Something has been bothering me recently with regards to compactness in finite vs. infinite dimensional vector spaces.
By theorem of compactness in finite dimensional normed space $(X,\left\lVert \right\rVert)$, any subset $M\subset X$ is compact iff $M$ is closed and bounded. However closedness and boundedness do not always imply compactness in infinite dimensions. This was demonstrated by taking the set of basis vectors, namely $(e_n)_{n\in\mathbb{N}}$ such that $e_n=(\delta_{in})_{i\in\mathbb{N}}$. Clearly, this set is bounded, since $\left\lVert e_n\right\rVert=1$. The argument given for closedness was that this is a point set and hence all points are isolated. Now for the questions:
1. Closedness of the set $(e_n)$
I was thinking of the following way to exactly prove closedness and am here wondering if the reasoning is correct:
Take an arbitrary sequence of these vectors $(e_n)$ and let us assume convergence to some $a, e_n\xrightarrow{n \rightarrow \infty} a$. If we prove $a=a_i=(\delta _{in})_{n\in\mathbb{N}}$, then $a$ will be in the same set and since the sequence choice was arbitrary, the set is closed.
For the convergence to hold, for $\forall \epsilon>0$ there must $\exists N\in\mathbb{N}$ such hat $\left\lVert e_n-a\right\rVert<\epsilon$ for $\forall n>N$. Let us just take the general p-norm in sequence spaces:
$$ \left\lVert e_n-a\right\rVert_p=(\sum_{i=1}^{\infty}\mid\delta_{in}-a_i \mid^p)^{1/p}<\epsilon \Rightarrow \mid\delta_{in}-a_i \mid^p<\epsilon, for \forall i $$
Then, for this to be arbitrarily small, when $i=n \Rightarrow a_i=1$, and when $i\ne n \Rightarrow a_i=0$. But this is exactly $a=a_i=(\delta _{in})_{n\in\mathbb{N}}$ and the set is closed.
2. Test of subset convergence in infinite dimension space
Now let us test compactness, namely, if $\forall$ sequence $(e_n)$ contains a converging subsequence $(e_{n_k})_{k\in\mathbb{N}}$.
A convergent sequence has to be Cauchy, which is to say that for $\forall \epsilon >0$ there $\exists N\in\mathbb{N}$ such that for $\forall m,n>N$:
$$\left\lVert e_n-e_m\right\rVert_p<\epsilon$$
But we know that the distance between any two elements is constant, namely $2^{1/p}$, since $n\ne m$, hence the set is not compact.
My problem here is the following: aren't the elements separated by the same amount even in the finite dimensional case (that is, $(e_n)_{n=1}^{N}$)? Then, by the before stated theorem, since the set $(e_n)$ is still bounded and closed even in the finite dimension, the set should be compact, yet the elements are a constant distance apart and hence cannot converge, which is a contradiction.
3. Generalization
This specific example got me thinking - what is that defining, the most general characteristic or property of finite dimensional spaces, which makes the bounded and closed sets always compact, but is missing in infinite dimensional case (because of which the bounded and closed sets are not necessarily always compact).
Is it the Bolzano-Weierstrass theorem, which states that only in the finite dimensions will the bounded sequences always contain converging subsequences?
I feel like I am missing something subtle here. Thanks for all the help and answers!
