I have some problems showing certain sets are topologies. Let X be any set, $B \subset X$ be a subset, there seems to be an extreme amount of confusion I am having as to how to actually show that a set is a topology. I see the examples, but I guess I still need more:
problem:
$\{X-A: A \text{ is a countable subset}\} \cup \{Ø\}$,
$\{U: U\subset B\} \cup \{X\}$,
$\{U: U \cap B=Ø \}\cup \{X\}$
These are problems from my textbook, and I will appreciate the help as I need alot of getting used doing these problems. Thank you.
My attempt at first part: I think yes with the hint given and the three conditions of being a topology, if I leave the null set cases as being trivial, then dealing with just the sets $X-A$, if we have some $A_i$ which is in the Topology $T$;then as mentioned by the given hint, using De-Morgans Law, we have, the topology $T$ is closed under finite intersections, that is $X-\bigcup^{n}_{i=1}A_i=\bigcap^{n}_{i =1}(X-A_i)$ and finite union of countable subsets are also countable $\bigcup^{n}_{i=1} A_i$, therefore $\bigcap^{n}_{i =1}(X-A_i)$ is closed in $T$.
In the case of checking arbitary unions, then we have the opposite $X-\bigcap_{i \in I}A_i=\bigcup_{i \in I}(X-A_i)$, since $\bigcap_{i \in I}A_i$ is also countable, then the arbitrary union of $X-A_i$ is in $T$.
I am not entirely sure if this is correct, but is this at least the correct path I am taking?
Regards to the third part, just to make sure, is $\bigcap_{i \in I}(U_i \cap B)=(\bigcap_{i \in I}U_i) \cap B = \emptyset$ and $\bigcup_{i \in I}(U_i \cap B)=(\bigcup_{i \in I}U_i) \cap B = U\cap B =\emptyset?$
The first one requires DeMorgan's rules which you are
expected to know and use.
For the second show the empty set and X are in that set.
Show if U and V are in that set, so is their intersection.
Finally show for any collection of sets in that set, that
the union of the collection is in that set.
For the third, do the same four things as you did for the second.