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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.
Maybe the question needs editing. At least clarify if the sequence is diverging in the sense that $x_n$ goes to (positive or negative) infinity or if its series is divergent in the sense that the partial sums diverges. Both cases are also different from oscillating, when the values for both the sequence and the series may be bounded, but never approach a constant.
If $x_k = (k \,mod \,3)-1$, then the sum oscillates between -1 and 0, the subsequence grabbing all zeros is convergent, but $lim_{k→∞}[xk−xnk]=lim_{k→∞}[xk] $ which does not exist.
