# Diagonalization method by Cantor (2)

I asked a while ago a similar question about this topic. But doing some exercises and using this stuff, I still get stuck. So I have a new question about this topic. (Here is the link for the previous question: Cantor diagonalization method for subsequences).

Now my new question: Suppose I have a sequence $(x_n)$, a set $K$ and a function $f$ and we define for all $l \in \mathbb{N}$ the set $M_l:= K\cap \{x|f(x)\le l\}$ with $f< \infty$ on $K$. I've proved that for a fixed $l$ there's a subsequence $(x_{n_k})$ converging on $M_l$, denote this by $(x^l_{n_k})$. First it's clear that $M_l \subset M_{l+1}$ and I want to show that there's a subsequence which converges on $\cup_{l\ge1} M_l = K$.

So I have different subsequences $(x^1_{n_k}),\dots,(x^p_{n_k}),\dots$ and define the diagonal sequence as ${x^{\phi(l)}_{n_{\phi(l)}}}$ where $\phi(l)$ is the $l$-th element of $n_k$ (just the diagonal sequence). Is it enough to say, since $(M_l)$ is increasing the sequence converges on $K=\cup_{l\ge1} M_l$?

I'm sorry, but I'm just confues about picking the right sequence and to prove that it is the right sequence using Cantor. Thank you for your help

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(1) What does "converging on $M_l$" mean? (2) What does $\phi(l)$ mean? You define it as the $l$-th element of $M_l$. So are the elements of $M_l$ integers? To me there are difficulties of exposition. In general, one can use diagonalization to produce subsequences with desired properties, and if your notation is clarified (fewer symbols?) I am sure an answer can be produced, indeed you can produc an answer. – André Nicolas Feb 10 '12 at 20:22
@ André Nicolas: I'm very sorry! That was absolutely my fault! I correct $\phi(l)$. This function just should take the "diagonal elements" and hence construct the diagonal sequence. Convergence on $M_l$ means, $x_n(t)$ converge for all $t\in M_l$ to $x(t)$. The sequence $(x_n)$ is a sequence of (measurable) functions. I didn't mention it, because I get stuck at the diagonal procedure. But sorry for the unclarity – user20869 Feb 10 '12 at 21:46
OK, I had assumed you meant a numerical sequence. Now that the indices are fixed, the answer of David Giraudo may settle things. You certainly want the subsequences to be nested. You may need some version of uniform convergence. – André Nicolas Feb 10 '12 at 21:57
Why should I need a version of uniform convergence? – user20869 Feb 10 '12 at 22:23
I do not know the details of the actual problem. However, although the $M_l$ are probably well-behaved (compact), that does not say that their union is. There are too many counterexamples around to be confident, without the details, that there are no issues. – André Nicolas Feb 10 '12 at 22:31

Denote by $(x_{\varphi_l(k)})$ a subsequence which works for $M_l$. In fact, you have to construct these subsequence by induction, in order to make $(x_{\varphi_{l+1}(k)})$ a subsequence of $(x_{\varphi_l(k)})$. Then we put $x_{n_k}=x_{\varphi_k(k)}$. Now we are sure that the sequence $(x_{n_k})_{k\geq N(j)}$ is a subsequence of $(x_{\varphi_j(k)})_{kgeq N(j)}$ for some integer $N(j)$.
It's important that the subsequences are nested, otherwise it may not work. For example, we assume that for $l$ even only a subsequence of the form $(x_{2k})$ work and for $l$ odd a subsequence of the form $(x_{2k+1})$. Then $x_{n_{\varphi(2l)}}^{\phi(2l)}=x_{4l}$ and $x_{n_{\varphi(2l+1)}}^{\phi(2l+1)}=x_{2l(2l+1)+1}$ so the sequence of indexes is not increasing.
@ Davide Giraudo: Thanks a lot for your answer. But I think I don't really understand the diagonal procedure. If I have choosen the diagonal sequence, I must show, that this sequence works, right? So generally do I have to choose, in a diagonal procedure, always $(x_{\varphi_{l+1}(k)})$ as a subsequence of $(x_{\varphi_{l}(k)})$? And if so, why? Thanks for your explanations – user20869 Feb 10 '12 at 21:51