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I strongly believe that a totally disconnected Hausdorff space has to have a topology that contains all the singletons. However, I am not sure how to go about proving it. I am especially unsure of how to use the totally disconnectedness in the argument.

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    You wrote that you thought that every such space must be discrete. Maybe the confusion was caused by similar reasons as for the poster of this question: [The rational numbers are totally disconnected but is not a discrete space?](http://math.stackexchange.com/questions/137793/the-rational-numbers-are-totally-disconnected-but-is-not-a-discrete-space)2012-05-06

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This would mean that such space is discrete. There is a lot of spaces that are totally disconnected but not discrete.

Let's have a look at the first example given at Wikipedia - $\mathbb Q$ as a subspace of real line.

Suppose that a subset $S\subseteq\mathbb Q$ is connected and it contains at least two different points $a$, $b$. Then there is an irrational number $c$ between $a$ and $b$. The subsets $(-\infty,c)\cap S$ and $(c,\infty)\cap S$ form a partition of $S$ into two non-empty open disjoint sets. This contradicts the assumption that $S$ is connected.

After a little searching I found the same proof posted in an answer to this question: What are all of the connected subsets of $\mathbb{Q}$?


Basically the same argument would work for any zero-dimensional Hausdorff space, i.e. a Hausdorff space that has a basis consisting of clopen sets.

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Your belief, as Martin remarks in the comments, is false.

First recall that a totally disconnected space is a space such that every connected component is a singleton. Also recall that connected components are always closed, but only when there are finitely many of them you can conclude that they are open. Discrete spaces, indeed are totally disconnected space and every singleton is both open and closed.

Example: The Rational Numbers, with the subspace topology from $\mathbb R$.

Suppose that $[a,b]\cap\mathbb Q$ is connected in $\mathbb Q$. The identity is a continuous function from $\mathbb Q$ into $\mathbb R$, so its image is connected. We know that connected subspaces of $\mathbb R$ are intervals, so the image is $[a,b]$. However this is a function from a countable set onto an uncountable set, unless of course $a=b$.

Other examples include the irrational numbers; the Cantor set; all countable metric spaces; and more.