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?

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    $\begingroup$ You might as well start with "Assume that $X$ is a Hausdrorff space." That should have made you doubt even more. $\endgroup$ – Hagen von Eitzen Jun 17 '13 at 16:54

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:

  1. $U \cap \{ a,b \} = \{ a \}$;
  2. $V \cap \{ a,b \} = \{ b \}$;

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:

  1. $U \cap A \neq \emptyset \neq V \cap A$;
  2. $A \subseteq U \cap V$; and
  3. $( U \cap V ) \cap A = \emptyset$.

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.

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  • $\begingroup$ "This only means that $\{a\}$ and $\{b\}$ are open subsets of the subspace $\{a,b\}$ of $X$." explains everything. Thank you! $\endgroup$ – Xena Jun 17 '13 at 16:58

Totally disconnected means that no two points are in the same connected component. So for $x,y\in X$ there is no connected set containing both points. In particular the set $\{x,y\}$ is not connected, thus is discrete, thus each point is open in the subspace topology of this set. That does not mean that it is open in $X$.

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