Prove that the Sierpiński space is a topology I need to prove that the Sierpiński space, $\mathcal{\tau} = \{\emptyset, \{1\}, \{0, 1\}\}$, is a topology.
I have only just started on toplogy, and so far just know the basic axioms.
To prove that the union of any collection of subsets from T is in T, do I need to list every single possible union of subsets and check that it is in T?
i.e. $\{1\} \cup \{0,1\} = \{0,1\} \in \mathcal{\tau}$,
$\emptyset ∪ \{0,1\} = \{0,1\} \in \mathcal{\tau}$ etc.
Or is there a much shorter and more concise method?
 A: The empty set adds nothing to a union, so we need not consider unions with the empty set.
Any union in which one of the sets is the universe $\{0, 1\}$ will just equal $\{0, 1\}$, so we can dispense with those.
Suddenly there's nothing left to check.
A: Checking every subset isn't too much work, but there is a way to do them simultaneously. You can observe that $T$ is totally ordered under the subset relation, i.e. $\emptyset \subseteq \{0\} \subseteq \{0,1\}$. Then for any subset $S \subseteq T$, the union of elements of $S$ is the maximal element of $S$ under inclusion, which is still in $T$. The same logic applies to intersections, where you instead get the minimal element.
How concise this argument really is depends on your standard of rigor. For instance, you'd have to prove that the union is the maximum and the intersection is the minimum. You may also want to specially consider the cases of the empty union and intersection. This isn't too hard, but it adds length. But regardless, if you were to do every union as a separate case, you'd end up using this same idea every time.
