# Proving the compactness using open cover definition

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!

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 $$1/m and then take $$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 $$x_j\not \in \cup_{k Then $$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.

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.

• 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. Apr 15 '19 at 11:18
• @bellcircle It's easily proven by definition Apr 15 '19 at 11:22