Okay so what the hell does that even mean and who cares? Let’s give a really simple example. So let’s just make up an abstract set of objects that happen to be some of the English alphabet letters. Here’s our set, $X = {a,b,c}$. So we have a collection of 3 distinct objects and we want to define a topology on that set. Our topology τ (tau) is simply going to be a set of sets, it’s a collection of subsets from X that satisfy the axioms of topology.
Let’s give it a try, maybe our topology τ should be this: ${{a},{b},{c}}$. So our topology τ is a collection of single element subsets from X. Notice the difference in notation. If I had simply written $\tau = {a,b,c}$ that would merely be the same as $X$, just an ordinary set with 3 elements. No, $\tau$ is a set whose elements are also sets (even if those sets contain one element).
Ok, anyway, does our $\tau$ satisfy the axioms? Does the empty set and $X$ itself belong to $\tau$? Uh no. The empty set is ${}$ (or $\emptyset$) and $X$ itself is ${a,b,c}$, our $\tau$ did not have those two sets as members, so already our attempted topology failed. Let’s try again. How about $\tau = {\emptyset,{a},{b},{c}, {a,b,c}}$. Now at least this τ satisfies the first axiom. The second axiom is less obvious, but take for example the union of ${a}$ and ${b}$, which yields ${a,b}$. Is ${a,b}$ in $\tau$? No it’s not, so this attempted topology also fails.
Alright here’s a legitimate topology on $X$… $\tau = {\emptyset, {a}, {a,b,c}}$. It satisfies the first axiom, it has the empty set ${}$ and the full $X$ set as members of $\tau$, and if you take the union of any combination of members of $\tau$, the resulting set is also a member of $\tau$.
For example, $ { } \cup { a } = { a } $ (read: the empty set union the set ${a}$ produces the set ${a}$). Obviously the union of the empty set and ${a}$ must produce ${a}$ which is in $\tau$. We must verify for all possible unions and intersections that the results are still in $\tau$.
$ {a} \cup {a,b,c} = {a,b,c} $, which is also in $\tau$. $ {a} \cap {a,b,c} = {a}$, which is also in $\tau$.
Hence, the union or intersection of any elements in $\tau$ is also in $\tau$, thus we have ourselves a valid topology on $X$.
I know this all seems rather academic at this point, but keep with me and we will eventually get to a point of using this new knowledge for practical purposes.