Question about subset of $ \ell^2 $ space

Question

I have some doubts regarding a problem that I found in notes about metric spaces.

We have the set A of sequences that satisfy $ \sum_{i=1}^{\infty} ({a}_{i})^2 \lt \infty $. This is the $ \ell^2$ space. The summation of each squared term of the sequence is a finite number. The metric is defined as $ \sum_{i=1}^{\infty} (({a}_{i} - {b}_{i})^2)^{1/2} $

The first part of the problem is : If we take the set of sequences $ X= \{ (1,0,0,…,0,0…); (0,1,0,0,…0,…); … ; (0,0,0,…,1,0,…);… \} $, we need to show that X belongs to $ \ell^2 $

Unless I am missing something, it seem pretty trivial : We need to show that, since any sequence in X has only one non zero term, the summation of only one non zero term squared ($1^2$) is finite.

The second part is : To show that X is bounded and closed, but not compact in $ \ell ^2 $ . Here I am more doubtful , but I try :

1. Boundedness : We know that the distance of any element of X is either zero or $\sqrt 2$. So there is a $ r \gt 0$ such that for every element $x,y \in X \space , d(x,y) \lt r$ . Therefore X is bounded
2. Closedness : Since the set X is a sphere with radius $ \sqrt 2 $, X is closed (I know this is true but I dont know the justification, so I would appreciate help here)
3. Compactness : Here I dont know how to show that X is not compact. I would appreciate some insight.

Thanks to all!

Answer 1

Hint:

1. Boundedness is clear since $\|\mathbf{e}_n\|_2=1$ for all $n$

2. As for closeness, notice that if $\mathbf{e}_{n_k}\xrightarrow{k\rightarrow\infty}\mathbf{x}$, then $\{\mathbf{e}_{n_k}:k\in\mathbb{N}\}$ is a Cauchy sequence. For $0<\varepsilon<1$, there is $K_0$ such that for $k',k\geq K_0$, $$\|\mathbf{e}_{n_k}-\mathbf{e}_{n_{k'}}\|_2<\varepsilon$$ Since $\|\mathbf{e}_{n}-\mathbf{e}_{m}\|_2=\sqrt{2}>1$ for all $m\neq n$, it follows that $\mathbf{e}_{n_k}=\mathbf{e}_{n_{K_0}}$ for all $k\geq K_0$. Therefore $\mathbf{x}=\mathbf{e}_{n_{K_0}}$. This proves that $\{\mathbf{e}_n:n\in\mathbb{N}\}$ is closed.

3. Lack of compactness can be proved as in (2). Another way to see this is to consider the open cover $\{B(\mathbf{e}_n;\tfrac{\sqrt{2}}{2}):n\in\mathbb{N}\}$, where $B(\mathbf{y};r)=\{\mathbf{z}\in\ell_2: \|\mathbf{y}-\mathbf{z}\|_2<r\}$. This open cover has no finite subcover.

Answer 2

Showing $X$ is in your space is indeed that trivial Bounded=it fits inside a sphere around a point, here every point is ON the sphere around $0$ of radius $\sqrt 2$, so it sits in any bigger sphere.

For closed, your points are all a fixed discrete distance apart from each other. So imagine you have a convergent sequence from $X, a_n$. That means you can pick an $\epsilon$ that eventually all the points are that close to its limit, whatever that limit is. So pick $\epsilon=.1$, or something else sufficiently small to use the triangle inequality to show that from some $N$ on the sequence is constant. That tells you the limit is that constant point, and hence in the space.

To show it is not compact, just show an infinite open cover with no finite subcover. Hint: Since your distinct points are all $\sqrt 2$ apart, make each one a member of its own open set and no other, thus there can be no finite subcover.

Edited to include the basic topological insight: Basically if you look at the induced topology on $X$ you are getting an infinite discrete metric space. The proofs for everything are identical for any infinite discrete metric space.