## Thoughts about connectedness (Totally disconnected spaces)

We now discuss the complete opposite of connected.

**Totally Disconnected:** Let be a topological space. We call *totally disconnected *if given any there is a separation of such that .

**Theorem:** If is a totally disconnected space then is Hausdorff.

**Proof:** Let be arbitrary. Then if is the separation of such that then clearly are the necessary disjoint open sets. .

*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 is a topological space with the order topology and is a subspace of such that is dense, then 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 and the subspace . It turns out that total disconnectedness is a hereditary property.

**Theorem:** Let be a totally disconnected space a subspace of , then is totally disconnected.

**Proof:** Let be distinct. Then, since is totally disconnected there exists some disconnection such that . Clearly and s is a disconnection.

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

**Theorem: **Let be an arbitrary collection of totally disconnected space, then is totally disconnected under the product topology.

**Proof:** Let be distinct. It follows that for some . Now, since is totally disconnected there exists a disconnection of it such that . Clearly then if we let be such that and we have that are non-empty disjoint open sets which contain respectively. It remains to show that , but this is obvious.

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

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

**Proof:** Let be distinct. Since is Hausdorff there exists disjoint open sets such that and . Clearly then there exists some open/closed such that . Clearly is the desired disconnection.

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

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

**Proof:** Let , and let be any neighborhood of . Clearly then is compact. Now, since is totally disconnected we know that for each there exists a separation of such that . Clearly is an open cover for and so by it’s compactness it has a finite subcover . Clearly then we have that and since is both open and closed so is their finite intersection. The conclusion follows.

We may now prove a neat little theorem:

**Theorem:** Let be a compact totally disconnected Hausdorff space, then is homeomorphic to a closed subspace of for some .

**Proof:** By the previous theorem we are able to state that there exists some open base whose elements are both open and closed. Note that given some fixed we have that the indicator function given by

is continuous. We claim that given by

is an embedding. Since is compact and Hausdorff we must merely show that is injective and continuous.

To see injectivity suppose that . Then, for every basic open neighborhood of . In particular, where is the set of all basic neighborhoods of . But, clearly since is we have that and so .

To see that it’s continuous let be subbasic open and let be the single subscript such that . Clearly then which by previous comment is open.

It follows that is an embedding. Noting that is a compact, and thus closed, subspace of completes the argument.

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