- #1

joshmccraney

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When proving a closed ball in ##L_1[0,1]## is not compact, I came across a proof, which states it is enough to prove that the space is not sequentially compact. Counter example: consider the sequence of functions ##g_n:x \mapsto x^n##. The sequence is bounded as for all ##n\in \mathbb N, \|g_n\|=1##. If ##(g_n)## would have a convergent subsequence, the subsequence would converge pointwise to the function equal to 0 on ##[0,1)## and to 1 at 1. As this function is not continuous, ##(g_n)## cannot have a subsequence converging to a map in ##L_1[0,1]##.

Why is it that because ##x=1## does not converge to the same thing as ##x\in[0,1)## this implies there are no subsequences converging to a map in this space?

Original proof here (second proof):

http://www.mathcounterexamples.net/a-non-compact-closed-ball/