I am working on a proof involving uniform continuity on functions with bounded domains. In particular $f:(a,b)\rightarrow\mathbb{R}$. I have an idea but it requires me to establish the following property.
Let $\{x_n\}$ be a bounded divergent sequence. Prove that there exists a convergent subsequence of $\{x_n\}$ call it $\{x_{n_k}\}$ such that $lim_{k\rightarrow\infty}[x_k-x_{n_k}]=0$
Can anybody find a proof or provide a counterexample?
This conjecture, as it stands, can be easily disproved. Consider $ a_n = (-1)^n $ It has 2 limit points, namely $1$ and $-1$. In other words, any convergent subsequence of $a_n$ must converge to either $1$ and $-1$. But that means that every subseqnece is eventually constant, ie. $$ \exists N \; \forall k>N a_{n_k} = 1 \vee a_{n_k} = -1 $$ But the whole sequence takes both values infinetely often,so the difference is always 1 somewhere.
Aside from the counterexample - if $$ \lim_{k\to\infty}(a_k - a_{n_k}) = a \; \wedge \; \lim_{k\to\infty} a_{n_k} = b $$ then $$ \lim_{k\to\infty} a_k = \lim_{k\to\infty} (a_k - a_{n_k} + a_{n_k}) = \lim_{k\to\infty} (a_k - a_{n_k}) + \lim_{k\to\infty} a_{n_k} = a+b $$ so the whole seqence would be also convergent, contradicting the assumption.