# Show that $\{x\}\cup\{x_n\mid n\in N\}$ is nowhere dense in $R$.

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?

• In the discrete metric every set is open. – amsmath Oct 28 '19 at 18:23
• 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. – Bungo Oct 28 '19 at 18:27
• @amsmath yes, I have used that in my examples, can you check if its correct – Abhay Oct 28 '19 at 19:15
• @Bungo thank you, I agree with you, can you please very my answer to there other part of question – Abhay Oct 28 '19 at 19:15
• @Abhay In the discrete metric no subset except $\emptyset$ is nowhere dense. So, you can take any converging sequence as a counterexample. – amsmath Oct 28 '19 at 19:25

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 $$\operatorname{int}(\overline{A})=\emptyset$$.