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?

  • $\begingroup$ Closed set in $\mathbb{R}$ is not necessarily a finite union of $[a,b]$. $\endgroup$
    – Jack Yoon
    Jun 23 '15 at 15:52
  • $\begingroup$ How was the exam overall ?How much did you score?@Prayagdeep $\endgroup$
    – Learnmore
    Jun 23 '15 at 16:16
  • $\begingroup$ @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..? $\endgroup$ Jun 23 '15 at 17:28
  • $\begingroup$ 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. $\endgroup$ Jul 5 '17 at 5:49

1 is not true:

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 $I$

Rest are fine


(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.


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