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.

  • $\begingroup$ With the metric on $P$ inherited from the usual metric on $\Bbb R^n$, the space $P$ is a complete metric space with no isolated points. We can show that a non-empty complete metric space $X$ with no isolated points has a subspace $Y$ which is homeomorphic to the Cantor Set. For the purposes of this Q it suffices to show there is a $Y\subset X$ which is a bijective image of the set of all binary sequences. $\endgroup$ Mar 20, 2019 at 3:34

2 Answers 2


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 $\{x_n: n\in\mathbb{N}\}$ and recursively define a sequence of sets $V_n$ ($n\in\mathbb{N}$).

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 $V_\eta\cap P$ 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 $(0,1)\supset(0,{1\over 2})\supset (0,{1\over 3})\supset ...$). 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.

  • $\begingroup$ Thank you so much. I have revised my question. $\endgroup$
    – Tengerye
    Mar 20, 2019 at 1:31

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 $F=\{P \setminus \{x\}: x\in P\}$ 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.

  • $\begingroup$ This A is unrelated to my comment to the Q regarding a subset of $P$ that's homeomorphic to the Cantor Set $\endgroup$ Mar 20, 2019 at 4:02

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