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?

  • 2
    $\begingroup$ 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.