Is the Closure of a Subset in l^{1} Compact?

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SUMMARY

The discussion centers on the compactness of the closure of the subset S in the Banach space l^{1}, defined as S={x ∈ l^{1} | ||x|| < 1}. It is established that S is not closed, leading to the inquiry about the compactness of its closure. The conclusion drawn is that the closure of S is not compact, as it fails to meet the criteria for compactness in the context of l^{1} space.

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  • Understanding of Banach spaces, specifically l^{1} space.
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  • Familiarity with closure properties in metric spaces.
  • Basic concepts of functional analysis.
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Homework Statement


Consider the Banach Space l^{1}. Let S={x \in l^{1}|\left\|x\right\|&lt;1}. Is S a compact subset of l^{1}? prove or Disprove.
 
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S isn't even closed. A more interesting problem is whether the closure of S is compact, and I suspect this is what you're supposed to work on.
 

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