# 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 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.

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 $$b_n:=a_{\lfloor n/2\rfloor}$$.

• 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.
– Bill
Dec 3, 2021 at 17:02
• What if we have $n=m$?
– Gerd
Dec 3, 2021 at 18:35
• @Bill I added a counter-example.
– Zuy
Dec 3, 2021 at 18:48

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 $$\|a_n-a_{n+1}\|=\frac{(1-1/(n+1))^n}{n+1}.$$