Is set of isolated points of S closed?

Question

In metric space, $(X,d)$

I know limit points of some set(S maybe) is closed in X

Then is set of isolated points of S is closed?

Want to make sure if closure of isolated points of S doesn't intersect with limit points of S

But i think i got it and i think i should leave it for people like me

Answer 1

It's not closed in general: take the set $\{\frac{1}{n} : n \in \mathbb{N} \}$.

Answer 2

In $(R, std metric) $

S={$\frac{1}{n}$ n$\in$N}

S is set of isolated points since $\forall \frac{1}{n}$$ \exists $r= $\frac{1}{n(n+1)} $

But closure of S has 0 as element

So set of isolated points may not be closed

Answer 3

Usually an isolated point in a metric space is defined to be a point that is an open set. Therefore any set of isolated points is in fact open, but it need not be closed. There was an example already given, and it is almost correct. Let $$X=\{0\}\cup\{\frac1n|n\in \mathbb Z^+\} $$ Then every point is isolated except for $0$, and the union of all isolated points is open, but it is not closed because a neighborhood of $0$ always contains an isolated point.