# Abstract Nonsense

## Crushing one theorem at a time

In this post we begin a series of discussions about one of the easiest to define concepts in topology. That said, it is one of the most important. We begin with some motivation.

If you were a calc II student and asked to visualize these two sets $(0,1)$ and $(-\pi,-e)\cup (\pi,e)$ and say what is the most obvious difference between them you most likely respond  “The first one is in one piece and the second one isn’t”.  This idea of something being in one piece is a fundamental concept. It intuitively fits with our idea of homeomorphisms that two spaces cannot be homeomorphic of one space is in two pieces and the other one.

There is also the vague notion of “reachability”. Imagine you took two people and placed one on a random spot in $(0,1)$ and the other in a random spot in $(-\pi,-e)\cup (e,\pi)$. Then, the first person could walk to any other point of the set he’s on, while the other cannot. This idea of when two things can be traveled between without leaps or jumps is a huge concept in more advanced topology, that we will eventually hit on.

So we want to formalize this idea of being in one piece. Using our original example again we may be able to model this idea, which is commonly called connectedness.

Connected: Let $X$ be a topological space. Then, $X$ is called connected if it cannot be expressed as $X=E\cup G$ where $E,G$ are open, non-empty and disjoint.

We begin by proving a couple of equivalent definitions.

Theorem: Let $X$ be a topological space. Then, $X$ is connected if and only if the only sets which are simultaneously open and closed are $X$ and $\varnothing$.

Proof:

Suppose that $X$ is connected, but there existed some $\varnothing\subset E\subset X$ which is both open and closed. Clearly then $E'$ is both open and closed. Thus, $E,E'$ are disjoint non-empty open sets such that $E\cup E'=X$. This is a contradiction.

Conversely, suppose that the only sets which are both open and closed are $\varnothing$ and $X$. Next suppose that $A,B$ were disjoint non-empty open sets such that $A\cup B=X$. Then since $A=B'$ (this is easy to check) and $B$ is open it follows that $A$ is closed. But, since $A\cup B=X$ and $A,B\ne\varnothing$ it follows that $\varnothing\subset A\subset X$ and in particular $A\ne \varnothing, X$. Thus, $A$ is a set that is both open and closed which is neither $X$ or $\varnothing$. Contradiction. $\blacksquare$

Theorem: Let $X$ be a topological space. Then $E\subseteq X$ is both open and closed if and only if $\partial E=\varnothing$.

Proof:

Suppose that $E$ is both open and closed but $\partial E\ne\varnothing$. Then, let $x\in \partial E$. We have two choices, if $x\in E$ then every neighborhood of $x$ intersects $E'$ and so $x\notin E^{\circ}$ but since $E=E^{\circ}$ it follows that $x\notin E$, contradiction. If $x\notin E$ then every neighborhood of $x$ must contain a point of $E$ different from $x$ and so $x\in D(E)$, but since $E$ is closed we know that $D(E)\subseteq E$ and so $x\in E$. Contradiction. It follows that $\partial E=\varnothing$

Conversely, suppose that $\partial E=\varnothing$. Then, clearly $\partial E\subseteq E$ and so from basic topology $E$ is closed.  But, let $x\in E$ then since $x\notin \partial E$ there exists some neighborhood $N$ of $x$ such that $N$ does not intersect both $E,E'$. But since every neighborhood contains $x$ itself it follows that there exists some neighborhood $U$ of $x$ such that $U\cap E'=\varnothing\implies U\subseteq E$. It follows that $x\in E^{\circ}$ and appealing once again to basic topology we see that $E$ is open. $\blacksquare$.

Corollary: A topological space $X$ is connected if and only if the only sets with empty interior are $X$ and $\varnothing$.

Before we continue with need a definition.

Separated sets: If $X$ is a topological space and $A,B\subseteq X$ are such that $A\cap\overline{B}=\overline{A}\cap B=\varnothing$ we call $A,B$ separated set.

Theorem: A topological space $X$ is connected if and only if it cannot be written as the union of none-empty separated sets.

Proof:

Suppose that $X$ is connected but $A\cup B=X$ where $A,B$ are separated and non-empty. Clearly, $A\cup B\subseteq A\cup\overline{B}$ and since $A\cup B=X$ it follows that $\overline{A}\cup B=X$. But, $\overline{A}\cap B=\varnothing$ and so $B=X-\overline{A}$ and so $B$ is open. But, it is also true that $A\cup\overline{B}=X$ and that $A\cap\overline{B}=\varnothing$ and so $A=X-\overline{B}$ and so $A$ is open. Thus, $A\cup B=X$, and $A,B$ are both non-empty and open. Lastly noting that $A\cap B\subseteq A\cap\overline{B}=\varnothing$ we arrive at our contradiction.

Conversely, notice that if $A,B$ are open and disjoint then $A\cap\overline{B}=B\cap\overline{A}=\varnothing$ (this is easy to prove). Thus, if $A,B$ are disjoint non-empty open sets, then they are disjoint separated sets and thus $A\cup B\ne X$. The conclusion follows. $\blacksquare$

This last one gives, in my opinion, the most elegant way of thinking about connectedness.

Theorem: If $X$ is a topological space, then it is connected if and only if the only continuous $\varphi:X\mapsto D$ where $D=\{0,1\}$ with the discrete topology is a constant map.

Proof:

Suppose that $X$ is connected but $\varphi:X\mapsto D$ is non-constant. Clearly then $\varphi^{-1}(\{0\}),\varphi^{-1}(\{1\})$ are open and non-empty. They are also clearly disjoint, for if  $x\in \varphi^{-1}(\{0\})\cap\varphi^{-1}(\{1\})$ then

$\varphi(x)\in\varphi\left(\varphi^{-1}(\{0\})\cap\varphi^{-1}(\{1\})\right)\subseteq \varphi\varphi^{-1}(\{0\})\cap\varphi\varphi^{-1}(\{1\})\subseteq \{0\}\cap\{1\}=\varnothing$

Thus, $\varphi^{-1}(\{0\}),\varphi^{-1}(\{1\})$ are disjoint non-empty open sets and

$\varphi^{-1}(\{0\})\cup\varphi^{-1}(\{1\})=\varphi^{-1}(\{0\}\cup\{1\})=\varphi^{-1}(\{0,1\})=X$

This of course contradicts that $X$ is connected.

Conversely, suppose that the only $\varphi:X\mapsto D$ which is continuous is a constant map, but $A,B$ are non-empty disjoint open sets such that  $A\cup B=X$. Define $\varphi:X\mapsto D$ by

$\varphi(x)=\begin{cases} 0 & \mbox{if} \quad x\in A \\ 1 & \mbox{if} \quad x\in B\end{cases}$.

It is clear by the assumption that these two sets are non-empty, disjoint, and their union equals $X$ that this mapping is well define. Also,

$\varphi^{-1}(\{0\})=A,\varphi^{-1}(\{1\})=B,\varphi^{-1}(\varnothing)=\varnothing,\varphi^{-1}(\{0,1\})=X$.

Thus, $\varphi$ is continuous. It follows that $\varphi$ is a non-constant continuous map from $X$ to $D$. This contradicts our assumption. $\blacksquare$

Remark: Clearly every two point discrete space is homeomorphic, so the usage of $0,1$ opposed to $a,b$ or $\Delta,\Sigma$ is merely notational. Also, notice, by virtue that $\text{card }D=2$, that we could have rephrased the above as saying “A space is connected if and only if every continuous map from $X$ to $D$ is not surjective”