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?
$\tau$ is the subspace topology induced from the usual topology on $\mathbb{R}$.
$\mathbb{Z}$ is compact in the topology $\tau$.
$\mathbb{Z}$ is hausdorff in the topology $\tau$.
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?
Best Answer
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