Does the compact subset of an infinite Banach have finite span?

[Fractal20]

Homework Statement

Hi all, I am struggling with getting an intuitive understanding of linear normed spaces, particularly of the infinite variety. In turn, I then am having trouble with compactness. To try and get specific I have two questions.

Question 1
In a linear normed vector space, is it true that limit point compactness, compactness, and sequential compactness are all equivalent? My understanding is the norm creates a metric space and these should all be true. But I had an earlier confusion that perhaps compact spaces of the infinite variety need not be bounded. So I'm unsure about this point.

Question 2
In this same direction, would a compact subset of an infinite banach space have to have a finite span? We proved in class that B* space is finite dimensional iff the unit sphere is compact so this is what makes me assume this would be true. It seems like if not, then we could create an infinite sequence of orthogonal vectors and it doesn't seem like this sequence would not have to have a convergent subsequence.

Homework Equations

The Attempt at a Solution

 

[Dick]

That all sounds right. Except I don't know in what sense you mean a subset has a 'span'. If a subset is a span of a set of vectors, then that subset is a subspace and subspaces aren't compact, right?

 

[Fractal20]

So would it be correct to say that a compact subset of a normed linear space (B* or Banach) must have a finite basis or has a finite dimension?

Thanks a lot!

 

[Dick]

So would it be correct to say that a compact subset of a normed linear space (B* or Banach) must have a finite basis or has a finite dimension?

Thanks a lot!

You are missing my point. Subsets don't have bases. Subspaces have bases. The unit sphere doesn't have a basis.

 

[Fractal20]

Hmm, can't all points in the unit sphere in say 3d be written as linear combinations of (1,0,0), (0,1,0) and (0,0,1)? Is your point about semantics or is it really erroneous to carry over these concepts to a general space? I thought that linear spaces have the same... I don't what to call it, properties/characteristics as vector spaces?

Okay, thinking about it I understand the differences in terms of things like closure. As in, a linear subspace should have additive closure etc, but if it is a subspace or a subset, can we still talk about it having some dimension? Thanks, I appreciate the help.

 

[Dick]

Hmm, can't all points in the unit sphere in say 3d be written as linear combinations of (1,0,0), (0,1,0) and (0,0,1)? Is your point about semantics or is it really erroneous to carry over these concepts to a general space? I thought that linear spaces have the same... I don't what to call it, properties/characteristics as vector spaces?

Okay, thinking about it I understand the differences in terms of things like closure. As in, a linear subspace should have additive closure etc, but if it is a subspace or a subset, can we still talk about it having some dimension? Thanks, I appreciate the help.

For a subset to have a 'dimension' it has to be similar in some way to a linear space that has a dimension. The concept you are looking for is 'manifold', I think. Look it up. Your 3d sphere is like a plane, except it's bent. Hence it's a two dimensional manifold. And, yes, I think in that sense you are right, that a manifold can only be compact if it's finite dimensional.

 

[micromass]

Question 2 is false. A compact subset of a Banach space doesn't need to have a finite-dimensional span.

An important counterexample for the statement is the Hilbert cube. Take in $\ell^2$ the subset

$$\{(x_n)_n\in \ell^2~\vert~0\leq x_n\leq 1/n\}$$

This can be shown to be compact. Yet, its span is entire $\ell^2$.

The statement that a compact manifold is finite-dimensional does seem correct. If a manifold is compact, then its coordinate charts are open sets of the manifolds which are then locally compact. But a locally compact Banach space is finite-dimensional.