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If a bounded sequence $\{a_n\}$ has no convergent subsequence, is it true that $|a_n−a_m|≥ \epsilon$ for some $\epsilon$ for all $n , m$?

If we do not require the sequence be bounded, then there are some counterexamples out there of sequence with no subsequence where the $\inf(a_n, a_m)=0$ but all those examples are unbounded sequences.

Edit: I should add that this is for general metric spaces. This is obviously true for $\Bbb R$ due to Bolzano.

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In the Euclidean space $\mathbb R^n$, any bounded sequence has a convergent subsequence. This is known as the Bolzano-Weierstrass theorem.

In this case, your conjecture is therefore vacuously true.

Your conjecture fails once your metric space possesses bounded sequences $(a_n)_{n\geq 0}$ with no convergent subsequence.

Indeed, in this case, consider the sequence $(b_{n})_{n\geq 0}$, given by $b_0=a_0$ and $b_{n+1}=a_n$ for $n\geq 0$. Then $(b_n)_{n\geq 0}$ is also bounded and has no convergent subsequence, yet $|b_1-b_0|=0$. Thus, your required condition cannot hold.

If you change your condition to be true for all $n\neq m$ starting from some rank $N$, the condition still fails in general: consider the sequence $(b_n)_{n\geq 0}$, given by $b_n:=a_{\lfloor n/2\rfloor}$.

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  • $\begingroup$ sorry I should have added I meant a general metric space and not just the reals. Is this still true in that case? I might also want to add the metric space is complete since the counterexamples are easy for incomplete spaces. $\endgroup$
    – Bill
    Dec 3, 2021 at 17:02
  • $\begingroup$ What if we have $n=m$? $\endgroup$
    – Gerd
    Dec 3, 2021 at 18:35
  • $\begingroup$ @Bill I added a counter-example. $\endgroup$
    – Zuy
    Dec 3, 2021 at 18:48
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Just another counterexample where $a_n \not= a_m$ $(n\not=m)$: Let $X=C([0,1],\mathbb{R})$ endowed with the maximum norm $\|\cdot\|$, and $a_n(t)=t^n$. Then $(a_n)$ has no convergent subsequence (each subsequence is pointwise convergent to the discontinuous function $b(t)=0$ $(t \in [0,1))$, $b(1)=1$). On the other hand $\|a_n-a_{n+1}\| \to 0$ $(n \to \infty)$. This can be seen by Dini's Theorem ($t^n-t^{n+1}=t^n(1-t)$ is pointwise decreasing to $0$) or directly by evaluating $$ \|a_n-a_{n+1}\|=\frac{(1-1/(n+1))^n}{n+1}. $$

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