Assume that $X$ is a totally disconnected space. Then every two-point set is disconnected, which implies that every singleton is open in the topology of $X$ (because the one-point subsets of two-point sets form a separation). Isn't the collection of those singletons a basis for the discrete topology? So isn't the topology discrete?
[Math] If totally disconnectedness does not imply the discrete topology, then what is wrong with the argument
general-topology
Best Answer
Note that being totally disconnected means that the only connected nonempty subsets are the singletons. To see that, for example, $\{ a , b \}$ is not connected, it suffices to find open subsets $U , V \subseteq X$ with the following properties:
This would not imply, however, that either $\{ a \}$ are $\{ b \}$ are open subsets of $X$. This only means that $\{ a \}$ and $\{ b \}$ are open subsets of the subspace $\{ a,b \}$ of $X$.
To give more details with regards to an actual example, consider the rationals $\mathbb{Q}$ as a subspace of the real line. Given any subset $A \subseteq \mathbb{Q}$ of size $> 1$ pick $p,q \in A$ with $p < q$. Then there is an irrational number $x$ such that $p < x < q$. Note that $U = ( - \infty , x ) \cap \mathbb{Q}$ and $V = ( x , + \infty ) \cap \mathbb{Q}$ are open subsets of $\mathbb{Q}$ which have the following properties:
This demonstrates that $A$ cannot be a connected subset of $\mathbb{Q}$; and in general, the only nonempty connected subsets of $\mathbb{Q}$ are the singletons.
Note, however all nonempty open subsets of $\mathbb{Q}$ are infinite.