# Closed subset of perfect complete metric space is homeomorphic to Cantor set

I previously had a question that was too long and should have probably been split into two questions. I was able to show that if $X$ was a complete seperable metric space with a subset homeomorphic to the Cantor set, then $\left[0,1\right]$ was the continuous image of $X$. I am having trouble showing this though:

If $X$ is a complete separable metric space and $\left\lvert X\right\rvert=2^{\aleph_{0}}$, then $X$ contains a subset homeomorphic to the Cantor set.

I know that since $\left\lvert X\right\rvert=2^{\aleph_{0}}$, then it has a set of condensation points, denoted as $\operatorname{cond}\left(X\right)$. I also know that $\operatorname{cond}\left(X\right)$ is perfect, so it is closed and thus is a complete perfect metric subspace of $X$ (also separable, since any subspace of a separable metric space is separable. Not sure if this is useful). My idea was to try and construct a homeomorphism $f$ from a closed subset of $\operatorname{cond}\left(X\right)$ to the Cantor set, but I've been having trouble coming up with one.

• Dont you just build a tree of closed sets of decreasing diameters ? – Rene Schipperus Dec 19 '17 at 3:01
• @ReneSchipperus, sorry I'm not familiar with the argument/proof you're referring to. – Jake Dec 19 '17 at 3:15
• Find two disjoint closed sets of power continuum. Inside each of those find two more. Repeat until you have a tree and take the intersections along each branch of the tree. This gives the cantor set. – Rene Schipperus Dec 19 '17 at 3:32