How is the Cantor set Hausdorff?

Question

For a proof I am working on, I assume the Cantor set is Hausdorff because that's what I recall from Analysis. However, now that I think about it, I don't know what an open subset of the Cantor set looks like. So how would I go about showing the Cantor set is Hausdorff?

Answer 1

I assume by "Cantor set" you mean the usual Cantor set, which is a subset of the unit interval $[0, 1]$ - in particular, the Cantor set is a subset of $\mathbb{R}$.

If I have a set $X$ with a topology $\tau$, then subsets of $X$ can be given the "subspace" topology: if $S\subseteq X$, an open subset of $S$ in the subspace topology is a set of the form $S\cap U$, where $U$ is an open subset of $X$ in the sense of $\tau$. So the natural topology on the Cantor set - indeed, the one which is used by default - is the subspace topology coming from $\mathbb{R}$.

For your problem, you don't really need a full description of the topology on the Cantor set- you just need to show that it's Hausdorff. This turns out to be straightforward: you already (presumably) know that $\mathbb{R}$ is Hausdorff; can you show that a subset of a Hausdorff space (with the subspace topology) is Hausdorff?

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EXERCISE: there is a natural way to assign to each $x$ in the Cantor set, an infinite sequence of zeroes and ones. There is a nice description of the topology on the Cantor set, in terms of these sequences - can you find it?