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This problem is from our analysis midterm exam.

Let $X$ be the set of all sequences $\{a_n\}$ with $\sup_n|a_n|\leq 1$, and the metric on $X$ is given as $$d(a,b)=\sup_{n}\left|\frac{a_n-b_n}{n}\right|$$ where $a=\{a_n\}$ and $b=\{b_n\}.$ Using the open cover definition of compactness, prove that $X$ is compact under this metric.

I already know that this metric induces the product topology of $[-1,1]^\mathbb{N}$, and so the compactness follows from Tychonoff's theorem. However, this problem requires me to prove the compactness using the open cover directly(since this is an undergraduate analysis exam). How should I do?

Any hints and advice are welcome!

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Here is a proof using sequences:

Let $A_s=(\{a_{n,s}\})_{s\in \Bbb N}$ be any sequence in $X .$

Take a sub-sequence $S_1$ of $\Bbb N$ such that the sequence $(a_{1,s})_{s\in S_1}$ converges to some $b_1\in [-1,1].$ Take $m_1\in S_1.$ Recursively,for $k\in \Bbb N$ take $S_{k+1}$ to be a sub-sequence of $S_k$ such that $(a_{n+1,s})_{s\in S_{k+1}}$ converges to $b_{k+1}.$ Then take $m_{k+1}\in S_{k+1}$ such that $m_{k+1}>m_k.$

Now $(m_k)_{k\ge n}$ is a sub-sequence of $S_n$ for each $n.$ So $(a_{n,m_k})_{k\in \Bbb N}$ converges to $b_n$ for each $n.$ Let $B=\{b_n\}\in X.$

For $\epsilon >0$ take $m\in \Bbb N$ with $1/m</2\epsilon,$ and then take $k^*\in \Bbb N,$ large enough that $\forall n\le m \,\forall k\ge k^*\,(|a_{n,m_k}-b_n|<\epsilon).$ Then $$\forall k\ge k^* \,(d(A_{m_k}, B)<\epsilon).$$

So $X$ is sequentially compact. For a metric space this implies being compact.

APPENDIX. Let $(X,d)$ be any sequentially compact metric space. Let $C$ be an open cover of $X$. For each $n\in \Bbb N$ let $D_n$ be the set of open balls $b$ of radius $1/n$ such that $b\subset c$ for some $c\in C. $

(i-a). If some $D_n$ is a cover of $X$ and $D_n$ has a finite sub-cover $E$ then for each $e\in E$ let $e\subset f(e)\in C.$ Then $\{f(e):e\in E\}$ is a finite subset of $C$ and is a cover of $X$.

(i-b). If some $D_n$ is a cover of $X$ but $D_n$ has no finite sub-cover then for $j\in \Bbb N$ let $b_j=B_d(x_j,1/n)\in D_n$ such that $x_j\not \in \cup_{k<j} b_k.$ Then $d(b_k,b_j)\ge 1/n$ for all distinct $j,k\in \Bbb N$ so the sequence $(b_j)_{j\in \Bbb N}$ has no convergent sub-sequence, a contradiction.

(ii). If no $D_n$ is a cover of $X,$ let $x_n\in X$ \ $\cup D_n$ and let $(x_{n_i})_{i\in \Bbb N}$ be a convergent sub-sequence of $(x_n)_{n\in \Bbb N},$ with limit point $x.$ There exists $c\in C$ and $m\in \Bbb N$ such that $B_d(x,1/m)\subset c$ and there exists $i\in \Bbb N$ such that $n_i>2m$ and $d(x,x_{n_i})<1/2m.$ But then $B_d(x_{n_i},1/n_i)\subset B_d(x,1/m)\subset c\in C,$ implying $x_{n_i}\in \cup D_{n_i},$ a contradiction.

Therefore a sequentially compact metric space is compact.

Remarks.(1). From the Appendix we also conclude that if $(X,d)$ is a compact metric space and $C$ is an open cover of $X$ then for some $n$ and some finite $E\subset D_n$ we have $X=\cup E,$ which is the Lebesgue Covering Lemma.

(2). The $\in$-order topology on the cardinal ordinal $\omega_1$ is sequentially compact but not compact, but this space is not metrizable.

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It's easy to construct a continuous surjection from $2^{\mathbb N}=\{0,1\}^{\mathbb N}$ to $[0,1]$. (Hint: $ [0,1]=\{\sum_{k \in N} \frac{b_{k}}{2^{k}} \text { where } b_{k} \in\{0,1\}\} $ ). Then there is a continuous surjection from $(2^\mathbb N)^{\mathbb N}$ to $[0,1]^\mathbb N$. Notice $(2^\mathbb N)^{\mathbb N}$ is homeomorphic to $2^{\mathbb N \times\mathbb N}$, which is homeomorphic to $2^{\mathbb N}$ (as there is a bijection from $\mathbb N \times\mathbb N$ to ${\mathbb N}$). Now it suffices to prove $2^\mathbb N$ is compact, as the continuous image of a compact set is compact. Let $C=\{\sum_{k \in N} \frac{a_{k}}{3^{k}}, \text { where } a_{k} \in\{0,2\}\}$. Again, it's easy to constuct a homeomorphism from $C$ to $2^\mathbb N$. But what is $C$? It's the Cantor set, being closed and bounded, is compact. Thus $2^\mathbb N$ is compact. Hence $[0,1]^\mathbb N$, being the continuous image of a compact set, is compact.


Remark: @DanielWainfleet proof is easier (+1!). But my proof doesn't use the axiom of choice. So I think it's worth leaving it here.

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  • $\begingroup$ I'm afraid the fact that $(2^{\mathbb{N}})^\mathbb{N}$ is homeomorphic to $2^{\mathbb{N}\times\mathbb{N}}$ does not appear in the analysis course. $\endgroup$
    – bellcircle
    Apr 15 '19 at 11:18
  • $\begingroup$ @bellcircle It's easily proven by definition $\endgroup$ Apr 15 '19 at 11:22

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