# Problem with null indexed family of subsets.

1. Oct 15, 2005

### inquire4more

I have been reviewing my set theory and topology and recently came across an assertion I was not familiar with, and frankly have trouble grasping. In words,
let I be a set (which is to serve as the set of indices), then for each $\alpha \in I$ let $A_\alpha$ be a subset of some set S. Now, assuming I to be the null set:

$$\cup_{\alpha \in \emptyset} A_\alpha = \emptyset$$
$$\cap_{\alpha \in \emptyset} A_\alpha =$$ S.

If someone could explain this to me, I would be grateful (only moderately grateful, mind you, so don't get any ideas). Or perhaps, point out the flaw in my reasoning, which follows. It seems to me that the union of, perhaps non-existent, subsets indexed by the empty set would be the empty set. However, assuming I followed that correctly, it seems to me that the intersection of these subsets would also be empty, yet apparently this is not the case, as the above asserts it is in fact S. Maybe I'm just horribly lost.

Last edited: Oct 15, 2005
2. Oct 15, 2005

### AKG

To me, the idea that the union of no sets is itself a set (the null set) seems a little strange. However, if a union is, by definition, a set (regardless of what it is the union of), then we can see why this is the case. If a union of a collection of sets is the set which contains every element that occurs in one of your sets in the collection, then you can see why the union would be the null set. On the other hand, I might think that taking the union of nothing, or a collection of sets indexed by the empty set, should not even be considered well-defined. What is the sum of zero numbers? It certainly shouldn't be zero.

Anyways, if you accept that $\cup _{\alpha \in \emptyset}A_{\alpha} = \emptyset$ then consider the following:

$$S = \emptyset ^C = \left (\cup _{\alpha \in \emptyset}A_{\alpha}\right )^C = \cap _{\alpha \in \emptyset}A_{\alpha}^C\right = \cap _{\alpha \in \emptyset}A_{\alpha}$$

The third equality follows from DeMorgan's Law. The fourth follows from the fact:

$$(\forall \alpha \in \emptyset)(A_{\alpha} = A_{\alpha}^C)$$

Last edited: Oct 15, 2005
3. Oct 15, 2005

### Hurkyl

Staff Emeritus
Actually, under most circumstances, we do define it to be zero!
As for the original post, really you just need to work through the definition -- it all traces back to vacuous truth. For any proposition P, $\forall x \in \varnothing: P(x)$ is true, and $\exists x \in \varnothing: P(X)$ is false.

4. Oct 15, 2005

### AKG

You could put it that way, defining a union of some sets indexed by I by:

$$(\forall x \in S)\left ((\exists \alpha \in I)(x \in A_{\alpha}) \Leftrightarrow x \in \bigcup _{\alpha \in I}A_{\alpha}\right )$$

or

$$\bigcup _{\alpha \in I}A_{\alpha} = \{x \in S| (\exists \alpha \in I)(x \in A_{\alpha})\}$$

and defining intersection by:

$$(\forall x \in S)\left ((\forall \alpha \in I)(x \in A_{\alpha}) \Leftrightarrow x \in \bigcap _{\alpha \in I}A_{\alpha}\right )$$

or

$$\bigcap _{\alpha \in I}A_{\alpha} = \{x \in S| (\forall \alpha \in I)(x \in A_{\alpha})\}$$

EDIT: fixed as per Hurkyl's suggestion.

Last edited: Oct 16, 2005
5. Oct 15, 2005

### inquire4more

AKG, Hurkyl, thanks both for your responses. They have given me several things to consider. I still have a few reservations/considerations.

AKG, yes, I should have seen that, assuming $\cup_{\alpha \in \emptyset} A_\alpha = \emptyset$ to be true, $\cap_{\alpha \in \emptyset} A_\alpha = S$ would follow by DeMorgan's Law. I feel like something of an idiot for missing that before posting. Thanks for pointing that out. Though the union being empty does make sense to me, I'm going to keep searching for some proper justification for this, I simply do not trust my own reasoning here, I may be making assumptions regarding the empty set and unions.

Hurkyl, I have two questions for you, if you do not mind. Firstly, as we are not dealing with elements of the empty set, but rather subsets indexed by elements of the empty set ($A_\alpha$ rather than $\alpha$ itself) does this change the validity of your statement regarding the vacuous truth? Secondly, assuming that it does not change, and all properties follow, may I then assert that $\cap_{\alpha \in \emptyset} A_\alpha = \emptyset$ and use DeMorgan's Law to arrive at $\cup_{\alpha \in \emptyset} A_\alpha = S$?

Let me know, either of you, what you think.

Last edited: Oct 15, 2005
6. Oct 16, 2005

### Hurkyl

Staff Emeritus
My claim is that when you write out the definition given in your book, it will involve phrases like $\forall \alpha \in \emptyset: P(\alpha)$, or equivalently, something like $\forall \alpha: \alpha \in \emptyset \implies P(\alpha)$. (This is also vacuous truth: recall that [False==>True]=True)

If your book doesn't have a formal definition, then you'll have to get one from someplace. AKG's looks good to me, except that he should have $\{ x \in S | \cdots$ instead of $\{ x | \cdots$, since these unions and intersections are presumably supposed to be living inside your space. (In the set theoretic case, the empty intersection isn't a set: it's the class of all sets)

By the way, an indexed family of things is just a fancy way of speaking about functions: A is simply a map from I to Set.

Last edited: Oct 16, 2005
7. Oct 16, 2005

### AKG

Do you know why it follows from DeMorgan's Law? Like I pointed out, one of the equalities in my proof used DeMorgan's Law, another used the "vacuous truth":

$$(\forall \alpha \in \emptyset)(A_{\alpha} = A_{\alpha}^C)$$

I could have put any sentence in the second brackets, and the above sentence would be true, which makes it vacuous.
Why would it? A sentence about the set indexed by $\alpha$ is still a sentence about $\alpha$, i.e. it can still be formalized in the form $P(\alpha )$.
Go back to the definitions of union and intersection. I beleive the ones I gave in my previous post are suitable definitions. By these definitions, you will find that you cannot just assert that the intersection is 0. In fact, you will be able to derive that the union is 0 and the intersection is S directly from the definitions, you don't need to derive one fact from the other like I did.

8. Oct 16, 2005

### inquire4more

Ok, you'll have to forgive my stubbornness or stupidity, I'm just having trouble with the idea of sets indexed by the empty set. I continue to go over the definitions, but to little avail. Proceeding simply from the definitions,

$$\bigcup _{\alpha \in I}A_{\alpha} = \{x \in S| (\exists \alpha \in I)(x \in A_{\alpha})\}$$

does indeed make sense to me. Allowing $\alpha \in \emptyset$ seems to imply that $\nexists x \in S$ s.t. $x \in A_\alpha$ for some $\alpha$. And yet, looking again at the definition of intersection,

$$\bigcap _{\alpha \in I}A_{\alpha} = \{x \in S| (\forall \alpha \in I)(x \in A_{\alpha})\}$$

seems to imply that $\forall x \in S, x \in A_\alpha$ for all $\alpha$. If in fact, the interestion is $S$.

It seems like something of a contradiction to me, even after considering the definitions. I apologize if this conversation makes you feel as though you are beating your head against a wall.

Last edited: Oct 16, 2005
9. Oct 16, 2005

### AKG

Yes, exactly. For all x in S, x is in Aα for all α in ø. This is true vacuously.

10. Oct 16, 2005

### inquire4more

Alright then, so as to clarify the situation, and thereby muddle my understanding, according to the definition of union, no x in S are elements of any A? And according to the definition of intersection, all x in S are elements of every A?

No x in S are elements of any A, according to $\cup_{\alpha \in \emptyset} A_\alpha = \emptyset$.
All x in S are elements of every A, according to $\cap_{\alpha \in \emptyset} A_\alpha = S$.

Does this not strike you as just a tad bit contradictory?

Frustration does not begin to describe my feeling at the moment.

11. Oct 16, 2005

### AKG

No, it's not contradictory. It seems contradictory because it seems to say, x is in every A but it is not in any A. Both these statements are true, they don't contradict each other. But they are true in a strange way. x is in every A, since there are no A to speak of. x is not in any A, because there are no A to speak of. For all A, x is not in A, and for all A, x is in A. So for all A, x is in A and is not in A. This would be contradictory ONLY IF there were some A. What we're really saying is not just "for all A" but "for all A in {Aa}a in {}" which is to say, "for all A in {}". Clearly, there are no A in {}, so the following is true:

$$(\forall A \in \emptyset )(x \in A \wedge x \notin A)$$

It may be easier to believe if you interpret the above as:

$$(A \in \emptyset )\ \Rightarrow \ (x \in A \wedge x \notin A)$$

You know that if the antecedent is false, then the whole conditional is true (even if the consequent is false), and the antecedent of this sentence is ALWAYS false (nothing is ever in {}).

In general, a sentence of the form:

$$(\forall v_1 \in \emptyset )(P)$$

is always true regardless of what P is. Sentences of the form:

$$(\exists v_1 \in \emptyset )(P)$$

are always false regardless of what P is.

Last edited: Oct 16, 2005
12. Oct 16, 2005

### inquire4more

AKG, gotcha. As the subsets we speak of are nonexistent all things may follow, the vacuous truth. That part I understand, I believe. The problem this presents for me then is that it seems I could assert, as I stated above, that:

$$\cap _{\alpha \in \emptyset} A_\alpha = \emptyset$$
$$\cup _{\alpha \in \emptyset} A_\alpha = S$$

assuming of course that all may follow from premises involving the empty set. And if this IS the case (and I do not insist that it is, it merely seems to me), then it seems the above would be just as true as the initial assertions:

$$\cup _{\alpha \in \emptyset} A_\alpha = \emptyset$$
$$\cap _{\alpha \in \emptyset} A_\alpha = S$$,

and then I wonder at the point of including this, or any statement regarding the empty set. Not to be a bother (I'm afraid I've already been that, harping on this point), but are these two sets of assertions equally valid, given the involvement, as a set of indices, of the empty set?

Thanks

13. Oct 16, 2005

### inquire4more

AKG, my apologies, I believe I was typing my response as you were editing yours, so that my last post was ill placed. Once edited, so as to include

$$(\exists v_1 \in \emptyset )(P)$$

as false, your point, and the truth of the initial statements, immediately became clear. I am contented and grateful to you for the explanation of what must have seemed to you to be a very basic point. Thank you so much.

14. Oct 17, 2005

### inquire4more

For what it's worth, and if anybody cares, I yesterday found my Munkres' topology text, and he had something interesting to say, somewhat justifying my initial dissatisfaction with the definition of the empty intersection. Regarding $\cap_{A \in \alpha} A = X$ for $\alpha = \emptyset$, Munkres states [p. 8]:

---
On the other hand, every x satisfies (vacuously) the defining property for
the intersection of the elements of $\alpha$. The question is,
every x in what set? If one has a given large set X that is specified at the
outset of the discussion to be one's "universe of discourse," and one
considers only subsets of X throughout, it is reasonable to let

$$\cap_{A \in \alpha} A = X$$

when $\alpha$ is empty. Not all mathematicians follow this
convention, however. To avoid difficulty, we shall not define the
intersection when
$\alpha$ is empty.

Munkres, James R. Topology: A First Course. Englewood Cliffs, NJ:
Prentice Hall, 1975.
---

His italicized emphasis, not mine. Just food for thought.

Last edited: Oct 17, 2005