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A topological space (X, T) is zero-dimensional iff it has a basis consisting of clopen sets.

A topological space (X, T) is totally disconnected iff every non-empty connected subspace is a singleton.

I am wondering how we can show the right to left direction of following:

A subspace of $\mathbb{R}$ is zero-dimensional if and only if it is totally disconnected.

I already know how to show the left to right direction, using the fact that zero-dimensional Hausdorff spaces are always totally disconnected. But I cannot find a way to prove the other direction.

Thanks!

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  • $\begingroup$ Yes, this equivalence holds in all ordered spaces. In general metric spaces these notions can differ a lot, though. There are even infinite-dimensional totally disconnected complete separable metric spaces. $\endgroup$ – Henno Brandsma Sep 13 '20 at 6:14
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HINT: Let $X\subseteq\Bbb R$ be totally disconnected. Show that $\Bbb R\setminus X$ is dense in $\Bbb R$, and use points of $\Bbb R\setminus X$ to define clopen sets in $X$.

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Suppose that $X \subseteq \Bbb R$ is totally disconnected, i.e. has no non-singleton connected non-empty subsets.

Let $O$ be open in $X$, so $O = U \cap X$ with $U$ open in $\Bbb R$, and $x \in O$ aritrary.

Let $(a,b)$ be an open interval containing $x$ and such that $(a,b) \subseteq U$. Then there are points in $(a,x)$ that are not in $X$ (otherwise $(a,x)$ would be a non-trivial connected subset of $X$ which cannot be), so pick $p \in (a,x)\setminus X$. Likewise pick $q \in (x,b)\setminus X$. Then $(p,q) \cap X = [p,q] \cap X$ and this implies that $(p,q) \cap X$ is a clopen (in $X$) subset of $O$ that contains $x$, showing that the clopen subsets of $X$ form a base, i.e. $X$ is zero-dimensional.

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