Definition of open set by sequence

Question

Let $(X, d)$ be a compact metric space. We know that if $U$ is an open set in $X$, $p\in U$ and $x_n\to p$ as $n\to \infty $, then $x_n\in U$ for every $n>N$ for some $N>0$.

In my research I need to show that a non-empty set $U$ is an open set in $X$.

Is it true if I show the following:

If for every $p\in U$ and every sequence $x_n\to p$, then we have $x_n\in U$ for every $n>N$ for some $N>0$

Please help me to know it.

Answer 1

Yes, while this is not true in general for an arbitrary topological space, it is true for metric spaces. Under the stated condition, you can show that for any $\ p\in U\ $ there is a positive integer $\ r_p\ $ such that the open ball $\ B\left(p, \frac{1}{r_p}\right)\ $$= \left\{x\in X\,\left\vert\,d(x,p) <\frac{1}{r_p}\right.\right\}\ $ is a subset of $\ U\ $. The proof of this by contradiction is not difficult, and it follows that $\ U\ = \bigcup_\limits{p\in U}B\left(p, \frac{1}{r_p}\right)\ $, being a union of open balls, must be open.

Answer 2

Yes. This is true for metric spaces (or more generally so-called sequential spaces):

The set $O \subseteq X$ is called sequentially open iff for all $p \in U$ and all sequences $(x_n)_n$ in $X$ such that $x_n \to p$, the sequence is eventually in $O$, i.e. there is some $N$ such that $x_n \in O$ for all $n \ge N$.

In any topological space $X$ an open set is sequentially open, just by the definition of what it means for $(x_n)_n$ to converge to a point.

The reverse does not hold in general spaces, but it does hold in the topological spaces that are "sequential" (this is in fact the definition of being sequential, in fact, at least one of the equivalent ways to define it) and in such spaces we can indeed describe open sets (and closed sets and continuity etc.) by only considering sequence convergence.

I'll show that first countable spaces are sequential, and metric spaces are first countable: the sets $B_d(x,\frac{1}{n})$ form a countable (decreasing) local base at $x$ for any $x$.

Suppose $O$ is sequentially open but not open. This means there is some point in $p \in O$ that is not an interior point of $O$, so for any open neighbourhood $U$ of $x$, $U \nsubseteq O$. Now assuming, as we do, that $p$ has a decreasing local base of neighbourhoods $B_n, n \in \Bbb N$, this means we can pick $x_n \notin O$ such that $x_n \in B_n$ (taking $U=B_n$, for all $n$). But then, as the $B_n$ form such a local base, $x_n \to p$, but no $x_n$ is in $O$ which is in blatant contradiction with $O$ being sequentially open. So the assumption of non-openness of $O$ leads to a contradiction and $O$ must be open.