Axiom of choice and Topology? I was just thinking if you need the axiom of choice to see if some topologies are actual topology.
$\cup_{\lambda \in \Lambda} U_{\lambda} \in \tau$
See if you have $\tau$ is uncountable and $\Lambda$ is uncountable. How do you know if you have a topology? You must need some axiom of choice to show that a set is topology or to show it's not a topology.
 A: First let us see when $\tau\subseteq\mathcal P(X)$ is a topology on $X$.


*

*$X\in\tau$, $\varnothing\in\tau$.

*For every $A\subseteq\tau$, $\bigcup A\in\tau$.

*For every $U,V\in\tau$ we have $U\cap V\in\tau$.


The second condition is to say, that no matter how many open sets I take (finitely many, countably many, continuum many, or any other number), the union of all the open sets will be an open set.
The fact that this union is at all a set we adhere to the Axiom of Union, suppose $A$ is a set of sets, then $\{u\mid \exists y\in A: u\in y\}$ is a set, this set is often denoted $\bigcup A$ and is known as the union of $A$.
The axiom schema of replacement says that if $\Lambda$ is a set, and $X_i$ is a set for every $i\in\Lambda$, then $\{X_i\mid i\in\Lambda\}$ is also a set. This is independent of the size of $\Lambda$. We only require that it is indeed a set.
From this we have that the requirement that a topology is closed under any unions is not dependent on the axiom of choice, the closure under finite intersections is also easily definable, as the property that the empty set and the entire space are in the topology.

Note that $\mathbb R$ with the standard topology is second countable, that means that every union of open sets can be reduced to a countable union of open sets; it means that every discrete subset is countable; and it means that there are at most continuum many open sets.
