# $\mathbb{Z}$ under finite-complement topology [CSIR-2015]

Consider the set $$\mathbb{Z}$$ of integers with the topology $$\tau$$ in which a subset is closed iff it is empty, or $$\mathbb{Z}$$, or finite. Which of the following is true?

1. $$\tau$$ is the subspace topology induced from the usual topology on $$\mathbb{R}$$.

2. $$\mathbb{Z}$$ is compact in the topology $$\tau$$.

3. $$\mathbb{Z}$$ is hausdorff in the topology $$\tau$$.

4. Every infinite subset of $$\mathbb{Z}$$ is dense in the topology $$\tau$$.

I think 1 is true since any closed set in $$\tau$$ is $$\mathbb{Z}\cap [a,b]$$(or a finite union of these sets of this type) a,b integers(for finite sets) or the entire real line when the closed set is entire $$\mathbb{Z}$$ .

Since any non-trivial open set is basically $$\mathbb{Z}$$-{a finite set of integers}.

3 is false because given two integers $$a,b$$ if i try to look for two disjoint open sets $$U_a$$,$$U_b$$ they will always have a infinite number of points in their intersection.

4 is true since if i take any open set in $$\tau$$(which is of the above type) it must intersect a infinite subset of $$\mathbb{Z}$$.

I am not able to say anything about compactness. Further are all of the above reasoning correct?

• Closed set in $\mathbb{R}$ is not necessarily a finite union of $[a,b]$. Jun 23 '15 at 15:52
• How was the exam overall ?How much did you score?@Prayagdeep Jun 23 '15 at 16:16
• @learnmore i guess i will end up with a tip over 100.Is this the right place to continue this discussion...i have some more doubts on the questions, is there another place where we can discuss..? Jun 23 '15 at 17:28
• A few years late, but in case anyone else is wondering: MSE has a chat room, which would be the best place for discussing contests after they've completed. Jul 5 '17 at 5:49

Since when $\mathbb Z$ is given subspace topology it becomes discrete i.e every one point sets are both closed and open which is not true with $\tau$.
$\tau$ is compact since if you consider the open cover $\{U_{\alpha}:\alpha \in I\}$ then $\mathbb Z\setminus U_{\alpha}$ is finite and hence the remaining points can be covered with finite number of elements from $I$
(1) is false. The subspace topology on $\Bbb Z$ induced by the standard topology on $\Bbb R$ is the discrete topology. So, $\{1\}$ is open in the subspace topolobgy, but not in the finite complement topology.