Compactness/convergence in Banach spaces

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Been doing exercises on compactness/sequential compactness of objects in Banach spaces and some of my solutions come down to whether
"every bounded sequence has a convergent subsequence"
holds in

a) arbitrary finite-dimensional Banach space
b) lp, 1 <= p <= infinity

Does it?
 
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No, this does not hold. Consider the sequence in \ell^2:

x_1=(1,0,0,0,...)
x_2=(0,1,0,0,...)
x_3=(0,0,1,0,...)
and so on.
 
Doesn't hold for finite-dimensional Banach space as well?
 
TaPaKaH said:
Doesn't hold for finite-dimensional Banach space as well?

It does hold for finite dimensional spaces since those are isomorphic to \mathbb{R}^n.
 
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Thank you!
 
In general, in metrizable spaces, compactness and sequential compactness are

equivalent. The unit ball is compact/seq. compact in a normed v.space V iff

V is finite-dimensional.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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