# Abstract Nonsense

## Thoughts about connectedness (Totally disconnected spaces)

We now discuss the complete opposite of connected.

Totally Disconnected: Let $X$ be a topological space. We call $X$ totally disconnected if given any $x,y\in X$ there is a separation $A\cup B$ of $X$ such that $x\in A,y\in B$.

Theorem: If $X$ is a totally disconnected space then $X$ is Hausdorff.

Proof: Let $x,y\in X$ be arbitrary. Then if $A\cup B$ is the separation of $X$ such that $x\in A,y\in B$ then clearly $A,B$ are the necessary disjoint open sets. $\blacksquare$.

Remark: A space is totally disconnected if and only if it’s components are its points.

Some examples of totally disconnected spaces are: any discrete space, the rationals, the cantor set, the irrationals, etc. In fact, although I will not introduce the concept, if $X$ is a topological space with the order topology and $D$ is a subspace of $X$ such that $X-D$ is dense, then $D$ is totally disconnected.

Notice that we never mentioned whether or not connectedness was a hereditary property, we didn’t because it was obvious that is not. For example, think about $\mathbb{R}$ and the subspace $(-\pi,-e)\cup (e,\pi)$. It turns out that total disconnectedness is a hereditary property.

Theorem: Let $X$ be a totally disconnected space $E$ a subspace of $X$, then $E$ is totally disconnected.

Proof: Let $x,y\in E$ be distinct. Then, since $X$ is totally disconnected there exists some disconnection $A\cup B$ such that $x\in A, y\in B$. Clearly $x\in A\cap E,\text{ }y\in B\cap E$ and s$\left(A\cap E\right)\cup\left(B\cap E\right)$ is a disconnection. $\blacksquare$

It is also not surprising that the product of totally disconnected spaces is disconnected.

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of totally disconnected space, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ is totally disconnected under the product topology.

Proof: Let $\bold{x},\bold{y}\in X$ be distinct. It follows that $\pi_\ell(\bold{x})\ne\pi_\ell(\bold{y})$ for some $\ell\in\mathcal{J}$. Now, since $X_\ell$ is totally disconnected there exists a disconnection $A\cup B$ of it such that $\pi_\ell(\bold{x})\in A,\pi_\ell(\bold{y})\in B$. Clearly then if we let $E,G\subseteq X$ be such that $\pi_\ell(E)=\begin{cases} A & \mbox{if }\quad j=\ell \\ X_j & \mbox{if} \quad j\ne \ell\end{cases}$ and $\pi_\ell(G)=\begin{cases} B & \mbox{if} \quad j=\ell \\ X_j & \mbox{if} j\ne \ell\end{cases}$ we have that $E,G$ are non-empty disjoint open sets which contain $\bold{x},\bold{y}$ respectively. It remains to show that $X=E\cup G$, but this is obvious.

It turns out there is a nice instance when Hausdorff spaces are disconnected.

Theorem: Let $X$ be a Hausdorff space with an open base whose elements are also closed, then $X$ is totally disconnected.

Proof: Let $x,y\in X$ be distinct. Since $X$ is Hausdorff there exists disjoint open sets $U,V$ such that $x\in U$ and $y\in V$. Clearly then there exists some open/closed $B$ such that $x\in B\subseteq U$. Clearly $B\cup B'$ is the desired disconnection.

It turns out that the converse is true if $X$ is compact. Namely:

Theorem: Let $X$ be a totally disconnected compact Hausdorff space, then $X$ has an open base which is both open and closed.

Proof: Let $x\in X$, and let $N$ be any neighborhood of $x$. Clearly then $N'$ is compact. Now, since $X$ is totally disconnected we know that for each $y\in N$ there exists a separation $A_y\cup B_y$ of $X$ such that $x\in A_y,y\in B_y$. Clearly $\left\{B_y\right\}_{y\in N'}$ is an open cover for $N'$ and so by it’s compactness it has a finite subcover $B_{y_1},\cdots,B_{y_n}$. Clearly then we have that $x\in A_{y_1}\cap\cdots\cap A_{y_n}\subseteq N$ and since $A_{y_j}$ is both open and closed so is their finite intersection. The conclusion follows. $\blacksquare$

We may now prove a neat little theorem:

Theorem: Let $X$ be a compact totally disconnected Hausdorff space, then $X$ is homeomorphic to a closed subspace of $\{0,1\}^\alpha$ for some $\alpha$.

Proof: By the previous theorem we are able to state that there exists some open base $\mathfrak{B}=\left\{B_a\right\}_{a\in\mathfrak{A}}$ whose elements are both open and closed. Note that given some fixed $B\in\mathfrak{B}$ we have that the indicator function $\chi_B:X\to\{0,1\}$ given by

$\chi_B(x)=\begin{cases} 1 & \mbox{if}\quad x\in B \\ 0 & \mbox{if} \quad x\notin B\end{cases}$

is continuous. We claim that $\varphi:X\to\{0,1\}^{\mathfrak{A}}$ given by

$\pi_a\varphi(x)=\chi_{B_a}(x)$

is an embedding. Since $X$ is compact and $\{0,1\}^{\mathfrak{A}}$ Hausdorff we must merely show that $\varphi$ is injective and continuous.

To see injectivity suppose that $\varphi(x)=\varphi(y)$. Then, $x\in B_a$ for every basic open neighborhood $B_a$ of $y$. In particular, $\displaystyle x\in\bigcap_{B_a\in\mathcal{B}}B_a$ where $\mathcal{B}$ is the set of all basic neighborhoods of $y$. But, clearly since $X$ is $T_1$ we have that $\displaystyle \bigcap_{B_a\in\mathcal{B}}B_a=\{y\}$ and so $x\in\{y\}\implies x=y$.

To see that it’s continuous let $O\subseteq \{0,1\}^{\mathfrak{A}}$ be subbasic open and let $\alpha$ be the single subscript such that $\pi_\alpha O\ne X_\alpha$. Clearly then $\varphi^{-1}(O)=\chi_{\alpha}^{-1}(O)$ which by previous comment is open.

It follows that $\varphi$ is an embedding. Noting that $\varphi(X)$ is a compact, and thus closed, subspace of $\{0,1\}^{\mathfrak{A}}$ completes the argument. $\blacksquare$