# Convergence in sequences of sequences

• October 18th 2010, 05:26 AM
Mimi89
Convergence in sequences of sequences
Hi,

I'm stuck on this problem:

Let $(X, \| . \|_{\infty} )$ be a normed vector space of complex sequences $( \alpha_{j} )$ that converge to zero, with the sup-norm:

$\| ( \alpha_{j} ) \|_{\infty} = sup \{ | \alpha | : j \in \mathbb{N} \}$

Let $Z= \{ (\alpha_{j} ) \in X : \alpha_{2j}=j^{2} \alpha_{2j-1}, j \in \mathbb{N} \}$

I need to show that Z is closed in X, but I'm getting a bit confused about sequences of sequences... here's what I got so far:

We will use the fact that Z is closed (in a metric space, as is a NVS) iff it contains all its limit points, i.e.

we want to show that whenever $( \alpha_{j} )_{n} \in Z$ is a sequence in Z, which converges to some $( x_{j}) \in X$ then we have that $(x_{j}) \in Z$.

So, we take an $( \alpha_{j} )_{n} \subset Z$, which converges to some $( x_{j}) \in X$. By definition, this means that

$\forall \epsilon > 0, \exists N \in \mathbb{N} \ s. \ t. \ \forall n>N, \ we\ have\ that\ \| (\alpha_{j})_{n} - (x_{j}) \|_{\infty} < \epsilon$

But how do I use this to show that $(x_{j}) \in Z$ ?

Possibly by contradiction - suppose $(x_{j})$ is not in Z, then at least one of its members will not satisfy $x_{2j}=j^{2} x_{2j-1}$, say, this is the k-th member; so we set $\epsilon = |x_{k}-k^{2} x_{2k-1}|$ and show that this contradicts convergence? But I'm still getting very confused in passing from the infinity-norm to the Euclidean norm...

Is this right, where do I go from here, and is there any way to prove this without contradiction?
• October 18th 2010, 06:41 AM
Mauritzvdworm
consider a sequence in $Z$, say $\{(\alpha_j)_l\}$ which converges to some $(\alpha_j)\in X$, we can write
$\lim_{l\rightarrow \infty}(\alpha_j)_l=(a_j)\in X$
now, is this $(\alpha_j)$ in $Z$?

Consider the following
$\lim_{l\rightarrow \infty}(\alpha_{2j})_l=\lim_{l\rightarrow \infty}j^2(\alpha_{2j-1})_l$
so
$\alpha_{2j}=j^2(\alpha_{2j-1})$
and $Z$ is closed
• October 19th 2010, 04:49 AM
Mimi89
Thanks a lot for that!

One more question on the same problem:

I have to show that the element

$x_{j}=(1,0,\frac{1}{4}, 0, \frac{1}{9}, o, ...)$ is in the closure of Y+Z but not in Y+Z.

Now, I've written it as the limit of:

$( 0, -1, 0, -1, 0, -1, ..., -1, 0, 0, 0, 0....) +
(1, 1, \frac{1}{4}, 1, \frac{1}{9}, 1, ...., \frac{1}{n^{2}}, 1, 0, 0, 0, ...)$

where the former is in Y (the last '1' occurs at the 2n-th place) and the latter in Z (the last nonzero entry is in the 2n-th place, again). So, I've shown (xj) is in the closure, as I've written it as the limit of a sequence which is in the sum, and closures are closed.

But how do I show it is *not* in the sum Y+Z?
• October 19th 2010, 06:46 AM
Mauritzvdworm
what is Y?
• October 19th 2010, 07:28 AM
Mimi89
Sorry - Y is the subset $\{ (\alpha_{j} ) \in X : \alpha_{2j-1} = 0, j \in \mathbb{N} \}$

And I've edited the above post; indeed, the first sequence in the sum does converge to 0 (just its limit doesn't).