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