# Walter Rudin's mathematical analysis: theorem 2.43. Why proof can't work under the perfect set is uncountable.

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.

• 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. Commented Mar 20, 2019 at 3:34

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.

• Thank you so much. I have revised my question. Commented 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.

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