I'm reading the definition of a topological space in terms of open sets, and am trying to figure out why the "union" property is stated the way it is.
The definition of a topological space in terms of open sets is the following:
A topological space is an ordered pair $(\Omega,\mathcal{T})$, where $\Omega$ is a set and $\mathcal{T}$ is a collection of subsets of $\Omega$ satisfying the following properties:
(1) The empty set and $\Omega$ are members of $\mathcal{T}$
(2) The intersection of any two members of $\mathcal{T}$ is itself a member of $\mathcal{T}$.
(3) The union of an arbitrary number of members of $\mathcal{T}$ is itself a member of $\mathcal{T}$.
My question pertains to property (3). Why is it not sufficient to say that the union of any two members of $\mathcal{T}$ is itself a member of $\mathcal{T}$? If the union $A\cup B$ of any two sets $A,B$ in $\mathcal{T}$ is itself a member of $\mathcal{T}$, it must be that this property holds for the case that at least one of either $A$ or $B$ is itself a union of two sets in $\mathcal{T}$. That is, if $A\cup B\in\mathcal{T}$, then $A\cup C\cup D\in\mathcal{T}$ for $B=C\cup D$. Accordingly, if $A\cup C\cup D\in\mathcal{T}$, then $A\cup C\cup E\cup F\in\mathcal{T}$ for $D=E\cup F$, and so on. Am I understanding this correctly?
If you assume that the union of any two elements of $\mathcal{T}$ is an element of $\mathcal{T}$, then the best you can say is (just as you wrote when saying so on) that a union of any finite number of elements of $\mathcal{T}$ is an element of $\mathcal{T}$. You can't however generalize this to an infinite union, while what is (or should be) meant in (3) by "arbitrary number" is "finitely or infinitely many".
In topology, even an infinite union of open sets is an open set. That's why a better way to state (3) is:
Here $\bigcup \mathcal F$ means the union of all the elements of $\mathcal F$.