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I'm confused about constructing a family of subsequence using a diagonalization procedure. Often I see the following argument during a proof: "using a diagonalization procedure we can assume....."

What exactly is meant by this? I just know Cantor's diagonalization method for proving some cardinality stuff. Unfortunately I don't see how this apply in this situation. I'm aware of asking a very general thing, but hopefully it's clear what I try to ask.

Just to be sure that there's no misunderstanding, here is an example:

Suppose $X$ is a separable space and $(\phi_k)_{k\in \mathbb{N}}$ is a bounded subset of the dual of $X$ (topological dual). Then there is a subsequence $\Omega \subset \mathbb{N}$ and an element $\phi \in X^* $ such that $ \phi_k $ converges weak$^*$ to $\phi$ for this subsequence.

Now in the proof, we argue: let $(x_n)$ be dense in $X$. Then looking at $ \phi_k(x_1) $. This is bounded sequence in $ \mathbb{R} $ therefore it exists a subset $\Omega_1 \subset \mathbb{N} $ such that $\phi_k(x_1)$ converges for $ k\in \Omega_1$. Repeating this for ever $n \in \mathbb{N}$ we get an decreasing family of subsets (subsequence) of $\mathbb{N}$,

$$\mathbb{N}\supset \Omega_1 \supset \dots \supset \Omega_l \dots \supset \Omega$$

Why can I do this? Should I choose a subsequence of $\Omega_l$ using Bolzano-Weierstrass to get $\Omega_{l+1}$? Why do I know there's a $\Omega$ ? Why do I know that $\lim_k\phi_k(x_n)$ converges for ever $n\in \mathbb{N}$ using $\Omega$? Are there any assumption to apply this diagonalization argument? How does the diagonalization argument works in this example?

I'm very thankful for your help. If you need further explanations on what I exactly want to ask, let me know.

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The sequence $\{\Omega\}$ is decreasing, not increasing. Since we can have, for example, $\Omega_l=\{l,l+1,\ldots,\}$, $\Omega$ can be empty. The idea of the diagonal method is the following: you construct the sets $\Omega_l$, and you put $\varphi(l)$ the $l$-th element of $\Omega_l$. Then show that this subsequence works. –  Davide Giraudo Dec 15 '11 at 17:48
    
First, after choosing $ \Omega_1 $ I look at the sequence $ \phi_k(x_2) $ for $ k\in \Omega_1 $ and choose again a convergent subsequence, right? If I put $ \varphi(l)$ as the l-th element of $ \Omega_l $ then all I know is, that going ahead, this constructed subsequence is not equal $ \Omega_k \forall k \in \mathbb{N} $. right? Now I get stuck, what more can I say? And again, is there a restriction for applying such arguments? –  user20869 Dec 15 '11 at 18:04
    
If you fix a $k$, then the sequence $\{\phi_k(x_{\varphi(j)})\}$ is for $j$ large enough a subsequence of $\{\phi_k(x_j)\}_{j\in\Omega_k}$. You can check that $\varphi$ is strictly increasing, –  Davide Giraudo Dec 15 '11 at 18:36

1 Answer 1

up vote 1 down vote accepted

$\phi_k(x_n)$ converges along $k\in\Omega_n$. now consider the sequence $\Omega=\Omega_n(n)$, the $n$th term of $\Omega_n$ (the diagonal sequence). check that $\phi_k(x_i)$ converges for every $i$ along $k\in\Omega$. this $\Omega$ is different from yours in that it isnt in $\cap\Omega_n$ (which might be empty), so this part of your understanding seems incorrect.

think about it this way: for any $n$ the tail of $\Omega$ is a subsequence of $\Omega_n$ so that $\phi_k(x_n)$ converges along this subsequence.

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By your last sentence, do you mean, that the last inclusion in $ \mathbb{N}\supset \Omega_1 \supset \dots \supset \Omega_l \dots \supset \Omega $ is wrong? –  user20869 Dec 15 '11 at 18:09
    
@hulik the inclusions $\Omega_{i+1}\subseteq\Omega_i$ are correct, but the diagonal sequence you want to work with (which I called $\Omega$ is not in the intersection; it comes from choosing the $n$th element of $\Omega_n$ –  yoyo Dec 15 '11 at 18:28

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