Enumerable Sets

Question

If $E$ is an enumerable set of $R^n$, is $R^n-E$ an open subset of $R^n$? May I conclude that $E$ is a discrete metric space for the euclidian metric?

Answer 1

No, because $E$ may not be closed: for example take a sequenve converging to a point and remove the limit point. This is not closed (by sequential criterion) and thus its complementary is not open.

To give an explicit example, let $n=1$ and consider $$E=\{ \frac{1}{n} \mid n\in \mathbb{N} \}$$ This is not closed, because if you take the obvious sequence in $E$ converging to $0$ in $\mathbb{R}$, the limit of this sequence (that is, $0$) is not in $E$.

On the other hand, if we consider now $E$ the same as before but including the $0$, then it is closed, but it is not discrete (because $0$ is an accumulation point).

You can also mix these two examples to produce non closed and non discrete countable subsets of any euclidean space.