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If $(x_n)\to x $ in R, show that $\{x\}\cup\{x_n\mid n\in N\}$ is nowhere dense in $R$. Is the same true of $R$ is replaced by arbitrary metric space? Is every countable set nowhere dense?

Clearly, $\{x\}\cup\{x_n\mid n\in N\}$ is closed and has empty interior since since this set is countable and so can't contain any open ball inside it which would make it uncountable

For the other part, replace $(R,|\cdot|)$ with $(R,d_{\text{discrete}})$

Now, sequence $(1,1,\ldots) \to 1$ but $(\overline{\{1\}})^{\circ}= \{1\}\neq \phi$. Is this good counter example?

Also $N\subset (R,d_{\text{discrete}}),$ is both closed and open and countable. But $\overline{N} =N$ and $N^{\circ}=N\neq \phi$. So $N$ is not nowhere dense. is this good example?

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    $\begingroup$ In the discrete metric every set is open. $\endgroup$ – amsmath Oct 28 '19 at 18:23
  • $\begingroup$ On a side note, why does density of $\mathbb R \setminus \mathbb Q$ imply that your set $\{x\} \cup \{x_n \mid n \in \mathbb N\}$ in the first part has empty interior? I would instead argue that the set is closed and therefore equals its own closure, and this closure cannot contain an interval because it's countable. $\endgroup$ – Bungo Oct 28 '19 at 18:27
  • $\begingroup$ @amsmath yes, I have used that in my examples, can you check if its correct $\endgroup$ – Abhay Oct 28 '19 at 19:15
  • $\begingroup$ @Bungo thank you, I agree with you, can you please very my answer to there other part of question $\endgroup$ – Abhay Oct 28 '19 at 19:15
  • $\begingroup$ @Abhay In the discrete metric no subset except $\emptyset$ is nowhere dense. So, you can take any converging sequence as a counterexample. $\endgroup$ – amsmath Oct 28 '19 at 19:25
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Usual topology:

A convergent sequence with limit is compact (in any space), so it's a closed subset in $\Bbb R$. And it has empty interior as any countable subset of $\Bbb R$ is (all non-empty open sets are uncountable). So it's nowhere dense as $\operatorname{int}(\overline{A})=\emptyset$.

In the discrete topology all subsets are open (and closed) and there is only one nowhere dense subset, the empty set. The constant sequence will do as a trivial counterexample, indeed.

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