# $\mathbb{Z}$ 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 $[a,b]$. – Jack Yoon Jun 23 '15 at 15:52
• How was the exam overall ?How much did you score?@Prayagdeep – Learnmore 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..? – Kayoken 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. – Eric Stucky 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.