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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?

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  • $\begingroup$ This is unanswerable unless you state what the metric $d_2$ is. $\endgroup$ – Umberto P. Feb 13 '18 at 16:29
  • $\begingroup$ By $d_2$ I mean the Euclidean distance. $\endgroup$ – Just_a_newbie Feb 13 '18 at 16:31
  • $\begingroup$ In that case there is no mistake. $\{5\}$ is an open set with no accumulation points. $\endgroup$ – Umberto P. Feb 13 '18 at 16:35
  • $\begingroup$ You should never try to prove something that is false. $\endgroup$ – Umberto P. Feb 14 '18 at 3:26
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Apparently, this was meant to be proved for $(M, d) = (\mathbb{R}^n, d_2)$.

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