Every perfect set has cardinality $2^{\aleph_0}$?

Question

It is well known that perfect sets in $\mathbb{R}^n$ are uncountable (e.g., baby Rudin states this). Recently I heard of this stronger result:

Every perfect set in $\mathbb{R}^n$ has cardinality $2^{\aleph_0}$.

This is easily proved if we assume the continuum hypothesis. However, this result does not rely on that. Is there a proof of this fact? And does this result hold for more general spaces (e.g., complete metric spaces)?

I understand that it's customary to show my effort here at math.SE, but honestly I have no idea how I should attempt at it...

Answer 1

Yes, it does not require the continuum hypothesis to prove.

Suppose $P$ is a perfect set. WLOG (if not, restrict to an appropriate closed interval) $P$ is bounded. Then we can find two disjoint closed subsets, $l(P)$ and $r(P)$, which are each also perfect. And we can iterate this process, building such things as $r(l(r(P)))$ and so on.

Now remember that the intersection of a descending sequence of closed and bounded sets is nonempty. This means that for every infinite binary sequence, the corresponding descending sequence of perfect sets has a point in the intersection. For example - identifying "$r$" with $1$ and "$l$" with $0$ - the sequence $$f=0,1,0,1,,...$$ yields the sequence $$l(P)\supseteq r(l(P))\supseteq l(r(l(P)))\supseteq r(l(r(l(P))))\supseteq...,$$ and we pick a point $p_f$ in the intersection of this chain of perfect sets.

Now just check that if $f\not=g$ we have $p_f\not=p_g$ (HINT: $r(X)\cap l(X)=\emptyset$ ...).

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It might seem like we used the axiom of choice here, in two places:

• Choosing $l(X)$ and $r(X)$, for a perfect set $X$.

• Choosing a point in the intersection of a decreasing sequence of closed and bounded sets.

However, we don't actually need AC here:

• If $X$ is perfect, we can in fact show that there is a pair of rationals $p<q$ such that $X\cap (-\infty, p]$ and $X\cap [q,\infty)$ are each perfect. But $\mathbb{Q}^2$ is well-orderable ...

• There is in fact an easily-definable choice function for nonempty closed sets. (HINT: if the set is bounded from below, just pick the smallest element; now do you see how to deal with the not-bounded-from-below case?) We could also require that $l(X)$ and $r(X)$ have diameter at most one half of that of $X$ (where the diameter of a set is the supremum of the distances between any two points in the set).