A possible vacuous logical implication in Topology

While reading "Introduction to Topology and Modern Analysis" by G.F Simmons, in the chapter on Topological Spaces, I came across the following statement that seemed unusual. I feel it may be a vacuous truth but I don't know why. The statement is the following ad verbatim:

Let $X$ be a non-empty set. Let $\mathcal{S}$ be an arbitrary class of subsets of $X$.

If $\mathcal{S}$ is empty, then the class of all finite intersections of it's sets is the single element class $\{X\}$ and the class of all unions of sets in this class is $\{\varnothing,X\}$.

I do not follow this argument. So I would appreciate it if someone could throw some light onto this issue. If it is a vacuous truth, I would like to see how.

If this question has been asked before, I'd like to know under what title it lies in as I did not find it.

Solution 1:

Both are instance of something being vacuously true, though one has to think a bit to see it.

Since $\mathcal{S}\subseteq\wp(X)$, for any $\mathscr{U}\subseteq\mathcal{S}$ we have

$$\bigcup\mathscr{U}=\{x\in X:\exists U\in\mathscr{U}(x\in U)\}\tag{1}$$

and

$$\bigcap\mathscr{U}=\{x\in X:\forall U\in\mathscr{U}(x\in U)\}\;.\tag{2}$$

Let’s take $(2)$ first. How could one conclusively demonstrate that some $x\in X$ did not belong to $\bigcap\mathscr{U}$? One would have to show that there was some $U\in\mathscr{U}$ such that $x\notin U$. If $\mathscr{U}=\varnothing$, it’s impossible even to find a $U\in\mathscr{U}$, let alone one that does not contain $x$, so it’s impossible to demonstrate that $x\notin\bigcap\mathscr{U}$. Thus, $x\in\bigcap\mathscr{U}$ for all $x\in X$ if $\mathscr{U}=\varnothing$, and hence $\bigcap\mathscr{U}=X$ when $\mathscr{U}=\varnothing$. In particular, if $\mathcal{S}=\varnothing$, then $\mathscr{U}\subseteq\mathcal{S}$ means that $\mathscr{U}=\varnothing$ and hence that $\bigcap\mathscr{U}=X$.

We deal with the union the same way. Given a particular $x\in X$, how do you show that $x\in\bigcup\mathscr{U}$? You show that there is a $U\in\mathscr{U}$ such that $x\in U$. If $\mathscr{U}=\varnothing$, you can’t even find a $U\in\mathscr{U}$, let alone one that contains $x$, so it must be that $x\notin\bigcup\mathscr{U}$. That’s true for all $x\in X$, so if $\mathscr{U}=\varnothing$, then $\bigcup\mathscr{U}=\varnothing$.

Note that in $(1)$ and $(2)$ I was careful to limit possible members of $\bigcup\mathscr{U}$ and $\bigcap\mathscr{U}$ to $X$. This didn’t matter much for $(1)$, since the argument given to show that no $x\in X$ belongs to $\bigcup\varnothing$ works equally well for any $x$ whatsoever, and we’d arrive at the conclusion that $\bigcup\varnothing=\varnothing$ even without the restriction. It matters a lot for $(2)$, however: without the restriction we end up concluding that $\bigcap\varnothing$ contains everything $-$ that it’s the ‘set’ of all sets, which doesn’t exist. By specifying that $\mathcal{S}$ is a class of subsets of $X$, Simmons is implicitly limiting the candidates for membership in $\bigcap\mathscr{U}$ (where $\mathscr{U}\subseteq\mathcal{S}$) to elements of $X$.

Solution 2:

This fact is relying on the tacit assumption that the computation of the intersections is performed inside the lattice $\mathcal P(X)$, the set of all subsets of $X$ where the lattice operations are intersection and union (the elements in $\mathcal (X)$ are then ordered by set inclusion).

Now, in $\mathcal P(X)$ there is a largest element, namely $X$. The intersection of no sets is equal to that largest element. To see, that think of the property of intersection, that if $U\subseteq V$ where $U,V$ each is a family of subsets of $X$ then the intersection of all sets in $V$ is contained in the intersection of all sets in $U$. That is, if you intersect more sets you get a smaller subset. So, by that rational, the intersection of no sets at all should be the largest possible thing, thus it is $X$. (More generally, in any lattice the meet of the empty set if the largest element in the lattice (e.g., the infimum in $\mathbb R$ of the empty set does not exist, but if you adjoin $\infty $ to $\mathbb R$ in the usual way then the infimum of the empty set in $\mathbb R \cup \{\infty \}$ is $\infty $).)

So, starting with $S$ and forming all intersections of its elements gives you $\{X\}$. Now, forming all unions of the class $\{X\}$ gives rise to $\{\emptyset, X\}$. It is clear why $X$ is there. $\emptyset $ is there too for the dual reason that $X$ is the intersection in $\mathcal P(X)$ of no sets. The union of no sets should be the smalest possible thing, namely the empty set. (Similarly, the supremum in $[0,\infty )$ of the empty set is $0$).

It should be noted that these can be taken as arbitrary definition, adopted since it agrees with what you want computations to give, but this is in fact strictly a consequence of the definitions of infimum and supremum in lattices. You just need to realize that you are computing in the lattice $\mathcal P(X)$.