Abstract Nonsense

Crushing one theorem at a time

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


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


March 28, 2010 - Posted by | General Topology, Topology, Uncategorized | , , , ,

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