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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 https://math.stackexchange.com/questions/2647211/rudin-2-43-every-nonempty-perfect-set-in-mathbbrk-is-uncountable – saulspatz Jun 23 '19 at 23:22
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    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. – Randall Jun 23 '19 at 23:25
  • Thank you very much! – Jia Cheng Sun Jun 24 '19 at 03:41

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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.

saulspatz
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  • I believe I understand Randall's comment. The statement that no xn lies in the intersection is somewhat similar to the statement that all natural numbers are finite. However, I don't get what you mean by V being the same sequence and why it matters. Can you explain? – Jia Cheng Sun Jun 24 '19 at 01:53
  • @JiaChengSun The first sentence was only intended to make the meaning of the rest of the answer clearer. If you understand the rest, just ignore the first sentence; it doesn't really matter. – saulspatz Jun 24 '19 at 03:08