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Rudin PMA p.157

I'm trying to prove;

"If $\{f_n\}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $\{f_n\}$ has a subsequence $\{f_{n_k}\}$ such that $\{f_{n_k}\}$ pointwise convergent on $E$"

It's clear that Rudin made dependent choice.

I'm trying to prove this withouc AC and want to know if there is a way to choose a subsequence of a sequence. That is when $\{n\}$ is a given sequence, then i want to choose a subsequence $\{n_k\}$, then again choose a subsequence of $\{n_{k_j}\}$ and again countable times.

Is it possible?

Or if one can prove this with a different argument, Or if it is unprovable, please let me know..

Thank you in advance

  • 0
    In other words: How much choice is needed for Tychonoff's theorem for a countable product of closed and bounded subsets of $\mathbb{C}$?2012-12-15
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    @Nate Yes exactly2012-12-15
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    It seems to me the key is: given a bounded sequence $\{a_n\}$ (say in [0,1] for simplicity) can we *canonically* choose a convergent subsequence? Here's a proposal: the set of subsequential limits of $\{a_n\}$ is closed and bounded, hence has a minimum element $a$. (So we canonically chose a subsequential limit.) Now choose the subsequence: let $n(k)$ be the least $n$ with $|a_n - a| < 1/k$. Then $a_{n(k)} \to a$.2012-12-15
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    Oh wait: "the set of subsequential limits is closed": we have to be a little careful not to use choice in proving this.2012-12-15
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    @Nate Yeah, there exists a limit point $p$ which is a limit point of every $\{f_n(x_i)\}$ where $x_i$ enumerate $E$.2012-12-15
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    Even if you can select your subsequences countably many times without AC, there is no guarantee that their intersection will be non-empty.2012-12-15
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    @TonyK Aren't they compact?2012-12-15
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    You seem to think that if you can produce a nested sequence of subsequences, then this will answer the question. But the intersection (or limit, if you prefer) of a nested sequence of subsequences can be empty.2012-12-15
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    For instance: the first subsequence is $\{1,2,3,4,5,...\}$, the second subsequence is $\{2,3,4,5,6,...\}$, the third subsequence is $\{3,4,5,6,7,...\}$,...2012-12-15
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    @TonyK I didn't mean 'that' intersection, but i meant the intersection of "the set of limit points of $\{f_n(x_i)\}_{n\in\omega}$" where $x_i$ enumerate $E$. Sorry for the ambiguity of my words..2012-12-15
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    @Katlus: Whatever _you_ meant, do you agree with my objection to your suggested argument? If you choose a subsequence countably many times, you may end up with nothing.2012-12-15
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    @TonyK Oh. yes i agree.. i thought you were talking about my comments at the first.. now i see. I'm sorry2012-12-15
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    @TonyK: That is easily repaired, though: The $n$th time you choose a nested subsequence to restrict to, decide always to keep the first $n$ points in the previous sequence unchanged. This will not change the limit of the subsequence you restrict to, and the eventual nest of subsequences is guaranteed to have infinite intersection.2012-12-15
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    (Or, equivalently, once you have chosen your sequence $f_{ij}$ of nested subsequences, just consider $(f_{nn})_n$. That is eventually a subsequence of each of the original sequences).2012-12-15
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    It is (maybe) off the topic, but if $a_{n_k} \rightarrow p$ and $a_{n_j} \rightarrow p$, then $a_{n_k \cup n_j} \rightarrow p$?2012-12-15
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    To be sure I understand the question, there is a countable set $E$ and a sequence of complex functions whose domains are all $E$, and at each $e\in E$ we have that the no function in the sequence maps $e$ beyond some disc; **then** we can find a subsequence which converges pointwise.2012-12-15
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    @Asaf That's right2012-12-15
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    Something is strange here, what if all the $f_n$'s map into the unit disc but the union of the ranges is a discrete set. How could you construct a p.w.-convergent subsequence?2012-12-15
  • 0
    Proof for Arzela-Ascoli's theorem in wikipedia is exactly the same as Rudin's and wikipedia says nothing about AC and the argument can be done by Induction. Why?2012-12-15
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    Typo: 'wikipedia says' the argument can be done by induction2012-12-15

2 Answers 2

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The answer is yes, here is a proof:

  • Every bounded sequence $\{x_n\}$ in $\mathbb{R}$ has a limit point.

Suppose not,let $K$ be a compact set containing the sequence, then $\{x_n\}$ must be infinite;no AC needed to prove this, thus for each $k\in{K}$ define $s(k)$ where $s(k)$ is the least positive integer $n$ such that $(B_{1/n}(k)-\{k\})\cap{\{x_n\}}=\emptyset$, then $\{B_{\frac{1}{s(k)}}(k)\}_{k\in{K}}$ is an open covering of $K$ with no finite subcover, since $\{x_n\}$ is infinite.

  • For every bounded sequence $\{x_n\}$ there exists a uniquely determined converging subsequence $\{f(\{x_n\})_m\}_{m\in{\mathbb{N}}}$ of $\{x_n\}$.

Let $E$ be the set of limit points of $\{x_n\}$, then $E$ is a bounded set since our sequence is bounded, let $a=infE$; $a$ exixts since $E$ is nonempty by what it was shown above, then every neighborhood of $a$ intersects the sequence; this is easily seen without using AC, then for each $m>0$ let $k(0)=0$ and for each $n$ let $k(m+1))$ be the least positive integer $r$ greater than all of $k(0),\ldots,k(m)$ and such that $|a-x_r|<\frac{1}{m+1}$. Define $f(\{x_n\})_m=x_{k(m)}$ for all $m\in{\mathbb{N}}$, it is clear that the series $\{f(\{x_n\})_m\}_{m\in{\mathbb{N}}}$ converges to $a$.

Now using the notation on page 157 in Rudin’s PMA define $S_0=\{x_n\}$, and for $m\in{\mathbb{N}}, S_{m+1}=f(S_m)$

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Let's confine ourselves to Real-valued functions.

That is, let $\{f_n\}$ be a sequence of pointwise bounded functions on $E$ such that $f_n$ maps from $E$ to $\mathbb{R}$.

Let $x_i$ enumerate $E$.

Let $S_1 = \{n_k \subset \mathbb{N} \times \mathbb{N} : f_{n_k}(x_1) \rightarrow \limsup_{n\to\infty} f_n(x_1) \}$

and $g_1:\omega \rightarrow \bigcup_{n_k \in S_1} rng(n_k)$ be the isomorphism.

$S_{i+1} = \{g_{i_{n_k}} \subset \mathbb{N} \times \mathbb{N} : f_{g_{i_{n_k}}}(x_{i+1}) \rightarrow \limsup_{n\to\infty} f_{g_{i_n}}(x_{i+1}) \}$

and $g_{i+1}:\omega \rightarrow \bigcup_{g_{i_{n_k}}\in S_{i+1}} rng(g_{i_{n_k}})$ be the isomorphism.

(Note that $S_i$ is nonempty and well-defined since $f_n$ is pointwise bounded on $E$)

By induction, $\{g_i\}$ is well defined and $g_{i+1}$ is a subsequence of $g_i$.

Define $h(n)=g_n(n)$ for all $n\in \mathbb{N}$

Then for any $i\in\mathbb{N}$, $f_{h(n)}(x_i)$ is convergent. Q.E.D.

This can be extended to $\mathbb{R}^k$.