MHB Is the Intersection of the Empty Set a Set?

[evinda]

Hi! (Cool)

I want to show that $\cap \varnothing$ is not a set.

That's what I have tried so far:

We suppose that $\cap \varnothing$ is a set.
Let $x \in \cap \varnothing$. Then, $\forall b \in \varnothing, x \in b$.
However, $\varnothing$ does not contain any element.
So, we cannot find a $b \in \varnothing$ such that $x \in b$.

Is it right so far? (Thinking)

How can I continue? Do I have to use maybe the following theorem?

If $\phi$ is a type and there is a set $Y$, such that:

$$\forall x(\phi(x)) \rightarrow x \in Y$$

then, the set $\{x: \phi(x) \}$ exists.

If so, how can I use this? (Thinking)

 

[Evgeny.Makarov]

I want to show that $\cap \varnothing$ is not a set.

Wikipedia has an explanation. In order to grasp it, you need to understand why $\forall x\in\varnothing\;P(x)$ is true regardless of the formula $P(x)$. This us because this statement stands for
\[
\forall x\;(x\in\varnothing\to P(x))
\]
and the implication whose premise is false is true by definition.

We suppose that $\cap \varnothing$ is a set.
Let $x \in \cap \varnothing$. Then, $\forall b \in \varnothing, x \in b$.
However, $\varnothing$ does not contain any element.
So, we cannot find a $b \in \varnothing$ such that $x \in b$.

The universal quantification does not require us to find any $x\in b$. That would be necessary if we had $\exists x\in b$.

 

[evinda]

Wikipedia has an explanation. In order to grasp it, you need to understand why $\forall x\in\varnothing\;P(x)$ is true regardless of the formula $P(x)$. This us because this statement stands for
\[
\forall x\;(x\in\varnothing\to P(x))
\]
and the implication whose premise is false is true by definition.

Could you explain me further why $\forall x\in\varnothing\;P(x)$ is true regardless of the formula $P(x)$ ? (Worried)

 

[Evgeny.Makarov]

Could you explain me further why $\forall x\in\varnothing\;P(x)$ is true regardless of the formula $P(x)$ ?

Hmm, I wrote that "the implication whose premise is false is true by definition". Could you say more explicitly what you don't understand?

 

[Nono713]

Could you explain me further why $\forall x\in\varnothing\;P(x)$ is true regardless of the formula $P(x)$ ? (Worried)

Some reading.

 

[evinda]

Hmm, I wrote that "the implication whose premise is false is true by definition". Could you say more explicitly what you don't understand?

Why when we have this:

$$
\forall x\;(x\in\varnothing\to P(x))
$$

and $x \in \varnothing$ is not true, then $P(x)$ is true? (Worried)

 

[Evgeny.Makarov]

I did not say that the fact that $x\in\varnothing$ is false implies that $P(x)$ is true. I said that it implies that the whole implication $x\in\varnothing\to P(x)$ is true.

You can check the truth table of implication. It agrees with common sense. For example, according to the US constitution, a president must be at least thirty-five years old. Thus, the statement "For any person $x$, if $x$ is a president of the USA, he or she is at least 35 years old" has been true since the Constitution is in action. In particular, it is true about Peyton Manning (a famous American football player). He is not a president, so the premise is false. However, he is over 35, so the conclusion is true. The complete implication is true. On the other hand, take Serena Williams (currently No. 1 in women's singles tennis). She is also not a president, so the premise is false, but she is younger than 35, so the conclusion is false. Yet the whole implication is again true. Thus, regardless of the truth of the conclusion, if the premise is false, the whole implication is true.

 

[evinda]

I did not say that the fact that $x\in\varnothing$ is false implies that $P(x)$ is true. I said that it implies that the whole implication $x\in\varnothing\to P(x)$ is true.

You can check the truth table of implication. It agrees with common sense. For example, according to the US constitution, a president must be at least thirty-five years old. Thus, the statement "For any person $x$, if $x$ is a president of the USA, he or she is at least 35 years old" has been true since the Constitution is in action. In particular, it is true about Peyton Manning (a famous American football player). He is not a president, so the premise is false. However, he is over 35, so the conclusion is true. The complete implication is true. On the other hand, take Serena Williams (currently No. 1 in women's singles tennis). She is also not a president, so the premise is false, but she is younger than 35, so the conclusion is false. Yet the whole implication is again true. Thus, regardless of the truth of the conclusion, if the premise is false, the whole implication is true.

A ok... (Thinking)

And, how do we use the fact that the implication $x\in\varnothing\to P(x)$ is true, in order to show that $\cap \varnothing$ is not a set? (Thinking)

 

[Evgeny.Makarov]

And, how do we use the fact that the implication $x\in\varnothing\to P(x)$ is true, in order to show that $\cap \varnothing$ is not a set? (Thinking)

Have you read the Wikipedia section linked in post #2?

 

[evinda]

Have you read the Wikipedia section linked in post #2?

Yes, I did.. (Nod) Is there also an other way, to explain why $\cap \varnothing$ is not a set? (Thinking)

 

[Evgeny.Makarov]

Well, there is the real reason and the technical one. The real reason is described in Wikipedia. The technical reason is that $\bigcap\varnothing$ cannot be defined in axiomatic set theory. Or, rather, the question should be addressed to the person who is asking why $\bigcap\varnothing$ is not a set: what do you mean by $\bigcap\varnothing$? In fact, it would be nice if you gave the context of this question: are you expected to give an informal explanation why defining $\bigcap\varnothing$ is a bad idea (which is what that article in Wikipedia does), or do you need to prove something precise, such as that $\bigcap\varnothing$ is a proper class?

Axioms in set theory provide ways of building sets. All definitions of sets must go through these constructions. As you said, if we simply have some property $P(x)$, we cannot conclude that $\{x:P(x)\}$ is a set. More formally, we cannot conclude
\[
\exists z\;\forall x\;x\in z\leftrightarrow P(x).\qquad(*)
\]
The statement you quoted in post #1 allows concluding (*) under the assumption that all objects satisfying $P(x)$ belong to something that has already been shown to be a set. Concerning $\bigcap\varnothing$, all we have is a property $P(x)$ that elements of $\bigcap\varnothing$ are supposed to satisfy; namely, $P(x)$ is $\forall y\;y\in\varnothing\to x\in y$. Nothing in set theory axioms allows us to conclude (*). Compare this with the axiom of union, which, given any $u$, allows concluding
\[
\exists z\;\forall x\;x\in z\leftrightarrow Q(x)
\]
where $Q(x)$ is $\exists y\;y\in u\land x\in y$. But the real reason the axiom of intersection is not added to set theory is that otherwise it would become contradictory and any statement at all would follow from it.

P.S.

Is there also an other way, to explain why $\cap\varnothing$ is not a set?

I was tempted to answer this question like professor N. in the following anecdote found here.

N. had the habit of simply writing answers to homework assignments on the board (the method of solution being, of course, obvious) when he was asked how to solve problems. One time one of his students tried to get more helpful information by asking if there was another way to solve the problem. N. looked blank for a moment, thought, and then answered, "Yes".

 

[evinda]

Well, there is the real reason and the technical one. The real reason is described in Wikipedia. The technical reason is that $\bigcap\varnothing$ cannot be defined in axiomatic set theory. Or, rather, the question should be addressed to the person who is asking why $\bigcap\varnothing$ is not a set: what do you mean by $\bigcap\varnothing$? In fact, it would be nice if you gave the context of this question: are you expected to give an informal explanation why defining $\bigcap\varnothing$ is a bad idea (which is what that article in Wikipedia does), or do you need to prove something precise, such as that $\bigcap\varnothing$ is a proper class?

After having done the following sentence and definition:

Sentence:

Let $A \neq \varnothing$. There is a unique set, of which the elements are exactly these one, that belong to all the elements of $A$.

Definition:

If $A \neq \varnothing$, then the set of the previous sentence is called generalized intersection of $A$ and is symbolized to $\cap A$.

the prof gave as the exercise:

Show that $\cap \varnothing$ is not a set.

Axioms in set theory provide ways of building sets. All definitions of sets must go through these constructions. As you said, if we simply have some property $P(x)$, we cannot conclude that $\{x:P(x)\}$ is a set. More formally, we cannot conclude
\[
\exists z\;\forall x\;x\in z\leftrightarrow P(x).\qquad(*)
\]
The statement you quoted in post #1 allows concluding (*) under the assumption that all objects satisfying $P(x)$ belong to something that has already been shown to be a set. Concerning $\bigcap\varnothing$, all we have is a property $P(x)$ that elements of $\bigcap\varnothing$ are supposed to satisfy; namely, $P(x)$ is $\forall y\;y\in\varnothing\to x\in y$. Nothing in set theory axioms allows us to conclude (*). Compare this with the axiom of union, which, given any $u$, allows concluding
\[
\exists z\;\forall x\;x\in z\leftrightarrow Q(x)
\]
where $Q(x)$ is $\exists y\;y\in u\land x\in y$. But the real reason the axiom of intersection is not added to set theory is that otherwise it would become contradictory and any statement at all would follow from it.

So, can't we use the definition of the generalized intersection, in order to show that $\cap \varnothing$ is not a set? (Thinking)

P.S.
I was tempted to answer this question like professor N. in the following anecdote found here.

N. had the habit of simply writing answers to homework assignments on the board (the method of solution being, of course, obvious) when he was asked how to solve problems. One time one of his students tried to get more helpful information by asking if there was another way to solve the problem. N. looked blank for a moment, thought, and then answered, "Yes".

(Bigsmile)

 

[Evgeny.Makarov]

First let me make a couple of general remark. First, what does it mean to show that an object $x$ satisfying property $\psi$ does not exist? It means that assuming $\exists x\;\psi(x)$ leads to a contradiction. Contradiction is often denoted by the symbol $\bot$, and it can be thought of, for example, as the conjunction of some formula and its negation: $\chi\land\neg\chi$. The important point is that a contradiction is always false and implies everything, as was discussed in post #7. So if some theory implies a contradiction, then it is useless because it implies every single statement. In ordinary mathematics, we would like to derive only true statements.

Second, once it was discovered that not all collections are sets (since if they were, that would lead to a contradiction), we must be careful which collections of objects we call sets. We could take an arbitrary formula $\varphi(x)$ with one free variable (maybe your textbook calls it a type judging by post #1) and consider all objects $x$ that satisfy it. These objects do not automatically form a set. How do we say formally that they do? We say that there is some object $z$ such that those $x$ that satisfy $\varphi(x)$ and only they are in the relationship $\in$ with $z$. That is, the collection of object satisfying $\varphi(x)$ is a set if
\[
\exists z\;\forall x\;x\in z\leftrightarrow\varphi(x).
\]
So from the standpoint of set theory, to form a set is to exist. The following two statements mean the same:

• There exists an object $z$ that contains all $x$ satisfying $\varphi(x)$
• All $x$ satisfying $\varphi(x)$ form a set.

(There are some variations of set theory where not everything is a set: there are also so-called classes.) Please think about this and make sure you understand this definition.

Combining these two remarks, what does it mean that the set of all sets does not exist, or, equivalently, that all sets do not form a set? It means
\[
(\exists z\,\forall x\;x\in z)\to\bot.\qquad(1)
\]
And indeed, assuming $\exists z\,\forall x\;x\in z$, we take this $z$ and using the subset axiom form a set $\{x\mid x\notin x\}$ and use it to derive a contradiction.

So, can't we use the definition of the generalized intersection, in order to show that $\cap \varnothing$ is not a set?

Yes. Let $A$ be fixed. We take the property $\varphi(x)$ from the definition of the generalized intersection: namely, $\varphi(x)$ is $\forall y\;y\in A\to x\in y$. We want to show that when $A=\varnothing$, all $x$ satisfying $\varphi(x)$ do not form a set, i.e.,
\[
(\exists z\;\forall x\;x\in z\leftrightarrow\varphi(x))\to\bot.\qquad(2)
\]
Well, suppose
\[
\exists z\;\forall x\;x\in z\leftrightarrow\varphi(x)\qquad(3)
\]
But $\varphi(x)$ is $\forall y\;y\in\varnothing\to x\in y$. Since $y\in\varnothing\leftrightarrow\bot$, we have $(y\in\varnothing\to x\in y)\leftrightarrow\top$, where $\top$ denotes a sentence that is always true ($\top$ can be thought of as $\neg\bot$). In other words, $y\in\varnothing\to x\in y$ and its universal closure are always true. So (3) says
\[
\exists z\;\forall x\;x\in z\leftrightarrow\top
\]
or simply
\[
\exists z\;\forall x\;x\in z.
\]
Then we use (1) to derive $\bot$. Thus, (2) is proved.

 

[evinda]

Combining these two remarks, what does it mean that the set of all sets does not exist, or, equivalently, that all sets do not form a set? It means
\[
(\exists z\,\forall x\;x\in z)\to\bot.\qquad(1)
\]
And indeed, assuming $\exists z\,\forall x\;x\in z$, we take this $z$ and using the subset axiom form a set $\{x\mid x\notin x\}$ and use it to derive a contradiction.

So, can we begin, saying that if $\cap \varnothing$ is a set, then it is the set of all sets? If so, why is it like that? (Worried)

 

[Evgeny.Makarov]

So, can we begin, saying that if $\cap \varnothing$ is a set, then it is the set of all sets?

That's a good way to put it.

If so, why is it like that?

Every answer post in this thread tries to explain that.

 

[evinda]

So, can we explain that $\cap \varnothing$ is not a set, like that? (Thinking)

From the definition, we have that:

$$\cap \varnothing = \{x: (\forall b \in \varnothing) x \in b\}$$

Let $x$ be an element.
We will show that $x \in \cap \varnothing, \forall x$.
The necessary and sufficient condition is:
$x \in b \text{ for all } b \in \varnothing$.
An other way to write this condition is the following implication:$$\forall b (b \in \varnothing \rightarrow x \in b)$$Let $A: b \in \varnothing$ and $B: x \in b$

So, we have to show that the implication $A \rightarrow B$ is true, for each $b$.
The implication can also be written as:

$$\neg A \lor B$$

In words: $\text{not } A \text{ or } B$.

$A$ is always false, because $b \in \varnothing$ does not hold for any $b$.
So, $\neg A$ is always true, therefore $\neg A \lor B$ is also always true.

So, $ V=\cap \varnothing $ is the set of all sets.

We define the identity $\phi: \text{ a set does not belong to itself, so : } x \notin x $.

From the axiom shema of specification, the set $V'=\{ x \in V: x \notin x\}$ exists.

So, we have $V' \in V' \leftrightarrow V' \notin V'$, Contradiction.

Therefore, $\cap \varnothing$ cannot be a set.

 

[Evgeny.Makarov]

You got the idea, though writing $\{x: (\forall b \in \varnothing) x \in b\}$ is not legitimate even if you are trying to prove by contradiction that this set does not exist. I showed a cleaner way in post #13. Also, this line:

We define the identity $\phi: \text{ a set does not belong to itself, so : } x \notin x $.

should be removed. It is not needed later, and $\phi$ is not an identity. I may add another remark later.

 

[evinda]

You got the idea, though writing $\{x: (\forall b \in \varnothing) x \in b\}$ is not legitimate even if you are trying to prove by contradiction that this set does not exist. I showed a cleaner way in post #13. Also, this line:

should be removed. It is not needed later, and $\phi$ is not an identity. I may add another remark later.

From the definition, we have that:

$$\cap \varnothing = \{x: (\forall b \in \varnothing) x \in b\}$$

Let $x$ be an element.
We will show that $x \in \cap \varnothing, \forall x$.
The necessary and sufficient condition is:
$x \in b \text{ for all } b \in \varnothing$.
An other way to write this condition is the following implication:$$\forall b (b \in \varnothing \rightarrow x \in b)$$Let $A: b \in \varnothing$ and $B: x \in b$

So, we have to show that the implication $A \rightarrow B$ is true, for each $b$.
The implication can also be written as:

$$\neg A \lor B$$

In words: $\text{not } A \text{ or } B$.

$A$ is always false, because $b \in \varnothing$ does not hold for any $b$.
So, $\neg A$ is always true, therefore $\neg A \lor B$ is also always true.

So, $ V=\cap \varnothing $ is the set of all sets.

We define the identity $\phi: \text{ a set does not belong to itself, so : } x \notin x $.

From the axiom shema of specification, the set $V'=\{ x \in V: x \notin x\}$ exists.

So, we have $V' \in V' \leftrightarrow V' \notin V'$, Contradiction.

Therefore, $\cap \varnothing$ cannot be a set.

I am confused now.. (Worried) Don't we suppose at the beginning that $\cap \varnothing$ is a set? How else could we write it, rather than using the definition? (Sweating)

Don't we use the identity $\phi$ at the axiom shema of specification? Or don't we have to write it, since we don't say how we use it? (Thinking)

 

[Evgeny.Makarov]

Don't we suppose at the beginning that $\cap \varnothing$ is a set? How else could we write it, rather than using the definition?

I answered this question in post #13. First, I wrote what it means for any collection of objects satisfying $\varphi(x)$ to be set.

the collection of object satisfying $\varphi(x)$ is a set if
\[
\exists z\;\forall x\;x\in z\leftrightarrow\varphi(x).
\]

Second, I wrote what it means that the set of all sets does not exist.

what does it mean that the set of all sets does not exist, or, equivalently, that all sets do not form a set? It means
\[
(\exists z\,\forall x\;x\in z)\to\bot.\qquad(1)
\]

Finally, to answer your specific question: what does it mean that $\cap \varnothing$ is a not set?

We take the property $\varphi(x)$ from the definition of the generalized intersection: namely, $\varphi(x)$ is $\forall y\;y\in A\to x\in y$. We want to show that when $A=\varnothing$, all $x$ satisfying $\varphi(x)$ do not form a set, i.e.,
\[
(\exists z\;\forall x\;x\in z\leftrightarrow\varphi(x))\to\bot.\qquad(2)
\]

I even asked you to think about this and to understand what it means, formally, for some collection of objects satisfying $\varphi$ to form a set.

Frankly, this ignoring of pretty detailed explanations and asking the same question is a bad tone. And you did it repeatedly in this thread. I understand that some explanations may seem incomprehensible at first, but the solution is not to ignore them, but to ask detailed additional questions, saying specifically what you do and don't understand.

That said, even though I believe it is formally incorrect to say, "Assume that $\{x:\forall y\;y\in\varnothing\to x\in y\}$ is a set" (without interpreting what you mean by this), it is probably a minor point and is acceptable at the level of your course.

Don't we use the identity $\phi$ at the axiom shema of specification? Or don't we have to write it, since we don't say how we use it?

The word "identity" in mathematics has two meanings that come to mind: an equality that always holds, and the identity function $f(x)=x$. Neither interpretation applies here. Besides, you don't use $\phi$ after you define it in post #16, so it is unnecessary.