An empty collection of sets

1. Jun 13, 2012

Useful nucleus

I read that an empty collection of sets, denote it by λ, is a little problematic when one considers $\bigcup$$_{A\inλ}$A and $\cap$$_{A\inλ}$A. I can see that the union should be ∅. However, for the intersection it was argued that if one considers a set X to be the universe of the discussion then the intersection indicated above would result vacuously in X. Then , why wouldn't the union also result in X vacuously?
Any help is appreciated!

2. Jun 13, 2012

theorem4.5.9

It sounds like the argument is heuristic and trying to appeal to intuition. In one sense you can view the union as only able to "see" the index set. This leads to the empty union being equivalent to the empty set.

Once you picked a line of reasoning, it's usually best to form other questions in terms of that reasoning in order to minimize the number of jumps of intuition you need to make. Suppose you have some superset $X$ where everything is taking place. Then one view you could take is that an intersection is a way to take things out of the universe. Very basically we see that $A = X\cap A$. Now from the viewpoint just mentioned, you can view this as $X\cap A = X - A^c.$ De Morgan's laws lets you apply this to general intersections, I show it for the easiest nontrivial case. $A\cap B = X\cap A \cap B = X - \left(A \cap B \right)^c = X-\left(A^c \cup B^c \right)$

In other words, $\bigcap _{A\in \Lambda}A = X - \bigcup _{A\in \Lambda} A^c$. If the intersection is empty, we might as well stick to what we know and insist this formula to still hold. Thus you may want to say the empty intersection is $X$ while the empty union is $\emptyset$.

Of course you could have started with the opposite assumption that the empty intersection is empty, but then the above formula would suggest that the empty union should be the universe. That's the problem with arguing heuristics at least.

3. Jun 13, 2012

micromass

In standard ZFC set theory, the empty intersection is not defined. Let's see why this is.

You'll to look at the definitions of union and intersection

For the union, you have $x\in \bigcup \mathcal{A}$ if there exists an A in $\mathcal{A}$ such that $x\in A$.

But if $\mathcal{A}$, then the sentence "there exists an A in $\mathcal{A}$ ..." is always false (whatever follows). So

$$x\in \bigcup\mathcal{\emptyset}$$

is always false. So the empty union has to be empty.

On the other hand, you have $x\in \bigcap \mathcal{A}$ if for all A in $\mathcal{A}$ holds that $x\in A$.

This sentence is always true for $\mathcal{A}$ empty. Indeed, you can find no A such that the sentence is false, so it must be true!!

So

$$x\in \bigcap\mathcal{\emptyset}$$

is always true, no matter the x. This lease us to the conclusion that $\bigcap \emptyset$ must be the set that contains all possible sets. But this set does not exist!! So the empty intersection does not exist (it is an undefined operation).

4. Jun 29, 2012

Useful nucleus

theorem4.5.9 and micromass, thank you very much for your explanations. That helped a lot! I can see more now why there is no general agreement between mathematician on this subject. However, I still have a question for theorem4.5.9.

In that last expression you wrote, how it would be possible to define Ac, when it is given that λ is an empty collection.

5. Jun 29, 2012

micromass

There is general agreement between mathematicians. The difference between theorem4.5.9 his answer and mine is that his answer supposes some kind of "universe".

That is: if you know that all sets are a subset of a certain set X. Then it is possible to define the empty intersection as X. This is sometimes a good definition.
If we are not working in a "universe", then the empty intersection is undefined.

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