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 $\mathbb R$

  • $\begingroup$ All is fine. You may want to provide additional details. (At this level, most instructors won't be satisfied with "clearly".) $\endgroup$ – 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 $(n,n+1)$


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