Is this a sloppy point in Brezis' construction of topology generated by a family of subsets? I'm reading Section 3.1 The Coarsest Topology for Which a Collection of Maps Becomes Continuous from Brezis's Functional Analysis:

Problem 2. Given a set $X$ and a family $(U_{\lambda})_{\lambda \in \Lambda}$ of subsets in $X$, construct the cheapest topology $\mathscr{T}$ on $X$ in which $U_{\lambda}$ is open for all $\lambda \in \Lambda$.

In other words, we must find the cheapest family $\mathscr{F}$ of subsets of $X$ that is stable$^1$ by $\cap_{\text {finite}}$ and $\cup_{\text {arbitrary}}$ and with the property that $U_{\lambda} \in \mathscr{F}$ for every $\lambda \in \Lambda$. The construction goes as follows.

*

*First, consider finite intersections of sets in $(U_{\lambda})_{\lambda \in \Lambda}$, i.e., $\cap_{\lambda \in \Gamma} U_{\lambda}$ where $\Gamma \subset \Lambda$ is finite. In this way we obtain a new family, called $\Phi$, of subsets of $X$ which includes $(U_{\lambda})_{\lambda \in \Lambda}$ and which is stable under $\cap_{\text {finite}}$. However, it need not be stable under $\cup_{\text {arbitrary}}$.


*Therefore, we consider next the family $\mathscr{F}$ obtained by forming arbitrary unions of elements from $\Phi$. It is clear that $\mathscr{F}$ is stable under $\cup_{\text {arbitrary}}$. It is not clear whether $\mathscr{F}$ is stable under $\cap_{\text {finite}}$; but indeed we have the following result:

Lemma 3.1. The family $\mathscr{F}$ is stable under $\cap_{\text {finite}}$.

The proof of Lemma 3.1-a delightful exercise in set theory-is left to the reader; see e.g., G. Folland [2]. It is now obvious that the above construction gives the cheapest topology with the required property.
${ }^{1}$ Meaning that a finite intersection of sets in $\mathscr{F}$ and an arbitrary union of sets in $\mathscr{F}$ both belong to $\mathscr{F}$.

Below is the cross reference from G. Folland [2]: Let $X$ be a set and $\mathcal S$ a collection of subsets of $X$. Let $\tau(\mathcal S)$ be the smallest topology on $X$ that contains $\mathcal S$.

Theorem:  Then $\tau (\mathcal S)$ consists of $\varnothing, X$, and all unions of finite intersections of members of $\mathcal{S}$.
Proof (by Folland): The family of finite intersections of sets in $\mathcal S$, together with $X$, is a base for some topology $\tau'$ on $X$, so the family of all unions of such sets, together with $\varnothing$, is ineed the topology $\tau'$. Obviously, $\tau'$ is contained in $\tau (\mathcal S)$, hence equal to $\tau (\mathcal S)$.


My question:
According to Folland, we have to "manually" include $X$ into $\tau (\mathcal S)$, i.e., it is possible that $X$ is different from the union of all finite intersections of sets in $\mathcal S$.
Let $\Gamma_0 = \emptyset \subset \Lambda$. By usual convention, we have $\cap_{\lambda \in \Lambda_0} U_{\lambda} = \emptyset$. However, it seems to me that $\mathscr{T}$ (obtained from Brezis' construction) does not necceassily contain $X$. Then $\mathscr{T}$ is not neccesasily a topology on $X$.

Could you confirm if my understanding is correct?

 A: Yes and no.
It's not sloppy because you don't actually need to define a topology to explicitly contain the empty set and the whole space. It is typical for authors to demand that within the definition but, strictly speaking, it's not necessary. The reason for that is due to the following formulas:

*

*$\bigcup \varnothing = \varnothing$.

*$\bigcap \varnothing = X$.

Now, there's a caveat with that second formula which I'll need to get to so just wait for that. The first formula should be quite easy to understand; if I know my proposed set $\tau(S)$ is closed under arbitrary unions, then I can just take an empty subset of $\tau(S)$ and take the union of that. This will automatically mean that $\varnothing \in \tau(S)$.
The issue with that second statement is that, implicitly, we're taking the intersection of the empty set within $X$ itself. It's only in this case that the formula actually makes sense. Otherwise, you actually have the following theorem.
Theorem:
Let $x$ be any set. Then, $x \in \bigcap \varnothing$.
Proof:
Otherwise, there exists a $y \in \varnothing$ such that $x \notin y$. But this is impossible. $\Box$
But if you make it clear that you're only considering the intersection of the empty set within $X$ itself, then it's all good. In that case, the fact that $X \in \tau(S)$ will just follow from the fact that it is closed under finite intersections.
With all of that being said, I do think that he is being sloppy because I don't think that the above is really what he might've been thinking of. I don't think that the above is what occurred to him when he didn't explicitly state that the entire set belongs to your collection.
