Is the Cantor set Hausdorff? Which separability axioms it has?
I know that $\mathbb R$ are Hausdorrf is it enough? Like the Cantor set is compact if it was Hausdorff then the Cantor set would be normal and regular, and it would be T1 for inherited? Or it has another axioms?
Thanks for your help
Being a subspace of the metric space $\mathbb R$, the Cantor set is metrizable. This implies that the Cantor set satisfies all the usual separation axioms --- Hausdorff, regular, completely regular, normal, hereditarily normal, perfectly normal.
"I know that R (real numbers) are Hausdor[f]f is it enough?"
Yes, it is. Suppose $X$ is a Hausdorff space and $Y\subseteq X$ has the subspace topology.
Given two points $x,y\in Y,$ there are open sets $U,V\subseteq X$ with $x\in U,$ $y\in V,$ and $U\cap V=\varnothing.$ The intersections of $U$ and $V$ with $Y$ are open subsets of $Y$ and contain $x$ and $y$ as members and do not intersect each other.
Given any two points $p,q\in \mathfrak{C}$ (the Cantor Ternary Set), we have that $p,q\in \mathbf{R}$ as well. Take open neighborhoods $U,V\subset \mathbf{R}$ such that $p\in U$, $q\in V$ and $U\cap V=\varnothing$. Then if we take intersections $\tilde{U}=\mathfrak{C}\cap U$ and $\tilde{V}=\mathfrak{C}\cap V$, we have that $p\in \tilde{U}$ and $q\in \tilde{V}$, with $\tilde{U}\cap \tilde{V}=\varnothing$. By definition of the subspace topology, $\tilde{U}$ and $\tilde{V}$ are open in $\mathfrak{C}$ and we are done.
Note that this construction works in general, for any subspace $Y$ of a Hausdorff space $X$.
