Problem Statement: $\mathcal{V}$ is an infinite dimensional version of $E^{n}$ with coordinate vectors expressed as infinite sequences $$\mathbf{x}=(x_{1},x_{2}, …, x_{n}, …).$$ Vector addition is defined by $$\mathbf{x}+\mathbf{y}=(x_{1}+y_{1},x_{2}+y_{2}, …, x_{n}+y_{n}, …),$$ and scalar multiplication is defined by $$\kappa\mathbf{x}=(\kappa x_{1},\kappa x_{2}, …, \kappa x_{n}, …), \forall \kappa\in \mathbb{R}.$$ Define the norm of $\mathcal{V}$: $$\lVert \mathbf{x}\rVert =[\sum_{n=1}^{\infty}(x_{n})^{2}]^{1/2},$$ where $\sum_{n=1}^{\infty}(x_{n})^{2}$ is finite, $\forall \mathbf{x}\in \mathcal{V}$.
(c) For $l=1, 2, …$, let $\mathbf{e}_{l}=(0, …, 0, 1, 0, …)$ where $l-1$ zeros precede the $1$. Let $A=\left\{\mathbf{e}_{1}, \mathbf{e}_{2}, …\right\}$ Show that $A$ is bounded and closed.
(d) Show that the set $A$ is not compact.
So I feel like I am understanding most of this proof, but the fact that we are dealing with an infinite dimensional vector space is making me question my approach to the problem.
(c) For boundedness, I am taking some sequence $\left\{\mathbf{x}_{n}\right\}_{n\geq 1}$ in $A$ and supposing that it is not bounded, i.e.
$\forall C>0$, $\exists N>0$ such that $\lVert \mathbf{x}_{n}\rVert \geq C$ $\forall n\geq N$.
Is the construction of the elements in $A$ enough to conclude that the norm of any element in the sequence is $1$? Since the sum of the squares of the coordinates of any vector in $\mathcal{V}$ must be finite, would it not be trivial that this space is bounded?
But relative to $A$, I would think that any element would have norm $1$ (since only one coordinate is $1$ and the rest are $0$). The fact that this is an infinite dimensional vector space is making me think that I cannot assume that the norm of any $\mathbf{e}_{n}\in A$ is $1$, even though my intuitions tell me it is.
Then to show $A$ is closed, I let $\left\{\mathbf{x}_{i_{n}}\right\}_{n\geq 1}$ be a sequence in $A$, converging to some $\mathbf{e}_{0}\in\mathcal{V}$. Taking $\epsilon >0$, $\exists N>0$ such that
$$\lVert \mathbf{x}_{n}-\mathbf{e}_{0}\rVert<\epsilon,\ \forall n\geq N.$$
Here is where I am mostly confused about the problem: I am not sure if I should assume $\mathbf{e}_{0}\notin A$? If I do, then I am also unsure of what I can assume from that other than the fact that either more than one coordinate of the vector is nonzero, or all but one coordinate $x_{i}$ of the vector is nonzero with $x_{i}=1$.
(d) For this part, the book [Fleming] says to let $U_{l}=\left\{\mathbf{x}\ \vert\ \lVert \mathbf{x}-\mathbf{e}_{l}\rVert <1 \right\}$, for $l\in \mathbb{N}$. So the set $\mathcal{U}=\left\{U_{1}, U_{2},…\right\}$ covers $A$.
I wanted to approach this proof by supposing that $A$ is compact, and thus, $A$ must have a finite subcovering $\mathcal{U}_{m}=\left\{U_{1}, U_{2},…,U_{m}\right\},$ for some $n$.
But then considering any $\mathbf{e}_{n}$ for $n>m$, then $\lVert \mathbf{e}_{m}-\mathbf{e}_{n}\rVert=\sqrt{2}>1$. So, $U_{m}$ does not cover $\mathbf{e}_{n}$ for all $m\neq n$, furthermore $\mathcal{U}_{m}$ does not cover $\mathbf{e}_{n}$, so the conclusion holds.
I feel like this would work in a finite dimensional space, but the fact that we are in infinite dimensions tells me that I should be very careful with my assumptions about these vectors.
Any tips/suggestions would be appreciated!
Best Answer
You are wise to be cautious when working over infinite dimensions, since not everything that's true for finite dimensional vector spaces is necessarily true for infinite dimensions. For example, this problem says that the Heine-Borel theorem can fail in infinite dimensions.
That being said, you're being a little too cautious. The slick arguments you were concerned about do work:
Given any $e_k\in A$, $(e_k)_n=1$ when $n=k$ and $0$ otherwise, so it is indeed true that $\|e_k\|=1$. This is sufficient to show $A$ is bounded.
$A$ is trivially closed because each point in $A$ is isolated. Indeed, simply consider the the open sets you define in (d). Put differently, you can say that there can be no sequence $\{a_n\}\subset A$ convergent to a point outside of $A$ since such a convergent sequence would also be Cauchy (we're working in a metric space). Since this sequence $\{a_n\}$ doesn't converge to a particular $e_j$, for any $N$ there is some $n> N$ such that $\|a_n-e_j\|=\sqrt{2}>1$ (since $a_n=e_i$ for some $i\neq j$). This is true for all $j$, so it is not hard to show from here that $\{a_n\}$ is not Cauchy: Let $\epsilon>0$ be smaller than $1$ and let $N$ be any positive integer. Then there is some $n$ with $\|a_n-a_N\|>1$, so it is not true that $n,m\geq N$ implies $\|a_n-a_m\|<\epsilon$.
This also provides a slick proof for (d), since a subset of a metric space is compact if and only if every sequence has a convergent subsequence. In particular, the sequence of the basis vectors $\{e_n\}$ has no convergent subsequences since such a subsequence would be Cauchy, but each point in the sequence is distance $1$ from each other.
That being said, your argument for (d) is mostly fine, but the notation can be improved. When you suppose for contradiction that $A$ is compact, we don't know which $U_i$ we're pulling into our finite subcover--we only know there are finitely many (writing $U_i$ means something! It is the particular open set containing $e_i$!). It would be better to say that our finite subcover has the form $\{U_{i_1},\dotsc, U_{i_m}\}$ for some positive integers $i_1,\dotsc, i_m$. Then take $n>\max(i_1,\dotsc, i_m)$, and note that $e_n$ isn't covered, as you have already done.