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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!

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Your proof of closure is confusingly written. You are using $e_n$ to denote both an arbitrary element of your set and an element of the sequence. You only really proved that in order for $a$ to be arbitrarily close to a particular element of the set, it must be equal to that element, which is, of course, true, but doesn't yet prove anything. $0$ is not arbitrarily close to any element of the sequence $\frac1n$, however it is there "in the limit".

I would change the proof to state that the only way a sequence of points from the set converges is if it is constant from some point on.


Your proof of non-compactness also needs more thought. The simple fact that the distance between any pair of elements of your set is $2^{\frac1p}$ that causes the set to be non-compact. The set $\{0,1\}\subseteq \mathbb R$ is compact, even though it has a similar property, and, as you noticed, finite dimensional bases also have this property.

The thing is that I can construct a sequence, in particular, $$e_1,e_2,e_3,\dots, e_n,\dots$$ for which, no matter what subsequence I take, any pair of elements of the subsequence will also be at a distance of $2^{\frac1p}$. This is a property not shared by finite dimensional bases, where I will inevitably have to repeat some basis vectors infinitely many times, and I can take the subsequence consisting of just that one infinitely repeated vector - and clearly, the subsequence will converge.

This should now let you see the error in your saying:

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.

What you are doing is assuming that all the elements are different, which means you are implicitly assuming infinitely many basis elements - a property explicitly prohibited in finite dimensions.

Note: Even in infinite dimension, it is possible to construct a sequence that does converge, by repeating an element infinitely many times. The catch is that in finite dimension, you will always have a convergent subsequence, but in infinite dimensions, you might not.

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  • $\begingroup$ Again, your explanation helped me majorly. It is so clear now. The only thing that I still have to think a little bit about is the closedness proof, but other than that, thank you kindly for your answer! $\endgroup$ – Nejc Kejzar Sep 25 '18 at 11:39
  • $\begingroup$ @NejcKejzar I think you have a good idea with the closedness proof, you just need to word it better. Pa srečno :) $\endgroup$ – 5xum Sep 25 '18 at 11:45
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In the finite dimensional case $\{e_n\}$ is a finite set. So any subsequence will have some $e_k$ repeated infinite number of times and there exists a constant subsequence. Of course, constant sequences are convergent. Bolzano Wierstrass Theorem holds in a normed linear space iff the space is finite dimensional, so this is a characterization of finite dimensional spaces.

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  • $\begingroup$ This is a beautiful argument (and indeed, rather subtle). With merely 3 sentences you managed to blow my confusion away. Thank you kindly! $\endgroup$ – Nejc Kejzar Sep 25 '18 at 8:05
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Let me answer this question:

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).

The defining property is whether the space is finite or infinite-dimensional. In fact, a normed space is finite-dimensional if and only if all bounded and closed sets are compact.

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