Compactness of closed unit ball

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SUMMARY

The closed unit ball of the space l∞, defined as B(0, 1) = {x ∈ l∞ ; ∥x∥∞ ≤ 1}, is proven to be non-compact. This conclusion is reached by demonstrating that there exists a sequence within B(0, 1) that lacks a convergent subsequence, specifically by considering sequences of the form (1/n). The discussion emphasizes the importance of recognizing that elements of l∞ are sequences themselves, which must be taken into account when constructing examples to illustrate non-compactness.

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Homework Statement


Let l∞ be the space of bounded sequences of real numbers, endowed with the norm
∥x∥∞ = supn∈N |xn | , where x = (xn )n∈N .
Prove that the closed unit ball of l∞ , B(0, 1) = {x ∈ l∞ ; ∥x∥∞ ≤ 1} , is not compact.

Homework Equations





The Attempt at a Solution


I'm thinking about using the notion of sequential compactness, since every sequence Xn has an upper limit here, but I'm not sure if that would help much. Could anyone please give me a hint? Any input is appreciated!
 
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Sure, use the sequential form. Find a sequence in B(0,1) with no convergent subsequence.
 
thanks for the reply, but I'm still not sure about which non-convergent sequence to choose, would something like 1/n work? I just don't know how to use the complete norm here.
 
rainwyz0706 said:
thanks for the reply, but I'm still not sure about which non-convergent sequence to choose, would something like 1/n work? I just don't know how to use the complete norm here.

Remember what the space is here. It's the set of bounded sequences. So a point in your space is itself a sequence. Thus a sequence in your space must in fact be a sequence of sequences. Give me some examples.
 

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