Let $\mathbb{N}$ denote the set of all positive integers. In $(\mathbb{R}, d)$, show that $E = \{\frac{1}{n} : n \in \mathbb{N} \}$ is not closed, but $F = E \cup \{0\}$ is closed.
A theorem in my notes say that: A subset $F$ of a metric space $(X, \rho)$ is called closed if its complement
$X \setminus F = \{x \in X : x \notin F \}$
is open.
If I use this theorem to prove the question, then I need to show that $\mathbb{R} \setminus E = \{x \in \mathbb{R}: x \notin E\}$ is open.
Since a subset U in a metric space $\{X, \rho\}$ is open if for every $x \in U$, there exists $r > 0$ (depending on x and U) such that $B(x, r) ⊆ U$.
Now I am stuck, finding a way of showing there always exist a $r>0$ s.t. $B(x, r) \subseteq \mathbb{R} \setminus E$ for any $ x \in \mathbb{R}\setminus E$.
Could someone enlighten me on how I would show this?
Edit: When I originally posted this question I mistakenly used / to denote the contrapositive. I have edited / as \, to clear to confusion of what I'm trying to say.
Any answers that show a full solution to the problem will be greatly appreciated.
Now I am stuck, finding a way of showing there always exist a $r>0$ s.t. $B(x, r) \subseteq \mathbb{R} / E$ for any $ x \in \mathbb{R}/E$.
You'll never succeed doing that for $x=0$!
Now take care!!! The contrary of not being closed is not to be open. So you can't apply the theorem you mention.
However, you can use contraposition. I.e. that the complement of $E$ is not open. And for that $0$ is your friend!