Hi,

I'm stuck on this problem:

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

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

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

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

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

$\displaystyle \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 $\displaystyle (x_{j}) \in Z$ ?

Possibly by contradiction - suppose $\displaystyle (x_{j}) $ is not in Z, then at least one of its members will not satisfy $\displaystyle x_{2j}=j^{2} x_{2j-1} $, say, this is the k-th member; so we set $\displaystyle \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?