Empty set disjoint with itself paradoxical?

Stoney Pete

Hi guys, I've got a philosophical question for you concerning the empty set.

I know that in axiomatized set theory the empty set is disjoint with itself. Because it has no members, the empty set cannot have any members in common with itself. This is common sense in set theory.

But if we say that set A is disjoint with set B, aren't we then implying that A is not identical with B? But how can the empty set be non-identical with itself? To me this has paradox written all over it, but to me surprise all textbooks about set theory treat the disjointness of the empty set with itself as something trivial and not very remarkable. Please let me know what you think!

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Erland

Science Advisor
But if we say that set A is disjoint with set B, aren't we then implying that A is not identical with B?

No, I don't see why it should be so. That A and B are disjoint means that there is no x which lies in both A and B, and it makes sense to say that there is no x which lies in A and A; this is then true only for A=∅.
I see no paradox.

Stoney Pete

Thanks for your reply. But I remain unconvinced.

Consider the following argument:

1. If A ∩ B = ∅
2. then obviously A ≠ B
3. so if ∅ ∩ ∅ = ∅
4. doesn't it necessarily follow that ∅ ≠ ∅?

This argument seems trivial to me. Yet you are telling me its not valid. So in which step is the mistake?

Erland

Science Advisor
Thanks for your reply. But I remain unconvinced.

Consider the following argument:

1. If A ∩ B = ∅
2. then obviously A ≠ B
3. so if ∅ ∩ ∅ = ∅
4. doesn't it necessarily follow that ∅ ≠ ∅?

This argument seems trivial to me. Yet you are telling me its not valid. So in which step is the mistake?
Well, step 2 is simply invalid. It holds only if it is not the case that A=B=∅.

Stoney Pete

I see what you mean. But it seems ad hoc to me. In effect you are saying: we have this general rule such that If A ∩ B = ∅ then A ≠ B, but this rule comes with a clause: it holds only if A ≠ B. Isn't this circular?

On the other hand, I understand that a formal system must avoid inconsistency. Hence the said clause to said rule. Can't we say that this clause is similar in status to the rule that forbids sets being members of themselves so as to avoid Russell-type paradoxes?

Erland

Science Advisor
I see what you mean. But it seems ad hoc to me. In effect you are saying: we have this general rule such that If A ∩ B = ∅ then A ≠ B, but this rule comes with a clause: it holds only if A ≠ B. Isn't this circular?
This "rule" is not an axiom. It is just a statement which can be proved, but only if A=B=∅ does not hold, and the latter is not the same as A ≠ B, so there is no circularity.

EDIT: The condition "not-A=B=∅" is equivalent to "A≠∅ or B≠∅ (or both)".
So the statement which can be proved is: "If A∩B=∅ and (A≠∅ or B≠∅), then A≠B". This maybe makes it more clear...

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Stoney Pete

Okay, I get it now. Thanks for your explanations!

I think what I'm after is something like paraconsistent set theory, if something like that exists...

PeroK

Science Advisor
Homework Helper
Gold Member
2018 Award
But if we say that set A is disjoint with set B, aren't we then implying that A is not identical with B?
As Erland pointed out, this is a paradox entirely of your own making. There was no logical reason to assume that disjoint => not identical.

You could equally create a paradox by declaring that 2x must be different from x. And then being stumped by 2x0 = 0

Stoney Pete

You say: There was no logical reason to assume that disjoint => not identical.

Are you serious? It is a self-evident, logical truth that if A is disjoint with B, then A is not B... Or isn't it? What am I missing here?

Stoney Pete

You say: There was no logical reason to assume that disjoint => not identical.

Are you serious? It is a self-evident, logical truth that if A is disjoint with B, then A is not B... Or isn't it? What am I missing here?
That is to say: it is a logical truth as long as A and B are not both ∅.

PeroK

Science Advisor
Homework Helper
Gold Member
2018 Award
You say: There was no logical reason to assume that disjoint => not identical.

Are you serious? It is a self-evident, logical truth that if A is disjoint with B, then A is not B... Or isn't it? What am I missing here?
You are missing logic!

Stoney Pete

Haha... yes I get that a lot, especially from people in exact sciences...

Well, don't get me wrong, I have the greatest respect for exact science... But I also think that sometimes logicians, mathematicians etc. get too much caught up in their own formal constructions. If you take a Platonic view of mathematics, then the formal system is just an approximation of an independent mathematical reality, and as they say: the map aint the territory...

Take for example zero or the empty set... In axiomatized set theory it is presupposed that ∅ is self-identical, just like any set is supposed to be self-identical (i.e. obeys the law of identity). In most of mathematics the same is presupposed about zero. But under closer scrutiny does this presupposition hold up? There is a heterodox tradition in mathematics which says that zero is the infinitely smallest numer, so that infinity times zero is not zero but one (of course you have to specify here what kind of infinity you re talking about here, but I hope you get what I mean in general). So on that assumption your litle sum 2x0 doesn't equal zero... In fact in calculus and other branches of mathematics operations like this are common...

So if mathematical Platonism is right, which view is the true one: is zero just zero or zero is the smallest infinitesimal? Suppose the last view were the correct one: in that case zero is not self-identical, since zero is not zero but the infinitely small. And then another pickle arises: is zero a positive or a negative number? Since zero is equally removed from 1 and -1, one would have to say that this infinitely small number is neither positive nor negative, or both...

What I am trying to get across is that zero is a troublesome concept, which can give rise to paradoxes, just like infinity.

In formal systems, these paradoxes are to be avoided. But on a Platonic view, the formal systems are not the mathematical reality itself, they are an approximation. So if mathematical Platonism is true, then perhaps there is a bit more to zero and the empty set than you think. Of course it is your prerogative to deny Platonism, but then that's also a philosophical choice and you have to disagree with people like Goedel, Cantor, Penrose, Tegmark etc.

I don't want to offend anybody, I 'just' want to know the truth about the philosophical foundations of mathematics... So let's be polite and have a discussion centered on content, not on ad hominem arguments.

To get back to our discussion, I would still like to know why it is incorrect to say: if A is disjoint with B, then A is not B (as long as A and B are not both ∅).

Fraktoz

So you are saying A ∩ B = ∅, but you are wrong because ∅ is only when you can say that a ≠ a, you cant obtain an empty set from the intersection

Stephen Tashi

Science Advisor
So if mathematical Platonism is right,
Right in what sense? - morally?

Mathematical Platonism is a philsophy and you are attempting to discuss philosophical issues. I myself am not adverse to Philosophy but the forum is rather sensitive about having philosophical discussions in the technical sections. You might try posting in General Discussions.

h6ss

To get back to our discussion, I would still like to know why it is incorrect to say: if A is disjoint with B, then A is not B (as long as A and B are not both ∅).
There are two cases to consider.

If $A ∩ B = ∅$, then either:

CASE 1 : $A = B = ∅$, because obviously $∅ ∩ ∅ = ∅$, or

CASE 2 : $A ≠ B$, because if they were equal then at least one element (and we know for sure that at least one element exists since we aren't in the first case so both sets can't be null) would be belonging to both of them.

So if you were to say that A is disjoint with B, you would need to talk about both cases to be exact.

It seems to be the problem here in the second step:

1. If A ∩ B = ∅
2. then obviously A ≠ B
3. so if ∅ ∩ ∅ = ∅
4. doesn't it necessarily follow that ∅ ≠ ∅?
I really like PeroK's analogy. Think about it Stoney Pete:

As Erland pointed out, this is a paradox entirely of your own making. There was no logical reason to assume that disjoint => not identical.

You could equally create a paradox by declaring that 2x must be different from x. And then being stumped by 2x0 = 0

TeethWhitener

Science Advisor
Gold Member
Take for example zero or the empty set... In axiomatized set theory it is presupposed that ∅ is self-identical, just like any set is supposed to be self-identical (i.e. obeys the law of identity). In most of mathematics the same is presupposed about zero. But under closer scrutiny does this presupposition hold up?
It's very difficult to conceive of a situation where it's useful to hold that $x \neq x$. But it's easily provable that the empty set and zero elements are unique, given their properties. Under the axioms of the theory, two sets are identical iff they have the same elements. Therefore, a fortiori the empty set is unique. Likewise, for zero, it can be easily proven that zero (as an additive identity) is unique. Proof: Assume that there are two things with the property of zero (call them $0_a$ and $0_b$). Then $x+0_a = x = x+0_b$ and $0_a = 0_b$. The uniqueness of the identity element is a well-known result in group theory.
There is a heterodox tradition in mathematics which says that zero is the infinitely smallest numer,
-1 < 0. Heterodox indeed. This brings up a larger point, though, which is that in math you start with precisely formulated definitions and axioms and prove theorems from them. If you don't like the definitions and axioms you're given, you're certainly welcome to use your own. Of course, you're still bound by logic as to what you can derive from them. But I think a lot of the trouble you're having is that you're trying to use your own definitions or preconceived notions of mathematical words to come up with results that don't make intuitive sense to you. So for instance, with your initial question, $A \cap B$ is defined as the set of elements which are members of both A and B. If that set has no elements, then (as per the definition of identity of sets--aka the axiom of extensionality) it is identical with the empty set. We can now define the notion of "disjoint" to mean that A and B are disjoint iff $A \cap B = \emptyset$. Of course, since the intersection of the empty set with itself has no elements, it is also identical to the empty set. If you want to say that the empty set is disjoint with itself, you're certainly welcome to, given the definition of "disjoint" above. If you don't like it, that's your prerogative--don't use the word--but regardless of what language you put on it, it doesn't change the fact that $\emptyset \cap \emptyset = \emptyset$.
so that infinity times zero is not zero but one
$\infty \times 0$ is what's known as an indeterminate form. It isn't zero or one as stated. Indeterminate forms arise in the study of limits and there are a number of ways to evaluate them. Often when they appear, their evaluation is context specific. So, taking three examples: $$\frac{x}{x}$$ $$\frac{x^2}{x}$$ and $$\frac{x}{x^2}$$ we see that as x goes to $\infty$, the first example evaluates to 1, the second evaluates to $\infty$, and the third evaluates to zero, even though they're all basically $\infty \times \frac{1}{\infty} = \infty \times 0$. So, yes, orthodox mathematics is quite familiar with these situations and can handle them readily.
So on that assumption your litle sum 2x0 doesn't equal zero
Why? You've given no evidence as to why this statement follows if you assume that $\infty \times 0 = 1$.
Of course it is your prerogative to deny Platonism, but then that's also a philosophical choice and you have to disagree with people like Goedel, Cantor, Penrose, Tegmark etc.

I don't want to offend anybody, I 'just' want to know the truth about the philosophical foundations of mathematics... So let's be polite and have a discussion centered on content, not on ad hominem arguments.
Setting aside the fact that the first sentence is an argument from authority, I think if you want to explore some of the issues of mathematical philosophy, you list a few people whose work you might want to consult. But it's important to keep in mind that in philosophy, just as in mathematics, precision of language is of the utmost importance. Imprecision can lead to confusion: witness yourself with the concept that the empty set is self-disjoint. Because you came into the discussion with a preconceived notion of the definition of "disjoint," you had trouble understanding how a set can be disjoint with itself. Once the language is made more precise, though, (hopefully) you've seen how such a situation can arise.

pwsnafu

Science Advisor
So if mathematical Platonism is right, which view is the true one: is zero just zero or zero is the smallest infinitesimal?
There are plenty of mathematical structures that contain infinitesimals. In all of them infinitesimals are non-zero.
This is because the word "infinitesimal" is defined that way. That's the entire point of introducing the concept of infinitesimals in the first place.

Stoney Pete

Okay guys, thanks for the constructive remarks. It is much clearer to me now. As you might have guessed, formal logic & math is not my forte (in fact I have dyscalculia so I really have to struggle to understand these things).

There is, however, still one remark I would like to make. TeethWhitener mentioned the axiom of extensionality and I think that underpins what I am trying to say concerning the empty set. Consider the following argument:

1. Axiom of extensionality: sets are identical iff they have all their elements in common.
2. The empty set has no elements in common with itself (i.e. ∅∩∅=∅).
3. Hence, given 1 en 2, ∅≠∅.

So what's wrong with this reasoning? Maybe I'm saying something really stupid... If that is the case, I'm sorry for imposing on your time, but could you please explain to me where the mistake is?

Stoney Pete

Okay, I've been frantically searching the internet to learn more about this issue.

It seems that the philosophet E.J. Lowe makes a similar point:

"It is not clear to me that the empty set has well-defined identity-conditions. A set has these only to the extent that its members do - but the empty set has none." (Lowe,E.J.: 'The Possibility of Metaphysics' [OUP 2001], p.254, n.8)

For Lowe this is a reason to reject the empty set. That's not a conclusion I would draw however.

TeethWhitener

Science Advisor
Gold Member
Okay guys, thanks for the constructive remarks. It is much clearer to me now. As you might have guessed, formal logic & math is not my forte (in fact I have dyscalculia so I really have to struggle to understand these things).

There is, however, still one remark I would like to make. TeethWhitener mentioned the axiom of extensionality and I think that underpins what I am trying to say concerning the empty set. Consider the following argument:

1. Axiom of extensionality: sets are identical iff they have all their elements in common.
2. The empty set has no elements in common with itself (i.e. ∅∩∅=∅).
3. Hence, given 1 en 2, ∅≠∅.

So what's wrong with this reasoning? Maybe I'm saying something really stupid... If that is the case, I'm sorry for imposing on your time, but could you please explain to me where the mistake is?
Again, this has to do with imprecision in the language. This time it's my fault. Formally, the axiom of extensionality states:
$$\forall A \forall B (\forall x ( x \in A \longleftrightarrow x \in B) \longrightarrow A = B)$$
But for the empty set, $x \in A$ and $x \in B$ are both false for all x. Since both members of the biconditional are false, the biconditional itself evaluates to true. And since the biconditional (plus quantifier) is the antecedent of the overall conditional, the only way for the axiom to evaluate to true is if A=B also evaluates to true. Hopefully an increase in precision helps out the understanding a little bit.

Incidentally, you might be wondering why the axiom of extensionality is taken as true at all. This is, in fact, a philosophical question, and one that philosophers (especially philosophers of language) have spent a good deal of time on. I won't violate forum etiquette by delving too deeply into a philosophical discussion here, but if you want more info, it may interest you to look into intensional logic and type theory as alternatives to set theory. Briefly (and with a considerable loss of precision), extensionality is, in a sense, the notion that two objects are equal if and only if all the properties that we can know about them are equal. This works well in math, but less well in language, where reference and meaning can be separated. But there are philosophers and logicians who have explored the implications of intensionality, that is, that objects can have unobservable internal properties that make them unequal to other objects to which they are extensionally equal. Ok, done with the philosophical discussion. Good luck.

Erland

Science Advisor
1. Axiom of extensionality: sets are identical iff they have all their elements in common.
2. The empty set has no elements in common with itself (i.e. ∅∩∅=∅).
3. Hence, given 1 en 2, ∅≠∅.

So what's wrong with this reasoning? Maybe I'm saying something really stupid...
Well, not stupid, but nonetheless a logical mistake, a common one...

The point is that the sentences "The empty set has all its elements in common with itself" and "The empty set has no elements in common with itself" do not contradict each other. Both are true! The reason is simply that the empty set has no elements, thus no elements which would make either of the sentences false.

This is an example of what we call vacuously true sentences. In effect: a sentence of the type "All X are Y" is regarded as true if there are no X at all, no matter what Y is. It can be reformulated as

"For all x, if x is X then x is Y".

If we assume that there are no X, then the open sentence "If x is X, then x is Y" is true for all x, because an implication P → Q is regarded as true if P is false, whether Q is true or not. This means that "All X are Y" is true if there are no X.

This may seem strange and overly formal, but it is the most logical way of doing things. Consider, for example, the sentence:

All real numbers greater than 3 has a square which is greater than 9. (*)

This is obviuosly true, right?

We can reformulate (*) thus:

For all real numbers x: If x > 3 then x2 > 9,

or with symbols:

∀x∈ℝ: x > 3 → x2 > 9

That this sentence is true means that the open sentence

x > 3 → x2 > 9

is true for all real numbers x. This includes that the following sentences are true:

1. 5 > 3 → 52 > 9
2. -5 > 3 → (-5)2 > 9
3. 2 > 3 → 22 > 9

Most people would probably agree that 1 is true, but many have difficulties to see that 2 and 3 are true. But we must accept that 2 and 3 are true if we accept that (*) is true.
So, an implication P → Q must be considered as true if P is false.

But this implies that a sentence such as:

All real numbers who are smaller then themselves are greater then themselves (**)

is true. It can be reformulated as

∀x∈ℝ: x < x → x > x

and since x < x is false for al real numbers x, the implication x < x → x > x is true for all real numbers x, so (**) is true. We say that (**) x < x → x > x is vacuously true.

Now, the sentence "The empty set has all its elements in common with itself" can be reformulated as

∀x (x ∈ ∅ → x ∈ ∅)

and "The empty set has no elements in common with itself" as

∀x (x ∈ ∅ → x ∉ ∅)

and they are both vacuously true, since x ∉ ∅ for all x.

Your conclusion, ∅ ≠ ∅, does not follow from this, for it can be reformulated as

not-∀x (x ∈ ∅ x ∈ ∅)

which, since we have ∅ on both sides, is equivalent to

not-∀x (x ∈ ∅ → x ∈ ∅)

and this is equivalent to

∃x not-(x ∈ ∅ → x ∈ ∅),

and, since the only way for an implication P → Q to be false is if P is true and Q is false (see above), this gives

∃x ( x ∈ ∅ and x ∉ ∅)

and this is certainly false, and it does not follow from the two true sentences above.

I fully understand if you think this unintuitive and overly formal, but it is logical, and the most practical way for mathematicians to use their notions.

Fraktoz

Okay guys, thanks for the constructive remarks. It is much clearer to me now. As you might have guessed, formal logic & math is not my forte (in fact I have dyscalculia so I really have to struggle to understand these things).

There is, however, still one remark I would like to make. TeethWhitener mentioned the axiom of extensionality and I think that underpins what I am trying to say concerning the empty set. Consider the following argument:

1. Axiom of extensionality: sets are identical iff they have all their elements in common.
2. The empty set has no elements in common with itself (i.e. ∅∩∅=∅).
3. Hence, given 1 en 2, ∅≠∅.

So what's wrong with this reasoning? Maybe I'm saying something really stupid... If that is the case, I'm sorry for imposing on your time, but could you please explain to me where the mistake is?
As I told before: Cannot obtain an empty set from the intersection

So you are saying A ∩ B = ∅, but you are wrong because ∅ is only when you can say that a ≠ a, you cant obtain an empty set from the intersection

Stoney Pete

Okay, so that's what I was looking for! Of course, I should have remembered my classes formal logic and the truth tables of the logical constants.... Well, the matter has been totaly clarified for me, I want to thank you all very much for your time and patience! I learned a lot.

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