# Given the metric space $(\mathbb{N}, d_2)$, what is the interior of $\{5\}$?

I'm trying to proof that, given a metric space $(M, d)$ and $A \subseteq M$, the interior of A is a subset of the set of accumulation points of A. However I have apparently come across with a counterexample, although I think that it is very likely that I have made a mistake in some part.

Let $(\mathbb{N}, d_2)$ be a metric space and $A = \{5\} \subset \mathbb{N}$. $5$ is an interior point because if we choose $r = 1/2$ then $B_r(5) \subset A$. On the other hand, $B_r(5) \setminus \{5\} \cap \ A = \emptyset$, which means that 5 is an isolated point and, therefore, is not an accumulation point.

Where is the mistake?

• This is unanswerable unless you state what the metric $d_2$ is. – Umberto P. Feb 13 '18 at 16:29
• By $d_2$ I mean the Euclidean distance. – Just_a_newbie Feb 13 '18 at 16:31
• In that case there is no mistake. $\{5\}$ is an open set with no accumulation points. – Umberto P. Feb 13 '18 at 16:35
• You should never try to prove something that is false. – Umberto P. Feb 14 '18 at 3:26

Apparently, this was meant to be proved for $(M, d) = (\mathbb{R}^n, d_2)$.