Prove that the complement of the range is an open set

Question

Let $x_{n}$ be a convergent sequence in a metric space $(X,d)$ such that it has no convergent subsequence in $X$. Show that the complement of its range is an open set.

What I understood was, if $a$ is a limit point then there would be a sequence, say, $(a_n)$ converging to $a$. Now if I take some other point (apart from $a$) in the complement of range, say $x$ such that it is a limit point then there would be another sequence which would converge to $x$.

I am not sure what should be my next step? or how I should write it down?

score 2 · Accepted Answer · answered Sep 12 '21 at 05:59

2

Suppose $x$ is on the complement of $\{x_n: n \geq 1\}$ and $x$ is not an interior point of $\{x_n: n \geq 1\}$. Then, for every $k \geq 1$. There exist $n_k$ such that $d(x_{n_k}, x ) <\frac 1k$. (We may suppose $n_i<n_{i+1}$ for all $i$). But then $x_{n_k} \to x$. But $(x_{n_k})$ has no convergent subsequence. This contradiction shows that every point of the complement of $\{x_n: n \geq 1\}$ is an interior point, so this set is open.

answered Sep 12 '21 at 05:59

Kavi Rama Murthy

311,013

What if $x_n$ was not convergent? – Sepia Glen Sep 12 '21 at 08:11
$(x_n)$ is not convergent. But $d(x_{n_k},x) <\frac 1k$ implies that $x_{n_k}$ converges to $x$ @SepiaGlen – Kavi Rama Murthy Sep 12 '21 at 08:25
That makes sense. Thank you! – Sepia Glen Sep 12 '21 at 09:47
Hi. I have another question. We showed that every point in complement is an interior point but we did so without constructing a subsequence. What if were to show that there doesn't exist a convergent subsequence? What would we do then? – Sepia Glen Sep 19 '21 at 22:04

Prove that the complement of the range is an open set

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