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Let $l^p$ be the space of $p$-summable sequences. von Neumann constructed a subset of $l^p$ space $$S=\{X_{mn}: m,n≥1\}$$ where $X_{mn}\in l^p$ are defined by $X_{mn}(m)=1, X_{mn}(n)=m$ and $X_{mn}(k)=0$ otherwise.

I am asked to show that this $S$ is closed in the strong topology. I tried to show the complement is open by trying to construct a contradiction, but no success. Could anyone help me ? Thanks in advance.

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Calculate the norm $\| X_{mn} - X_{m'n'} \|$. When can this tend to $0$? What can a Cauchy sequence in $S$ look like? –  Tom Cooney Oct 30 '12 at 10:32
    
@Norbert : why did you delete your answer? –  Davide Giraudo Nov 27 '12 at 23:13

1 Answer 1

Hints:

  1. Prove the following general claim. The set $S$ in metric space $M$ with the property $$ \exists C>0\quad \forall x'\in S\quad\forall x''\in S\quad (x'\neq x''\implies d(x',x')>C)\tag{1} $$ is always closed.

  2. Prove that the set $S$ in your problem satisfy this condition.

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I dont know how to prove 1 . But following the comment , any cauchy sequnce in such a space must eventually be constant and thus converges in S. The space should be complete and thus closed . Just wondering how to show 1 by topological argument ? –  needhelp Oct 30 '12 at 15:12
    
@needhelp, Consider $T:=M\setminus T$ - the complement of $S$. Take arbitrary $y\in T$, and prove that $B(y, C/3)$ contains at most one point of $S$, say $x$. Define $r=\min(d(x,y), C/3)$, then $B(y,r/2)$ doesn't contain points of $S$ (i.e. $B(y,r/2)\subset T$). Since $y$ is arbitrary $T$ is open, hence $S$ is closed. –  Norbert Oct 30 '12 at 15:26

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