Showing the unit ball in $l^{\infty}$ is not compact.

Asked Feb 24 '14 at 14:57

Active Feb 24 '14 at 14:57

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If $l^{\infty}$ is the set of bounded sequences of real numbers with norm $||x||_{\infty}$.

To do this I have tried to use the fact that a metric space is compact iff it is sequentially compact. So now I am trying to find a sequence with no convergent subsequence, perhaps by choosing it so that $||x_m-x_n|| > 1/2$ (so that it is not cauchy), however I am struggling to see what is actually going on and how to pick a sequence with this property.

Thanks

asked Feb 24 '14 at 14:57

Wooster

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Let $x_n$ be $1$ in its $n$'th coordinate and $0$ in the others. – David Mitra Feb 24 '14 at 15:00
Doesn't that mean that $||x_m - x_n|| = 0$ for any m,n? – Wooster Feb 24 '14 at 15:04
1

No, for $n\ne m$, $x_n-x_m$ has exactly two non-zero coordinates, the $n$'th with value $1$ and the $m$'th with value $-1$. – David Mitra Feb 24 '14 at 15:13
So in this case, we can't find a convergent subsequence because the sequence itself isn't convergent? – man_in_green_shirt Sep 21 '15 at 16:54

Showing the unit ball in $l^{\infty}$ is not compact.

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