Gfrktyl

I found several discussions about this theorem, like this one. I understand the proof adopts contradiction by assuming the perfect set $P$ is countable.

My question is if the assumption is $P$ is uncountable, the proof seems remains the same, i.e., the $P$ can't be uncountable either. In other words, I think whatever the assumption is, we can draw the contradiction in any way.

I don't understand in which way the uncountable condition could solve the contradiction in the proof.

First, there's a typo in your question: the proof proceeds by assuming for contradiction that $P$ is countable (not uncountable, as you've written).

More substantively, countability is used right away: we write $P$ as ${x_n: ninmathbb{N}}$ and recursively define a sequence of sets $V_n$ ($ninmathbb{N}$).

If $P$ were uncountable, we couldn't index the elements of $P$ by natural numbers. We'd have to index them by something else - say, some uncountable ordinal. So now $P$ has the form ${y_eta:eta<lambda}$ for some $lambda>omega$.

We can now proceed to build our $V$-sets as before, but at the "first infinite step" we run into trouble: we need $V_etacap P$ to be nonempty for each $eta$, but how do we keep that up forever? In fact, our $V$-sets might disappear entirely: while at each finite stage we've stayed nonempty, but we could easily "become empty in the limit" (consider the sequence of sets $(0,1)supset(0,{1over 2})supset (0,{1over 3})supset ...$). The recursive construction of the $V_n$s - which is the heart of the whole proof - relies on always having a "most recent" $V$-set at each stage, that is, only considering at most $mathbb{N}$-many $V$-sets in total. That this is sufficient follows from the countability of $P$. As soon as we drop this, our contradiction vanishes.

The Baire Category Theorem: If $P$ is a complete metric space and $F$ is a non-empty countable family of dense open subsets of $P$ then $cap F$ is dense in $P.$

Suppose $P$ is a non-empty closed subset of $Bbb R^n.$ Let $P$ inherit the usual metric from $Bbb R^n.$ Then $P$ is a complete metric space. Now suppose $P$ is countable and is a perfect subset of $Bbb R^n.$ Then $F={P setminus {x}: xin P}$ is a non-empty countable family of dense open subsets of the space $P,$ so $cap F=emptyset$ is dense in $P,$ which is absurd.

(If $P$ were not assumed to be perfect then not all members of $F$ could be assumed to be dense in $P.$)

Aside: The proof of the Baire Category Theorem is direct and simple. Some students seem to be uncomfortable about this theorem, perhaps because it is unlike anything they've ever seen.