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I would be grateful if you can give me some hints for the following homework problem.

Let $C$ be a subset of $l^\infty$ (with uniform norm) such that $C = \left\{(x_n) \mid |x_n|\leq \frac1n \,\forall n\geq 1\right\}$ Is $C$ a compact set or not?

Honestly, I am stuck. I tried to use sequentially compactness and attempted to construct different sequences which may serve as counter examples but all of them failed to do so.

I also thought that if $y_n$ is a constant sequence equal to $0$ and if we take the open ball $B(y_n,2)$ with radius $2$, does that count as a finite open cover of $C$?

Thanks in advance.

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    As to your last comment, remember that isn't how compactness works: to show a set is compact, you have to show that *every* open cover whatsoever has a finite subcover. You don't get to pick the cover you start with. Yes, $B(0,2)$ is an open cover of $C$, but that doesn't help with this problem (except that it shows $C$ is bounded, which is useful to know).2011-12-01

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Since C is normed, it is compact iff it is sequentially compact. Let $\{x_n\}\subset C$ be a sequence, then by the Cantor diagonal trick, it has a pointwise convergent subsequence \{x'_n\}, i.e. x'_n(i)\to y(i) as $n\to\infty$ and every $i$. Clearly, $y\in C$, and for any $\epsilon>0$, choose $N>2/\epsilon$. Then, for all $i, |x'_n(i)-y(i)|<\epsilon for all $n$ large enough, while for $i\ge N$ we have |x'_n(i)-y(i)|\le 1/N+1/N <\epsilon. Thus \{x'_n\} converges in norm, and thus C is sequentially compact.

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    +1. The "Cantor diagonal trick" can also be thought of as a special case of the Banach-Alaoglu theorem (giving compactness in the weak-* topology, viewing $\ell^\infty$ as the dual of $\ell^1$) or the Tychonoff theorem (giving compactness in the product topology). For bounded subsets of $\ell^\infty$ these all coincide with the topology of pointwise convergence, which is first countable, and so you have a pointwise convergent subsequence. Then you can use the definition of $C$ to see that the pointwise convergence is actually uniform.2011-12-01
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First, we show that $C$ is closed. If $x=\{x_n\}\notin C$ then we can find $N$ such that $|x_N|>N^{-1}$, and the ball of cented $x$ and radius $\frac{|x_N|-N^{-1}}2$ is contained in the complement of $C$. Now, we use the result that a pre-compact subset in a complete space has a compact closure (hence here $C$ will be compact).

Fix $\varepsilon>0$ and $N$ such that $N^{-1}\leq \varepsilon$. Since the product $\prod_{j=1}^{N-1}\left[-j^{-1},j^{-1}\right]$ is precompact, we can find an integer $p$ and $x^{1},\ldots,x^{p}\in \prod_{j=1}^{N-1}\left[-j^{-1},j^{-1}\right]$ such that for each $v\in \prod_{j=1}^{N-1}\left[-j^{-1},j^{-1}\right]$, we can find $1\leq i\leq p$ such that $\max_{1\leq j\leq p}|v_j-x_j^i|\leq \varepsilon.$ Now put for $1\leq j\leq p$, $y^j$ the sequence defined by $y^j_k=\begin{cases}x_k^j&\mbox{if }k For all $x\in C$, we can find $1\leq i\leq p$ such that $\lVert x-y_i\rVert\leq \varepsilon$, and you can conclude.

A little more generally, you can show that if $\{a_n\}$ is a sequence which converges to $0$ then the set $C:=\left\{\{x_n\}\mid\forall n, |x_n|\leq a_n\right\}\subset \ell^{\infty}$ is compact.