# If a diverging sequence has a convergent subsequence, does it always have another subsequence that converges to a different limit?

I have a question about subsequences of a sequence in a metric space. let $$X$$ be a metric space and $$(x_n)$$ be a diverging sequence in $$X$$. Suppose $$(x_n)$$ has a subsequence with a limit $$a \in X.$$ Does it hold true that there exists a subsequence of $$(x_n)$$ such that it has a limit $$b \in X$$, $$b \neq a$$?

• No, unless $X$ is compact. For example, the sequence $0,1,0,2,0,3,0,4,\dots$ has only one limit point $0\in\mathbb{R}$ (but of course it has the other limit point $+\infty\in\bar{\mathbb{R}}$ that is not in $\mathbb{R}$). – user10354138 Apr 3 at 10:51

That assertion holds if and only if $$X$$ is compact. In fact:
• If $$X$$ is compact, then every sequence has a convergent subsequence. So, every divergent sequence has a convergent subsequence.
• If $$X$$ is not compact, then it has a sequence $$(x_n)_{n\in\Bbb N}$$ with no convergent sequence. So, if $$x\in X$$, the sequence$$x,x_1,x,x_2,x,x_3,\ldots$$has a convergent subsequence, but it has no subsequence that converges to a different limit.