# Question about empty sets in set theory

1. Mar 17, 2015

### xwolfhunter

His response is that no $x$ fails to meet the requirements, thus, all $x$es do. He reasons that if it is not true for a given $x$ that $x \in X~ \mathrm{for ~ every} ~X~ \mathrm{in} ~ \phi$, then there must exist an $X$ in $\phi$ for which $x \notin X$ is true. He then says that since there are no $X$es in $\phi$, this result is absurd. So no $x$ fails to satisfy the stated condition, thus every $x$ does.

Isn't this mind-numbingly faulty reasoning? The answer is "no $x$es are specified by this sentence."" You don't even have to try and quibble with the details of the sentence, the quibbling of which is faulty the way I see it anyway. Neither the $X$es nor the $x$es specified in the sentence exist.

Besides the issues inherent with trying to perform logical operations on literally nothing, his logic seems to be vacuous anyway - sure, if an $x$ is not a member of all $X$es in a set then there must be some $X$ which does not contain that $x$, but why does he act as though this sentence is logically useful in this context? While it's true that all the $x$es or none of the $x$es in the framework of the sentence meet the conditions, either way no $x$es are specified, because there are no $x$es in the framework. The operations are on unknown $x$es - his result is legitimate for an unspecified set rather than $\phi$. But since we know the set is $\phi$, we know that there are no $x$es in existence in this reference frame, so the argument he used is rendered moot. His result is entirely of unspecified $x$es of unspecified $X$es of an unspecified set. He's talking about a hypothetical scenario, but the one we have is defined, and then he's applying the results of the hypothetical scenario to the actual scenario.

My question is what the heck is the point of this? He implies that you need to take caution to avoid scenarios like this - my question is why? Why would you avoid scenarios like this? It's cut and dry, there are no $x$es specified by the sentence. There's no mathematical uncertainty here that I see.$$\{ x \in X~ \mathrm{for ~ every} ~X~ \mathrm{in} ~ \phi \} = \phi$$

2. Mar 17, 2015

### Svein

No. It is standard contrapositive reasoning in first-order language.

3. Mar 18, 2015

### xwolfhunter

Yes, but it is incomplete, and in this case wrong. "If x∈X for all X in ∅ is not true, then there must be some X in ∅ for which x∉X" is only try if there is some X at all in ∅ for which x∈X, and in this case there is not, because there are no Xes of which an x might be an element. It's pretty simple, how can this be in a math book?

4. Mar 18, 2015

### jbriggs444

Have you ever encountered the notion of something being vacuously true?

"I stopped at every red light on the way to work this morning".
"I ran through every red light on the way to work this morning".

If there were no red lights on the way to work this morning, both statements are vacuously true. There is no counter-example to either.

5. Mar 18, 2015

### xwolfhunter

Defining $\mathcal{E}$ as an arbitrary set, the union of $\phi$ (when restricted by the constraints of $\mathcal{E}$) does not equal $\mathcal{E}$ but rather vacuously equals $\mathcal{E}$; and so the union is meaningless? So that's why they avoid situations like this, not because there's a problem with them but rather because they provide no useful information.

6. Mar 18, 2015

### DrewD

A union of empty sets is not equal to an arbitrary set and "vacuously equals" is not a phrase I have heard nor has Google.

I wish I could remember, but there is a story about a mathematician doing his/her thesis work proving some important theorems about such-and-such. The theorems essentially said such-and-such is true for all $x$ with some property. Unfortunately, the set of elements with that property was empty. Anything you want is a true statement about the elements of an empty set. That isn't interesting math. I am sure that this has happened a number of times, but I had a professor who told it with specifics that helped make it more real.

7. Mar 19, 2015

### xwolfhunter

It means that its appearance of being equal is vacuous. I'm just using English words to express my meaning.

Paul Halmos does state that when restricted by arbitrary set $\mathcal{E}$, the intersection of $\phi$ is $\mathcal{E}$. This he states directly, in a "this is absolutely true" fashion:
But that's not true, is it? As you said, anything that can be said about the intersection is true, but only vacuously. So it is false that $\cap_{X\in\phi} X$ is equal to $\mathcal{E}$, right?

Last edited: Mar 19, 2015
8. Mar 19, 2015

### DrewD

Somewhere things are getting confused. The intersection of empty sets is equal only to the empty set. The statement that the intersection of empty sets is equal to another set is not true (vacuously or otherwise) unless that set is the empty set. If you are using $\phi$ to be $\emptyset$ then what you wrote does not seem true unless I am misunderstanding something you wrote. If you are not using $\phi$ to mean $\emptyset$, then I am very confused by your posts and I am sorry.

$\cap_{X\in\phi} X$ is very different from the intersection of empty sets. This is an intersection of nonexistent sets rather than empty sets.

9. Mar 19, 2015

### xwolfhunter

Yes, it's the intersection of nonexistent sets. Haha, I didn't know that there was the symbol "$\emptyset$." I just chose the closest-looking symbol I could find, and that was $\phi$.

His argument is that $\cap_{X \in \emptyset} X = \mathcal{S}$, where $\mathcal{S}$ is the set of all $x$ which are tested for $x \in \cap_{X \in \emptyset} X$.

His argument is chiefly that
But that argument is only valid if there is any $X$ in $\emptyset$ for which $x \in X$ is true. But since we can't tell if $x \in X$, because it's the equivalent of asking whether or not $x \in \mathrm{cow}$, then while it is true that we can't say that an $x$ fails to satisfy $x \in X ~ \mathcal{for ~ all} ~ X \in \emptyset$, it is equally true that we can't say that it does satisfy it. So basically I'm saying that $\cap_{X \in \emptyset} X = \{ x \}$ is no more valid than $\cap_{X \in \emptyset} X = \{ x ~ \mathrm{that ~ are ~ blue} \}$ or whatever else we want to say, because the answer is indeterminable.

10. Mar 19, 2015

### jbriggs444

It is the intersection of the all the sets in an empty collection of sets. One need not invoke the questionable phrase, "nonexistent set".

Suppose you have an empty collection of marbles. What attributes hold of every member of the collection? They are all red. They are all green. They are all chipped. They are all smooth.

Suppose you have an empty collection of vintage automobiles. They are all Fords. They are all Chevys. They have all rusted out. They are all in mint condition.

Suppose you have an empty collection of sets. They are all empty. They are all non-empty. No matter what x you name, they all contain x. No matter what x you name, they all fail to contain x.

Any predicate that you can come up with holds for all members of an empty collection.

11. Mar 20, 2015

### DrewD

Since I used it first, I just wanted to mostly agree with you. It looked like xwolfhunter was confusing the sets in an empty collection of sets with an empty collection of sets. In retrospect making up a term was probably not necessary.

xwolfhunter, I think the issue may be that you are confusing logical statements in English with formal logical statements. There are very precise rules that apply when manipulating logical statements. If one statement is true, then so is the contrapositive. If the original statement seems confusing to you but the contrapositive is clearly true (or false), then the original statement must be true (or false). In this case I also struggle to accept the original statement on face value. Maybe going through the contrapositive carefully will help.

12. Mar 20, 2015

### xwolfhunter

That's what I meant when I said
Basically $\cap_{X \in \emptyset} X$ is undefined.

13. Mar 20, 2015

### xwolfhunter

I will present my logical argument and then ask you to logically critique it:
First I define collection $\mathcal{C}$ as represented in this diagram:
$$\mathcal{C} \begin{cases} \mathcal{Q} \begin{cases} 1 \\ 2 \\ 3 \\ 4 \\ \end{cases} \\ \mathcal{R} \begin{cases} 1 \\ 2 \\ 3 \\ \end{cases} \\ \mathcal{S} \begin{cases} 1 \\ 2 \\ \end{cases} \\ \mathcal{T} \begin{cases} 1 \\ \end{cases} \\ \end{cases}$$
So it contains sets $\mathcal{Q}$, $\mathcal{R}$, $\mathcal{S}$, and $\mathcal{T}$, which contains their elements as shown in the diagram.

Now we have the statement

$\mathcal{F} =$ "If '$x \in X$ for all $X$ in $\mathcal{C}$' is false, then there must be some $X$ in $\mathcal{C}$ such that $x \notin X$."

Let us examine the statement $\mathcal{F}$.

$\mathcal{F}$ is clearly true.

According to $\mathcal{F}$, when $x = 2$, since $x$ is not in all $X$ in $\mathcal{C}$, there must be some $X$ in $\mathcal{C}$ for which $x \notin X$. This is clearly true - $2$ is not found in the set $\mathcal{T}$.

Conversely, now I define

$\mathcal{G} =$ "If $x \notin X$ for all $X$ in $\mathcal{C}$ is false, then there must be some $X$ in $\mathcal{C}$ such that $x \in X$."

$\mathcal{G}$ is also clearly true - if an $x$ is not in no $X$, then it must be in some $X$.

So far, through $\mathcal{F}$ and $\mathcal{G}$, we have established that if $x$ is not in all $X$, then there must be some $X$ that $x$ is not in, and that if $x$ is not in no $X$, then there must be some $X$ that $x$ is in.
The argument of Halmos is that $\cap_{X \in \emptyset} X = \{ x \}$ where $\{ x \}$ is the set of all $x$ that are tested for $x \in \cap_{X \in \emptyset} X$. He uses $\mathcal{F}$ to prove this - he says that since there are no $X$ in $\emptyset$, the "then" clause in $\mathcal{F}$ cannot be true, thus the "if" clause must not be true either.

I'm saying that $\cap_{X \in \emptyset} X = \{ \emptyset \}$ (where $\{ \emptyset \}$ is the set of none of the $x$ that are tested) can be "proved" using the same logic with $\mathcal{G}$.

As you can "prove" both cases, neither can be true. Or both must be true. Or whatever else you'd like to prove about the set. That's what I'm saying - Halmos' argument, which he treats as mathematical truth in his book, leads to an untrue result.

14. Mar 20, 2015

### jbriggs444

Let me rephrase the $\mathcal{F}$ and $\mathcal{G}$ above.

$\mathcal{F}$: If x is not present in every X then it must be absent in at least one.
$\mathcal{G}$: If x is not absent from every X then it must be present in at least one.

Now let us try to make the respective arguments more explicit. Assume that X is empty

Halmos: Since X is empty, x cannot be absent from any X. So it must be present in every X. By definition, this means that x is present in the intersection.

Xwolfhunter: Since X is empty, x cannot be present in any X. So it must be absent from every X. This means that x is absent from the intersection.

The problem is the leap from x being absent from every member of the collection to x being absent from the intersection. That leap is reasonable for non-empty collections. It fails for empty collections.

15. Mar 20, 2015

### xwolfhunter

A prelude - please address all three points of the post. People usually just pick which part they want to respond to and respond to that. That bugs me.

Firstly $X$ is not empty, it's nonexistent, right? If $X$ is actually an empty set, I want to see a proof of that, because it doesn't make sense to me.

Secondly, could you expound upon and defend your last statement? Also explain why if my statement is true up until the last sentence, then this does not render Halmos' argument inert. Even if the leap is unjustifiable, which I'm hazy on, if the statements "$x$ is in all $X$" and "$X$ is in no $X$" are both true, doesn't this demonstrate that results obtained from the one cannot be accepted?

Lastly, the definition of the intersection of sets is, as found in Halmos' book,
$$A \cap B = \{ x \in A : x \in B \} .$$
So, by definition, the set $\{ x \notin A ~ \mathrm{or} ~ x \notin B \}$ is the set of all $x$ which are not in $A \cap B$. Thus if $x \notin X$ for any $X$ where $X$ is a set being intersected with another set, then $x$ is in the set of all $x$ which are not in the aforementioned intersection. Right?

So if $x \notin X$ for some $X$ in $\emptyset$, which is a provable result (for if "$x \notin X$ for all $X$ in $\emptyset$", then "$x \notin X$ for some $X$ in $\emptyset$" is true as well), then $x$ is an element of the set of all $x$ which are not in the intersection of $\emptyset$. Thus, it is not in the intersection. And since it can indeed be proved that $x \notin X$ for all $X$ in $\emptyset$, the intersection is empty.

I'm not asserting that this is absolutely true; I'm saying it's as true as what Halmos said, and therefore neither are true - both are vacuously true.

Last edited: Mar 20, 2015
16. Mar 20, 2015

### jbriggs444

The union is taken over an empty collection of sets.

Defend what statement against what attack?

The definition of a union of a collection of sets that I am more familiar with is that it is the collection of all elements present in all members of the collection.

That is not what the definition says. It says nothing about what is not in a set. Only about what is.

I cannot even parse that.

17. Mar 21, 2015

### xwolfhunter

Just ensuring you read and respond to all I had to say, which, incidentally, you did not.

$X$ is nonexistent.

No attack, just defend it. You simply said some words and did nothing to make me think they were true.

Talkin' bout intersections.

So if I give you the set {2,3,7}, that set doesn't necessarily not have 4, 9, and 1,000,000 as members? Man math is confusing. Please explain the rules regarding exclusions, because if the parameters of a set don't indicate what elements are not in that set, then what does? What's real, what isn't? How can we tell?

And even if we pretend that the set $\{ x \in A:x \in B \}$ isn't unambiguous, then the axiom of specificity or whatever dictates that we can make a set which is unambiguous, contains only the elements found in the intersection, and clearly displays all the elements that are not in that set. So all in all, we can define a set which contains all elements which are not in an intersection, by definition(s).

Considering that the sentence in question is a result of unaddressed mathematical reasoning, I will try to make clear my reasoning, and the results thereof, and hopefully in doing so I will discover some error I made or some unclear thing I said, those things which I guess can come of mathematical reasoning when done in english (for I lack the dictionary and grammar to say these things in mathematical language); and then we can be on the same page either because I will see what I've done wrong or I will make myself unignorably clear. Yes, "unignorably" is not a word. No, I will not change it.

$\mathcal{I} = \cap_{X \in \mathcal{C}}X$
$\mathcal{G} = \{x \notin \mathcal{I} \}$

If we can establish $\mathcal{G}$ then it will follow that $\mathcal{I}$ will contain all $x$ that are not in $\mathcal{G}$. This is all by definition. (As in, if we can make $x$ fail to fulfill $\mathcal{I}$'s requirements, it becomes a member of $\mathcal{G}$; and if we can establish the parameters of $\mathcal{G}$, then what's left over is the set $\mathcal{I}$.)

Halmos determines that $\mathcal{I}$, when $\mathcal{C} = \emptyset$, contains all $x$ that are tested for $x \in \mathcal{I}$. Thus $\mathcal{G} = \emptyset$.

My aim is to prove that $\mathcal{G}$ contains all $x$ that are tested for $x \in \mathcal{I}$ (where $\mathcal{C} = \emptyset$), and if I can do so with the same rigor that Halmos uses to prove his case, then I have proved that Halmos' result is vacuously true, because I will have proved that $\mathcal{I} = \emptyset$.

The first step is to examine the statement that is the foundation for Halmos' argument:
This is clear enough. Now what about a different sentence:
This is equally clear.

Now we establish $\mathcal{C} = \emptyset$.

The reasoning from here is simple and straightforward. This is the reasoning that I want critiqued. And the words "that doesn't work" hold no sway with any logical argument, true or otherwise. You need to show why it doesn't work.
As I said before, if an argument here fails, please explain why explicitly. I'm not here to pretend that I know all math and am right all the time, I'm here to learn about how math works, so please show me why I'm wrong if I am.

18. Mar 21, 2015

### jbriggs444

I reject the attempt to shift the burden of proof. However, the core error of your argument is as was identified previously.

Halmos: If x is an element of every member in a collection then x is an element of the intersection of that collection.

This is true by the definition of the intersection of a collection of sets.

Xwolfhunter: If x absent from every member in a collection then x is absent from the intersection of that collection.

That does not follow. What would follow is:

Corrected: If x is absent from every member in a collection C then x is not an element of the union of that collection.

For non-empty collections, a collection's intersection is a subset of that collection's union and the result that you hope for follows. For empty collections, the result that you hope for is not justified.

[Note the wording "absent" rather than "not an element" to avoid the parsing ambiguity between "not ( an element of every member )" versus "( not an element ) of every member"]

19. Mar 21, 2015

### xwolfhunter

The burden of proof for my statement falls on me. I provided what I consider proof of my statement. You say what you think is wrong with my statement, but provide no support for it. In your statement, the burden of proof falls on you. What you say may be true, but if I don't have the reasons why, then why should I think it so?
Actually it does follow, just another step behind. It's still a necessary result. If $x$ is not an element of the union of a collection, then it is not an element of the intersection of that collection. This is elementary absolute truth, is it not?

So if my proof is that $x \notin \cup_{X \in \emptyset} X$ for all tested $x$, then it is a necessary result that it is also a proof that $x \notin \cap_{X \in \emptyset} X$ for all tested $x$. Where is the logical fallacy? Where does my proof break down, and what prevents Halmos' proof from breaking down in the corresponding spot?

And if you say "because we're talking about empty sets" then I respond with the equally rigorous "Halmos' argument is also false because we're talking about empty sets."

Am I mistaken in remembering that anything can be said about the $x$ in the intersection of an empty set and it will be true and false, because there is no reference to falsify whatever claim you have? How is $x \in \cap_{X \in \emptyset} X$ any different from $x \in \{ \text{three-sided squares} \}$?

20. Mar 21, 2015

### jbriggs444

No. It is not. See if it follows from the definitions for the union and intersection of a collection, as applied to the case of an empty collection. As I pointed out in post #4 above, this has everything to do with the notion of vacuous truth.

21. Mar 21, 2015

### xwolfhunter

And you ignore everything else I said.

I know my proof is vacuously true. That's the whole point of this - Halmos' proof is vacuously true, and I prove that by proving the opposite result to his.

22. Mar 21, 2015

### jbriggs444

Why harp on the irrelevant when the key concept can be addressed simply?

I suggest you look up the definition of "vacuous truth". You are being stupid.

23. Mar 21, 2015

### xwolfhunter

I looked up the definition of vacuous truth, and then I looked up the proof on proofwiki.

I learned a few things while doing so.

Halmos' proof is not the proof of this. As far as I can tell, Halmos' proof is not that great. At least, it is by no means convincing anywhere on its face. The actual proof is extremely simple and doesn't mention any of the stuff we've been talking about.
You tell me that $\mathcal{R}$ does not follow from the definitions of union and intersection. Well, it's clear that it does in existent sets. Why does this change in nonexistent sets? Which part of the definition does not allow for this? Spell it out so even a person who's being stupid can't argue with it, and then I can't argue with it. I am incapable of coming to that conclusion on my own - I think that neither of the definitions touch on nonexistent sets, so how could they possibly change when presented with them?

24. Mar 21, 2015

### jbriggs444

There are no non-existent sets involved in this argument. There are collections without members. That is not at all the same thing. Existence is not a predicate.

Edit: Let me back up and try to motivate a reason for defining the intersection of an empty collection as we do. [I'm am going to be a bit sloppy here and not work very hard to avoid Russell's Paradox].

If one has a collection of sets one can take its "intersection". As one adds additional sets to the collection, the "intersection" can be reduced but can never be increased. So the intersection of an empty set should, naively, be larger than the intersection of a singleton. Let's firm that up a bit...

We would like for the intersection of a singleton set to be simply that singleton. That is to say we want:

$\cap\{A\} = A$

We also want:

$\cap(A \cup B)$ to be $(\cap A) \cap (\cap B)$.

If both of these are to hold, then $\cap\emptyset$ must have as members every element in our domain of discourse.

Similarly, we want:

$\cup \{A\} = A$

And we also want:

$\cup(A \cup B)$ to be $(\cup A) \cup (\cup B)$

If both of these are to hold then $\cup\emptyset$ must be $\emptyset$.

This is not a proof that the intersection and union are defined in this way for the empty collection. It is merely a motivation for why we might want to define them that way. [We do, in fact, conventionally define them in this way].

Last edited: Mar 21, 2015
25. Mar 22, 2015

### xwolfhunter

So I've decided that before I try and wield my mathematical faculty I should learn how logic works. So I'm reading a book on logic. Please tell me I'm not insane:
If Elliot is unhappy, then he is not happy. If Elliot is not happy, then it is not the case that Elliot is happy. Elliot is not happy, and he is unhappy. He is not between. Sentence seven is H, and sentence eight is ¬H. If this is not true then I have no idea what to do with myself. I mean, it's in the definition of 8 - "Elliot is unhappy." How could it be that he is between unhappy and happy if he is unhappy? The dictionary definition of "unhappy" is "not happy" for God's sake. You can't be between "happy" and "not happy."

Last edited: Mar 22, 2015