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Let $A \subset \ell^p$, where $1 \le p \lt \infty$. Suppose the following conditions are true:

1) $A$ is closed and bounded

2) $\forall \epsilon \gt 0, \: \exists \: N \in \mathbb{N}$ such that $\forall x \in A$, we have $\sum_{n \ge N}|x_{n}|^{p} \lt \epsilon$.

Then show that $A$ is compact. Also, show that the converse is true.

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Any closed subset of a complete space is complete. Also, any complete and totally bounded subset is compact. So would it suffice to show that A is totally bounded? –  Heisenberg Oct 22 '12 at 2:36
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Yes that would suffice. –  Isaac Solomon Oct 22 '12 at 2:51
    
Thank you. How would I prove the converse, i.e, if A is compact, then the two conditions hold? –  Heisenberg Oct 22 '12 at 3:13
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up vote 4 down vote accepted

The properties 1) and 2) are quite obvious when $A$ is a finite set. So we see that a compact set "almost behaves as a finite set". Here is a formal proof.

  • As $\ell^p$ is complete, $A$ is complete whenever it's closed. So we just have to show pre-compactness. Fix $\varepsilon>0$, and apply 2) with $\varepsilon/2$. This gives an integer $N$ such that for all $x\in A$, $\sum_{j\geq N}|x_j|^p<\varepsilon/2$. As $A$ is bounded, we can find $M>0$ such that $\lVert x\rVert\leq M$ for all $x\in A$. Since $[-M,M]^N$ is pre-compact, we can find an integer $K$ and sequences $x^{(0)},\dots,x^{(K)}$ such that for all $v\in [-M,M]^n$, there exists $i\in \{0,\dots,J\}$ such that $\sum_{j=0}^{N-1}|v_j-x^{(i)}_j|^p\leq \frac{\varepsilon^p}{2^p}$. Define $y^{(j)}:=(y^{(j)}_0,\dots,y^{(j)}_{N_1},0,\ldots,0)\in \ell^p$ to see that $A$ is pre-compact.

  • Conversely, we assume $A$ compact. A compact subset of a Hausdorff space is closed. It's bounded, as we can extract from the open cover $\{B(x,1)\}_{x\in A}$ a finite subcover $\{B(x,1)\}_{x\in F}$, where $F$ is finite. Then for all $x\in A$, $\lVert x\rVert\leq 1+\max_{y\in F}\lVert y\rVert$. Fix $\varepsilon>0$, then by pre-compactness, we can find an integer $K$ and $x^{(1)},\dots,x^{(K)}$ such that $\bigcup_{j=1}^KB(x^{(j)},\varepsilon/2)\supset A$. For each $j\leq K$, take $N_j$ such that $\sum_{i\geq N_j}|x_i^{(j)}|^p<\frac{\varepsilon^p}{2^p}$. Then take $N:=\max_{1\leq j\leq K}N_k$.

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Why the downvote? –  Davide Giraudo Oct 25 '12 at 20:19
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