# If every convergent subsequence converges to the same limit, does the whole sequence converge to that limit?

$$(E,d)$$ be a metric space and $$(x_n)_{n\in\mathbb N}\subseteq E$$ be relatively compact.

Assume that there is a $$x\in E$$ such that every convergent subsequence converges to $$x$$. Does it already follow that $$(x_n)_{n\in\mathbb N}$$ is convergent and converges to $$x$$?

The convergence of $$(x_n)_{n\in\mathbb N}$$ to $$x$$ is equivalent to the assertion that every aribtrary subsequence is convergent and converges to $$x$$; but we only know that every subsequence, for which we a priori assume that it is convergent, converges to $$x$$.

Let the original sequence be $$\sigma$$, and let $$K=\operatorname{cl}\{x_n:n\in\Bbb N\}$$; by hypothesis $$K$$ is compact. If $$\sigma$$ does not converge to $$x$$, there are an $$\epsilon>0$$ and a subsequence $$\sigma_1=\langle x_{n_k}:k\in\Bbb N\rangle$$ such that $$d(x_{n_k},x)\ge\epsilon$$ for all $$k\in\Bbb N$$. Clearly $$\sigma_1$$ is a sequence in the compact set $$K$$, so it has a convergent subsequence $$\sigma_2=\langle x_{n_{k_i}}:i\in\Bbb N\rangle$$. But $$\sigma_2$$ is also a subsequence of $$\sigma$$, so by hypothesis it converges to $$x$$, contradicting the fact that $$d(x_{n_{k_i}},x)\ge \epsilon$$ for each $$i\in\Bbb N$$. Thus, $$\sigma$$ must converge to $$x$$.

• I was thinking of counterexamples in $l^2(\Bbb R)$ – weierstrash Dec 27 '20 at 6:13

What about the following example in the interval (0,2): $$1,\frac{1}{2},1,\frac{1}{3},1,\frac{1}{4},1,\frac{1}{5},1,\frac{1}{6},1,\frac{1}{7},\dots$$

Every convergent subsequence converges to 1, but the entire sequence does not converge

• the sequence set is not compact. – weierstrash Dec 27 '20 at 5:59
• right, fixed (if I understand the correctly the definition of relatively compact, then (0,2) is relatively compact in [0,2] ) – Igor Shinkar Dec 27 '20 at 6:06
• it needs to be only relatively compact, i.e. its closure needs to be compact. en.wikipedia.org/wiki/Relatively_compact_subspace – Igor Shinkar Dec 27 '20 at 6:08
• the subsequence ${\frac{1}{n}}$ converges to $0$ – weierstrash Dec 27 '20 at 6:08
• but 0 is not in the (0,2) – Igor Shinkar Dec 27 '20 at 6:08