# Sequentially compact metric space is totally bounded.

I want to prove this: " If for any sequence $$(x_n)$$ from a metric space $$(E,d)$$ we can extract a convergent subsequence then for any $$r>0$$, we can cover $$E$$ by a finite number of open balls of radius $$r$$"

I star by the construction of the sequence

Let $$x_0\in E$$.

If $$E=B(x_0,r)$$ then we are done.

If not, there exists an $$x_1\in E$$ such that $$x_1\notin B(x_0,r)$$. If $$E=B(x_0,r)\cup B(x_1,r)$$ we are done.

If not .... there exists $$x_n\in E$$ but $$x_n\notin B(x_0,r)\cup \ldots\cup B(x_{n-1},r)$$

Then there exists a sequence $$(x_n)\in E$$ such that $$d(x_n,x_{n-1})>r,\, \forall n\in \mathbb{N}^*$$, but how to continue ?

thank you

The constructive condition $$x_n\notin \bigcup\limits_{i=0}^{n-1} B(x_i,r)$$ doesn't just imply that $$d(x_n,x_{n-1})\ge r$$ for all $$n>0$$, but moreso that $$d(x_n,x_m)\ge r$$ for all $$n>m$$ (and thus for all $$n\ne m$$). Therefore, this sequence has no Cauchy subsequences, because $$\inf\limits_{n\ne m} d(x_n,x_m)\ge r$$.
• It's the contrapositive of what you are trying to prove: "If we can't cover $E$ with finitely many $r$-balls, then not every sequence will have a convergent subsequence."
• @Vrouvrou. If $S=(x_{n(i)})_{i\in \Bbb N}$ is any sub-sequence of $(x_n)_{n\in \Bbb N}$ then $S$ is not Cauchy (because $i\ne j\implies d(x_{n(i)},x_{n(j)})> r)$ so $S$ certainly cannot be convergent Apr 25 '19 at 7:32