Open union in the finite-complement topology.

Question

In the second chapter of an introductory topology text, I stumpled upon the following example concerning the finite-complement topology.

Edit. In a previous version, I accidently defined $\mathcal T$ as a topology.

Let $X$ be an infinite set. We, firstly, define the set $\mathcal T$ by $$\mathcal T = \{U\subseteq X\ |\ U = \emptyset\ \text{ or }\ X - U\ \text{ is finite} \}. $$

Consider an arbitrary collection of open sets $\{U_\alpha,\ \alpha \in J\}$.

Edit. We now continue, to establish the second requirement of a topology as mentioned in the supplement.

By DeMorgan's law we have

$$X- \bigcup_{\alpha \in J} U_\alpha = \bigcap_{\alpha \in J}(X-U_\alpha)\ . $$ Each $X-U_\alpha$ is finite or all of $X$, so we have $X-\bigcup_{\alpha \in J} U_\alpha$ is finite or all of $X$.$\quad (*)$

$(*)$ should implicate that $\bigcup_{\alpha \in J} U_\alpha$ is open. I fail to see why.

I could argue that the union of a collection of open sets is open. I also know that a set is closed if its complement is open.

Edit.

Or is $(*)$ only used to establish that the collection $\bigcup_{\alpha \in J}U_\alpha$ is contained in $\mathcal T$?

So that we seperately use that, indeed, the union of a collection of open sets is open?

Supplement.

This question concerns proving the second requirement of a topology: the union of an arbitrary collection of members of $\mathcal T$ is in $\mathcal T$.

McCleary's definition of a topology.

Let $X$ be a set and $\mathcal T$ a collection of subsets of $X$ called open sets. The collection $\mathcal T$ is called a topology on $X$ if the usual three requirements are met.

Answer 1

As others have pointed out in the comments, this result you are looking at seems to be about proving that the finite complement topology is indeed a topology. So we are given a set $\mathcal{T}$. We do not yet know that it is a topology. As of now, it is just a collection of subsets of $X$, and we must show that $\mathcal{T}$ satisfies the four requirements to be a topology. In particular, one step is to show an arbitrary union of elements in $\mathcal{T}$ is itself an element of $\mathcal{T}$. That is why it is important to show that $\bigcup_{\alpha \in J} U_\alpha \in \mathcal{T}$ without assuming ahead of time that the finite complement topology is a topology.

So, we begin with any collection of subsets, $\{U_\alpha\}_{\alpha \in J}$, such that $U_\alpha \in \mathcal{T}$.

As you say, we have by DeMorgan's law that $$X- \bigcup_{\alpha \in J} U_\alpha = \bigcap_{\alpha \in J}(X-U_\alpha)$$ Notice that $X-U_\alpha$ is finite or empty. We know this because we chose each $U_\alpha$ in our arbitrary collection from $\mathcal{T}$, and by definition $U_\alpha$ is only an element of $\mathcal{T}$ if $X-U_\alpha$ is finite. This means that $$\bigcap_{\alpha \in J}(X-U_\alpha)$$ is an arbitrary intersection of finite sets. Hence, the intersection itself can contain no more than finitely many elements. This should be intuitive, but if it's not, you should prove it formally. Once you have established that $\bigcap_{\alpha \in J}(X-U_\alpha)$ is a finite set or empty set, we know it's equivalent $$X- \bigcup_{\alpha \in J} U_\alpha$$ is also a finite or empty set. Since $\bigcup_{\alpha \in J} U_\alpha$ is a subset of $X$ and $X- \bigcup_{\alpha \in J} U_\alpha$ is finite or empty, this satisfies the necessary criteria for $\bigcup_{\alpha \in J} U_\alpha$ to be an element of $\mathcal{T}$. This establishes that an arbitrary union of elements of $\mathcal{T}$ is itself an element of $\mathcal{T}$, and we did it without assuming any set was open, or that $\mathcal{T}$ was already a topology.

From here, it would be great practice to prove that $\emptyset \in \mathcal{T}, X \in \mathcal{T}$ and a finite intersection of elements of $\mathcal{T}$ is itself an element of $\mathcal{T}$.