Prove this set is compact

Question

Let $\{a_{n}\}_{n \in \mathbb{N}}$ be a sequence with the property that $\{a_{n}\}$ converges to $0$ when $n \rightarrow \infty$. Now let's consider this set: $$K=\{\{x_{n}\}_{n\in\mathbb{N}} \in l_{\infty} : |x_{n}|\leq|a_{n}| \text{ } \forall \text{ } n\}$$ Prove that $K$ is compact in $(l_{\infty},d_{\infty})$.

Where $l_{\infty}=\{\{x_{n}\}_{n\in\mathbb{N}} \subseteq \mathbb{R} : \{x_{n}\}_{n\in\mathbb{N}} \text{ is bounded } \}$, and $d_{\infty}(\{x_{n}\}_{n\in\mathbb{N}},\{y_{n}\}_{n\in\mathbb{N}})=sup_{n\in\mathbb{N}}|x_n-y_n|$

I think I should try to prove that K is complete and totally bounded.

a) Using that $(l_{\infty},d_{\infty})$ is complete, I will prove K is a closed subset of $(l_{\infty},d_{\infty})$ so it's complete. Let's consider a sequence $\{x^{k}\}_{k\in\mathbb{N}}$ in $K$ that it converges to the sequence $u$. Let's prove $u \in K$.

Let $\epsilon_l=1/l>0$ ($l \in \mathbb{N})$ so there is a $k_{0l}\in \mathbb{N}$ such $|x_{n}^{k}-u_n|\leq d_{\infty}(\{x^{k}\}_{k\in\mathbb{N}},u)<1/l$ for all $k \geq k_{0l} $, for all $n \in \mathbb{N}$

So: $|u_n|\leq|x_{n}^{k_{ol}}-u_n|+|x_{n}^{k_{ol}}|<1/l+|u_n|.$ So when $l\rightarrow \infty$ we have $|u_n|\leq |a_n|$ for all $n \in \mathbb{N}$ so $u\in K$ so $K$ is closed and then complete.

Is this correct?

b) When i want to show that $K$ is totally bounded, the only thing that comes to mind is the next: let's fix $b\in K$. So if $y\in K$ then given $\epsilon>0$ there is a $n_0 \in \mathbb{N}$ such $|y_n-b_n|\leq2|a_n|\leq\epsilon$ for all $n \geq n_{0}$. And this $n_{0}$ is only a function of $\epsilon$ and it doesn't depend on the sequence $y$. But I don't know what to do with the finite terms of any other sequence in K.

Thanks a lot for reading!!

score 2 · Accepted Answer · answered Dec 10 '13 at 21:12

2

Fix $\epsilon>0$. There is an $m\in\Bbb N$ such that $|a_n|<\epsilon$ for all $n\ge m$. Let $K_0=\prod_{n<m}[-a_n,a_n]$. $K_0$ is a compact subset of $\Bbb R^m$, so there is a finite $F_0\subseteq K_0$ such that every point of $K_0$ is less then $\epsilon$ from some point of $F_0$ in the $\sup$ metric. Let $F$ be the subset of $K$ obtained by extending each point of $F_0$ with zeroes. Show that every point of $K$ is within $\epsilon$ of some point of $F$.

answered Dec 10 '13 at 21:12

Brian M. Scott

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Thanks a lot for your help! – Maxi Dec 10 '13 at 22:02
@Maxi: You’re welcome! – Brian M. Scott Dec 10 '13 at 22:04
Sorry Brian to bother you. I notice you answer this post http://math.stackexchange.com/questions/367194/cardinality-of-the-set-of-bijective-functions-on-mathbbn. But I can't comment there because I don't have the reputation. I don't understand where the set P is. P is in $\mathbb{N} x \mathbb{N}$ ? or in the power set of $\mathbb{N}$? or somewhere else? Thanks in advance. – Maxi Dec 10 '13 at 22:43
@Maxi: Each member of $P$ is a set ${2n,2n+1}$ for some $n\in\Bbb N$, so each member of $P$ is a subset of $\Bbb N$, and $P\subseteq\wp(\Bbb N)$. – Brian M. Scott Dec 10 '13 at 22:46
Thank you. Now I understand! – Maxi Dec 10 '13 at 22:57
@Maxi: You’re welcome! – Brian M. Scott Dec 10 '13 at 22:58

Prove this set is compact

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