# Prove or disprove : $\mathbb Z$ is closed?

Let $\tau_1$ be the usual topology on $\mathbb R$ . Define another topology $\tau_2$ on $\mathbb R$

$$\tau_2 = \{U \subseteq \mathbb R \ \ | \ \ U^c \ \ is \ \ countable \ \ or \ \ empty \ \ or \ \ \mathbb R \}.$$

Then $\mathbb Z$ is

1. closed in $\tau_1$ but not in $\tau_2$.

2. closed in $\tau_2$ but not in $\tau_1$.

3. closed in both $\tau_1$ and $\tau_2$.

4. closed in neither $\tau_1$ nor $\tau_2$.

My attemt is :

clearly $\mathbb Z$ is closed in $\tau_2$ and $\tau_1$ is induced by the metric d(x,y) = |x-y|. In this metric space $\ \ \mathbb Z \ \$ is closed in $\mathbb R$

• All is fine. You may want to provide additional details. (At this level, most instructors won't be satisfied with "clearly".) – David Mitra Feb 28 '14 at 16:23

As $\tau_2$ has precisely the countable sets as its closed sets, $\Bbb Z$ is closed.
Note that in the definition of $\tau_2$ it is not necessary to mention the empty set explicitly, as it is already among the countable sets.
As for the closedness of $\Bbb Z$ in the metric space $\Bbb R$, one could argue that its complement is a union of open intervals $(n,n+1)$