0

Let $(X, d)$ be a metric space where $X$ is infinite and let $A$ be the set of isolated points of $X$, that is, $A = \{x \in X : x \text{ is an isolated point of } X\}$. Assume that $A$ is infinite. Is it true that $A$ is an open set in $X$?

I think it is an open set. My (very informal) argument is like this. We know that $x \in X$ is an isolated point of $X$ if $\{x\} \subseteq X$ is open. So we know there is an infinite number of these $x$'s since $A$ is infinite. Well then, $A$ can be written as the union of an arbitrary number of $\{x\}$'s, each of which is open and we know the union of an arbitrary collection of open sets is open, so $A$ is open.

If my informal argument is "correct", can someone please formalize it and make a formal argument?

user40333
  • 985
  • How many there are is quite irrelevant. – Henno Brandsma Jul 25 '17 at 07:59
  • "very informal"? Well, I hardly see possibilities to do that better. You could add the notation $A=\bigcup_{x\in A}{x}$ . Also note that the cardinality of $A$ and $X$ is irrelevant. – drhab Jul 25 '17 at 08:00

1 Answers1

3

The argument is quite formally correct and can be summarised as:

$$A = \bigcup\{\{x\}: x \in A\}$$ is a union of (by definition) open sets so open.

Henno Brandsma
  • 242,131