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 $\inf\limits_{n\ne m} d(x_n,x_m)\ge r$.

  • $\begingroup$ and so it has not a convergent subsequence, so we can't cover E ? I don't understand how to conclude?? @Saucy $\endgroup$
    – Vrouvrou
    Apr 24 '19 at 20:53
  • $\begingroup$ Correct. "No Cauchiness" implies implies no convergence. $\endgroup$
    – avs
    Apr 24 '19 at 20:57
  • $\begingroup$ and then this is a contradiction or what ? I don't understand how to conclude @avs $\endgroup$
    – Vrouvrou
    Apr 24 '19 at 20:58
  • 1
    $\begingroup$ 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." $\endgroup$
    – avs
    Apr 24 '19 at 21:01
  • $\begingroup$ @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 $\endgroup$ Apr 25 '19 at 7:32

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