Totally Disconnected and Locally Connected Space is Discrete
Theorem
Let $T = \struct {S, \tau}$ be a topological space which is both totally disconnected and locally connected.
Then $T$ is the discrete space on $S$.
Proof
So, let $T = \struct {S, \tau}$ be that topological space which is both totally disconnected and locally connected.
As $T$ is totally disconnected, every point is a component and therefore closed.
As $T$ is locally connected, there exists a basis $\BB$ of $T$ such that every element of $\BB$ is a component of $T$.
In order for $T$ to be covered by $\BB$, every singleton subset of $T$ must be in $\BB$.
From the definition of topology, the union of any number of these singleton sets is an open set of $T$.
That is, every subset of $S$ is open in $T$.
That is, every element of the power set of $S$ is open in $T$.
This is precisely the definition of the discrete space on $S$.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness