Unions and Intersections

Question

Will I be correct in stating the following?

The intersection of a collection of sets need not be a subset of its union.

This is what I’m thinking while positing the above: $\bigcup \varnothing =\varnothing,$ but $\bigcap \varnothing$ is not even a set.

But I think that my proposition will be false for a nonempty collection of sets, correct?

If $\alpha\in A_n$ for every $n$, is $\alpha\in A_k$ for some $k$? — Simply Beautiful Art, Jun 27 '19 at 08:36
You probably mean $\emptyset$ (i.e. the empty set) and not $\phi$. — Mark Kamsma, Jun 27 '19 at 08:38
@SimplyBeautifulArt Read the question carefully. I'm talking about empty sets in particular. — Atom, Jun 27 '19 at 08:49
Still, my first comment is relevant here, and highlights the cases of $\forall x\in\varnothing$ and $\exists x\in\varnothing$. — Simply Beautiful Art, Jun 27 '19 at 08:55
Looking at your profile, you seem to be very bad at giving titles to questions. "Unions and intersections" means nothing. "Are my logical translations correct?" and "What does Paul Halmos mean here?" are also very bad titles. You have 150 characters to use for a title. Please use them. You can use math symbols, and you should when it makes sense. People shouldn't read your question in order to get an idea what your question is about, the title should be doing that already (for the most part). — Asaf Karagila, Jun 27 '19 at 09:20
See https://math.stackexchange.com/questions/348668/intersection-of-the-empty-set-and-vacuous-truth — John Douma, Jun 27 '19 at 09:21
Some books (I think Kunen, but don't have it right now to verify) may suggest the convention that $\cap\emptyset=\emptyset$ just so the operation is defined (for the usual interpretation could lead to $\cap\emptyset=$ the class of all sets, which is not a set). It is only a convention (with the virtue that $\cap A$ is always a well-defined set, for every set $A$). — Mirko, Jun 28 '19 at 16:10

score 1 · Answer 1 · edited Jun 27 '19 at 10:13

Facts first:

If $A\ne \varnothing$, then $\bigcup A$ and $\bigcap A$ are both sets and $\bigcap A\subseteq \bigcup A$.
If $A =\varnothing$, then $\bigcup A=\varnothing$, and $\bigcap A$ does not exist (as a set).

Whether your statement is a reasonable expression of this state of facts is more a question of language and communication, than one of hard mathematical truth. You certainly have a defensible position that it is not wrong. But still it sounds pretty odd, because in order to be right, it has to be speaking of "the" intersection of a collection of sets in a situation where there is no such intersection at all.

Other than winning bar bets because the claim tricks the listener into not considering the empty collection, I have doubts that your claim serves a useful communicative purpose, standing alone.

We could imagine uttering that claim while trying to develop an automatic proof verification system, where someone had implemented a rule that $\bigcap A\subseteq \bigcup A$ for general $A$. But then it would be much better communication to actually point at the concrete problem for $A=\varnothing$ than to merely deny that the rule is valid.

In any case, this is not a problem that is specific to set theory. We could get mostly the same discussion out of considering a statement in arithmetics such as

$\frac 1 x \cdot x$ is not necessarily $1$.

score 0 · Answer 2 · answered Jun 27 '19 at 10:43

If we're talking about the subsets of a fixed "universal" set (these form a kind of structure called a complete atomic Boolean algebra), then every collection of sets has a union and an intersection, and the intersection is contained in the union, except when the collection is empty.

More generally, in a complete lattice, every set has a supremum and an infimum, and $\inf X\le\sup X$ holds except when $X=\emptyset$. For example, every set of real numbers has a supremum and an infimum in the extended real line, and $\inf X\le\sup X$ when $X\ne\emptyset$, but $\sup\emptyset=-\infty$ and $\inf\emptyset=+\infty$.

Unions and Intersections

2 Answers2