Consider a complete metric space E with the following property:
If $x_n$ is a bounded sequence, then $\forall \epsilon > 0$, $\exists i,j , i \neq j$ such that $d(x_i,x_j) < \epsilon$.
Question:
Given any such bounded sequence in $E$, can we always find a convergent subsequence ?
I am unable to construct a convergent subsequence for an arbitrary bounded sequence in E - the given property seems too weak! Am I missing something?
Let $(x_n)_{n=1}^\infty$ be a bounded sequence in $E$. We claim that $X := \{x_n\}_{n=1}^\infty$ is totally bounded. Indeed, suppose this were not the case. Then there is some $\epsilon >0$ with the following property:
We construct a subsequence $(x_{n_k})_{k=1}^\infty$ as follows. Put $n_1 :=1$. Suppose $n_1<...< n_{k-1}$ were obtained so that $d(x_{n_i},x_{n_j})\geqq \epsilon$ whenever $1\leqq i,j <k$, $i\neq j$. Since we assumed $X$ is not totally bounded, the balls
$$B(x_{n_1};\epsilon),...,B(x_{n_{k-1}};\epsilon)$$
must fail to cover $X$, and there must be an infinite amount of points of $X$ outside their union. Let $n_k>n_{k-1}$ be so that
$$x_{n_k} \not\in B(x_{n_1};\epsilon)\cup...\cup B(x_{n_{k-1}};\epsilon).$$
Thus, the bounded sequence $(x_{n_k})_{k=1}^\infty$ is such that $d(x_{n_i},x_{n_j})\geqq \epsilon$ whenever $i \neq j$, and the stated property of $E$ fails.
It follows that if $E$ has the property, $X$ must be totally bounded. Since $E$ is complete, $X$ must be relatively compact (its closure must be compact). Hence, the sequence $(x_n)_{n=1}^\infty$ in $X$ has a convergent subsequence.