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This is the part of proof in Rudin PMA p.41

Let $P(\subset \mathbb{R})$ be a perfect set. Since $P$ has limit points, $P$must be infinite. Suppose that $P$ is countable. Then, we can denote the points of $P$ by $x_1, x_2,...$. Let $V_1$ be any neighborhood of $x_1$.(i.e. open ball). Suppose $V_n$ is constructed. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ of some point $x_m \in P$ such that (i) $\overline {V_{n+1}}$ $\subset V_n$ and (ii) $x_n \notin \overline {V_{n+1}}$ and (iii) $V_{n+1} \cap P ≠ \emptyset$. Then form a sequence $\{V_n \subset \mathbb{R}^k | n\in \omega \}$.

Here, Axiom Of dependent choice is used. I have tried some other ways, but ,informally speaking, proof by 'squeezing' region requires AC. (Forming a decreasing sequence)

I want a proof without AC. Help..

Katlus
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  • What are you asking? Are you looking for some way to prove the result without choice? – KReiser Jul 30 '12 at 18:09
  • @KReiser Yes. I'll add it – Katlus Jul 30 '12 at 18:11
  • You actually use DC since the choice of $V_n$ depends on $V_{n-1}$. DC is stronger than countable choice (it just happens that I am studying the proof of that... weird, huh?) – Asaf Karagila Jul 30 '12 at 18:13
  • @Asaf Edited. It's off the topic, but it is hard to find where AC is used at the first time, then it is reallly hard to notice what kind of choice is used.. – Katlus Jul 30 '12 at 18:17
  • @Katlus: Yes. It's not an easy task to see the precise form of choice used in a proof. It requires a lot of practice and knowledge on the various forms. Dependent choice is like the name suggests: the next choice depends on the previous ones. Essentially when defining by induction you often use DC. Sometimes this use is excessive, but this is not a big deal most of the time. – Asaf Karagila Jul 30 '12 at 18:22
  • Without choice there is another kind of problem: what exactly do you mean by uncountable? – tomasz Jul 30 '12 at 18:29
  • @tomasz: Perfect sets are required to be closed. Closed sets have CH. It doesn't really matter, though. If the proof fails because the set is D-finite then the set is still uncountable. – Asaf Karagila Jul 30 '12 at 18:31
  • @tomasz I mean 'a set that is neither finite nor countable', which doesn't mean that cardinality of uncountable set is greater than $\aleph_0$. (I don't know whether this definition is correct in ZF though.. If you know a better definition let me know :)) – Katlus Jul 30 '12 at 18:33
  • @Katlus: Uncountable simply means not countable or finite. Strange non-well ordered sets are also uncountable by that definition. Your definition is fine. – Asaf Karagila Jul 30 '12 at 18:34
  • @Katlus: this in turn provokes the question: what do you mean by finite? The proof that I know of closed->CH uses injection of a Cantor set by a Cantor scheme (so it depends on AC). Maybe it doesn't make much of a difference in this particular case (I don't really know, I'm not familiar with descriptive set theory without choice), but I think you should make it clear what you mean in questions like this. – tomasz Jul 30 '12 at 18:38
  • @tomasz: You can prove in ZF that every $G_\delta$ set is countable or of size continuum. Beyond that some choice beginning to appear. See Arnold Miller's paper about D-finite Borel sets. – Asaf Karagila Jul 30 '12 at 18:40
  • @AsafKaragila: Interesting. Are there any good books (or fairly comprehensive papers) to read about this kind of thing (descriptive set theory sans choice)? I suspect it's going to be so messy I'll quickly lose patience, but I'd like to take a look, just out of curiosity. – tomasz Jul 30 '12 at 18:42
  • @tomasz: Not much that I know of. You can find in Fremlin's big book a chapter about measure theory without (or with very little) choice; a few papers here and there but not much more, I believe. – Asaf Karagila Jul 30 '12 at 18:44
  • @tomasz: I apologize, I just looked at Miller's linked paper and we can get up to $G_{\delta\sigma}$, at $F_{\sigma\delta}$ we need some choice, though. – Asaf Karagila Jul 31 '12 at 01:52
  • @Katlus: How to construct $V_n$: Let $V_1$ be neighbourhood of $x_1$. Since $x_2$ is a limit point. choose $2 < |x_2 - x_1 |$, the $V_2 = S{r_2}(x_2)$, so $ \overline{V_2} \subset V_1$. Similarly choose $r_3 < |x_3 - x_2|$, we obtain $V_3$ such that $V_3 \cap P \neq \phi$ and $\overline V_3 \subset V_2$ and $x_2 \not \in V_3$ . Similarly we can construct $V_n$ . Am I right or not. – user120386 May 23 '14 at 05:43

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The trick is simple. Use the fact the $P$ is well-ordered and that the rationals are well-ordered.

In the induction step, instead of taking "some $x_m\in P$" take $x_m$ such that $m$ is the least $k$ for which $x_k$ has an open neighborhood etc. etc.

Also require that the open neighborhoods are open balls of rational radius. Well ordering the rationals we can require the radius to be of the least rational in a fixed enumeration such that a ball around $x_m$ with the wanted properties exists.

Now we have a canonical choice of the open neighborhoods, and we can show that their intersection is non-empty because it is equal to the intersection of the closures - which is compact (by a previous question of yours) and therefore contains a point. This point is not in $P$.

Asaf Karagila
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    I don't get the part 'rationals are well-ordered' – Katlus Jul 30 '12 at 19:05
  • @Katlus: The rationals are countable, therefore they are well-ordered. Fix a bijection with $\omega$ and you have proved that. We use this to choose a rational for the radius of $V_{n+1}$, we simply take the least rational in this enumeration such that $B(x_m,q_j)$ has the properties we want from $V_{n+1}$. – Asaf Karagila Jul 30 '12 at 19:07
  • Oh got it, it's a different ordering of $\mathbb{Q}$ from the usual. Thank you – Katlus Jul 30 '12 at 19:13
  • I'm curious why baby Rudin and Munkres don't exercise similar tricks in their proofs to avoid such confusion about the use of AC... – YuiTo Cheng Dec 18 '18 at 15:02
  • @YuiToCheng: Maybe because choice is generally accepted, and many point find it more confusing from a pedagogical point of view to dwell on these subjects. – Asaf Karagila Dec 18 '18 at 15:10