There is a more general Brouwer fixed point theorem which works for any dimension, although we cannot prove it yet. The beginning of the proof is very similar to the above proof, though, so let's get as far as we can go. (Recall the $\partial A$ denotes the boundary of a set $A$.)
Brouwer Fixed Point Theorem. Let $B^n$ denote the closed unit ball in $\R^n$. Then every continuous function $f:B^n\to B^n$ has a fixed point.
Beginning of Proof. Again, we proceed by contradiction and suppose that $f$ has no fixed points, i.e., $f(x)\ne x$ for every $x\in B^n$. We define a new function $g:B^n\to\partial B^n$ by $$g(x)=\frac{f(x)-x}{d\big(f(x),x\big)},$$ where $d$ is the standard metric.
We can visualize this function as follows: since every point $x$ is distinct from its image $f(x)$, there is a unique line segment in $B^n$ starting at $f(x)$ and passing through $x$ which intersects the boundary $\partial B^n$ in exactly one point, $g(x)$.
Since $f(x)\ne x$ for every $x\in B^n$, $d\big(f(x),x)>0$ for every $x\in B^n$ and thus $g$ is continuous. Furthermore, $g(x)=x$ for every $x\in\partial B^n$ This should lead to a contradiction, but we can't show this right now.
The contradiction is that there is no continuous function which maps the ball to its boundary while keeping the boundary fixed. You can probably visualize why this is in your head - any such "retraction" would require tearing a hole in the ball. Unfortunately, we need the idea of homology in order to establish that there is no retraction to the ball's boundary, and we are a long way from that.
I think there's time for one last proof. Basically, we will prove that fixed point properties are preserved by homeomorphism.
Theorem. Suppose $X$ is a topological space for which every continuous function $f:X\to X$ has a fixed point, and that $Y$ is a space homeomorphic to $X$. Then every continuous function $g:Y\to Y$ has a fixed point.
Proof. Let $h:X\to Y$ be a homeomorphism, and let $g:Y\to Y$ be continuous. By hypothesis, $h$ and $h^{-1}$ are continuous, so $h^{-1}\circ g\circ h:X\to X$ is continuous since it is the composition of continuous functions. Thus, $h^{-1}\circ g\circ h$ has a fixed point. That is, there exists some $x\in X$ for which $\big(h^{-1}\circ g\circ h\big)(x)=x$. But then $\big(g\circ h\big)(x)=h(x)$, and thus $h(x)$ is a fixed point of $g$.
I'll leave you with the following interesting fact that might occasionally occur to you as you stir hot beverages: the Brouwer fixed point theorem, together with the theorem above, tells us that some particle of your drink will end up in the same place it began. This is because the space occupied by the liquid (presumably a cylinder) is homeomorphic to the three-dimensional closed unit ball.