# Baby Rudin 2.43

Theorem 2.43 states that any non-empty perfect set in the real vector space is uncountable.

Rudin uses basic induction to show that no point of $$P$$, where the points of $$P$$ are $$x_1, x_2, \dots$$ lies in the intersection of countably infinite sets. But as far as I know, basic induction can only prove a case for any natural number n and not infinity. And I'm pretty sure, in this case, Rudin is proving that for countably infinite points $$x_n$$, none of them lies in the intersection. A similar case is when induction alone can't prove that the countably infinite union of countably infinite sets is countable.

• For those with the book handy, the proof is detailed at math.stackexchange.com/questions/2647211/… Jun 23, 2019 at 23:22
• He inducts to construct $V_n$. Hence you know that every $V_n$ has said property. Now take unions or intersections of them all or whatever. He isn't claiming that some "$V_\infty$" has this property. Jun 23, 2019 at 23:25
• Thank you very much! Jun 24, 2019 at 3:41

It's the same sequence $$V$$ at every step. If he were proving that for every $$n$$, there exists a sequence $$V_1, V_2,\dots,V_n$$ such that the sequence has some property, an it were a different sequence at every step, then yes, we wouldn't be able to conclude that there was an infinite sequence $$V_1, V_2, \dots$$ with that property. However, Rudin explicitly constructs infinitely many sets.