Why does $\bigcup \emptyset = \emptyset$ but $\bigcap \emptyset$ is not defined?

If I had to guess, I'd say it's also equal to $\emptyset$.

  • 2
    $\begingroup$ I agree with you. $\endgroup$ – user122283 Mar 17 '14 at 23:16
  • $\begingroup$ Who is saying it isn't defined? $\endgroup$ – Ian Coley Mar 17 '14 at 23:17
  • $\begingroup$ As far as my teacher concerned, it is not defined. $\endgroup$ – AnnieOK Mar 17 '14 at 23:19
  • $\begingroup$ I'm fairly sure it's defined. $\endgroup$ – recursive recursion Mar 17 '14 at 23:19
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    $\begingroup$ math.stackexchange.com/questions/660656/… $\endgroup$ – mle Mar 18 '14 at 18:23

That's fine. Some authors define it that way. The idea is that, if we do not modify the usual definition, we would have $\bigcap\emptyset=\{y\mid \forall A\in\emptyset\,(y\in A)\}$. Since there are no sets $A$ in the empty set, the requirement $\forall A\in\emptyset\,(y\in A)$ is satisfied for any $y$ (it is vacuously true), so $\bigcap \emptyset$ would be the universal set. In some set theories, the universal set does not exist (it is a proper class). Even if you work in a set theory that allows the existence of universal sets, this seems exceedingly wasteful.

A good compromise is to redefine things a bit: Note first that (for nonempty $\mathcal B$), if $y\in\bigcap \mathcal B$ then for all $A\in\mathcal B$ we have $y\in A$. Thus, $y\in\bigcup \mathcal B$. Hence, we have the identity $$ \bigcap \mathcal B=\{y\in\bigcup\mathcal B\mid\forall A\in \mathcal B(y\in A)\}. $$ The right hand side expression makes sense even if $\mathcal B=\emptyset$: In that case, $\bigcup\mathcal B=\emptyset$, so the right hand side is empty. Since this seems more desirable than the set of all sets, some authors define $\bigcap \mathcal B$ by the displayed identity. This makes no difference in general, and gives us $\bigcap \emptyset =\emptyset$.

That said, this is mainly a manner of convention, and whether we call the empty intersection empty, the universal set, or undefined, makes no serious difference and adds no more than a couple of minor footnotes here and there.

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    $\begingroup$ If we want that $\bigcap (A\cup B)=\bigcap A\cap\bigcap B$, then $\bigcap\emptyset=\emptyset$ is a really bad idea. $\endgroup$ – Carsten S Mar 17 '14 at 23:33
  • $\begingroup$ Yes, of course. All the change in definition accomplishes is to move the usual caveats or remarks to different places. I find the effort silly anyway: $X=\bigcap A$ is simply an abbreviation, in the language of set theory we do not have brackets or symbols for unions or for the empty set or for intersection, etc. $\endgroup$ – Andrés E. Caicedo Mar 18 '14 at 0:13
  • $\begingroup$ I can see that you do not want to assign too much importance to a minor definition. Btw, did you time at Boise overlap with Stefan's? I studied with him in Berlin. Good times! $\endgroup$ – Carsten S Mar 18 '14 at 22:16
  • $\begingroup$ Yes, we coincided for a short while. I haven't seen him for a couple of years or so now. Hopefully soon. $\endgroup$ – Andrés E. Caicedo Mar 18 '14 at 22:17

The problem with $\bigcap\emptyset$ is that it should be everything, since every $x$ is an element of every element of the empty set. Now there is no set of all sets, but it can be taken to be the class of all sets. However, that is often inconvenient. So if there is a set $X$ in which everything takes place, then it is customary to set $\bigcap\emptyset=X$. For example, if we are talking about a topological space $X$ and require that the set of open sets is closed under finite intersections, then it makes sense to understand this in such a way that it implies that $\bigcap\emptyset= X$ is open. (Surely the empty set is finite.)


The problem with defining $\bigcap \emptyset = \emptyset$ is that it fails to satisfy the otherwise universal rule that $$\left(\bigcap \mathcal A\right) \cap \left(\bigcap \mathcal B\right) = \bigcap \left(\mathcal A \cup \mathcal B\right).$$

In particular, applying this rule with $\mathcal A = \emptyset$, we see that, if $\bigcap \emptyset$ were defined and satisfied this rule, it would have to be the neutral element for the set intersection operator, i.e. the universal set.

Unfortunately, standard ZFC set theory does not have such a universal set. There do exist other formulations of set theory that do include a universal set $\mathbf V$, and, in those set theories, defining $\bigcap \emptyset = \mathbf V$ is a natural choice.

  • $\begingroup$ Excellent opening point. That sort of thing should be more wideley appreciated. Another example is that the number of partitions of the empty set is $1$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Mar 17 '14 at 23:52
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    $\begingroup$ In addition to this, another problem with defining $\bigcap \emptyset = \emptyset$ is the property $A \subseteq B \implies \bigcap B \subseteq \bigcap A$ no longer holds. $\endgroup$ – Mathemanic Nov 26 '16 at 8:29

In some contexts $\displaystyle\bigcap_{\alpha\in\varnothing} S_\alpha$ is defined to be the universal set --- the set that contains everything as a member. The whole space.

The reason it's undefined in Zermelo–Fraenkel set theory and similar contexts is that there is no set containing the whole space.

The idea behind this definition is that in expressions like $$ A\cap B\cap C\cap\cdots\cdots, $$ every term imposes further limits one what can be a member of the set that is the intersection. If there are no terms, then there are no such limits, so everything is a member.

This is like the fact that $\inf\varnothing=+\infty$. Every member of a set $S\subseteq\mathbb R$ imposes a limit on how big the infimum can be: it cannot be bigger than that number. If there are no members of $S$, then there are no such limits. The infimum is always the biggest it can be subject to those constraints, so when all such contraints vanish, then the infimum is infinite. This is useful in one context I can think of: the distance between two points in a manifold is the infimum of the set of all lengths of paths connecting them. Hence if no paths connect them, then the distance is $+\infty$.

I heartily endorse Ilmari Karonen's point.


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