A question about intersection of system of sets

[mahmoud2011]

The book I am reading says that $\bigcap \phi$ because every $x$ belongs to $A \in \phi$(since there is no such A ) , so $\bigcap S$ would have to be the set of all sets. now my question is why every $x$ belongs to $A \in \phi$.In other word I don't completely understand what this statement mean.

sorry if my question is silly.I began reading set theory for only 12 hours and I am new to the subject , I don't know if I had to take a course in logic first or not. If I must know logic , please recommend me to a book in logic .

 

[micromass]

The book I am reading says that $\bigcap \phi$ because every $x$ belongs to $A \in \phi$(since there is no such A ) , so $\bigcap S$ would have to be the set of all sets. now my question is why every $x$ belongs to $A \in \phi$.In other word I don't completely understand what this statement mean.

sorry if my question is silly.I began reading set theory for only 12 hours and I am new to the subject , I don't know if I had to take a course in logic first or not. If I must know logic , please recommend me to a book in logic .

Well, can you give me a $A\in \emptyset$ such that $x\notin A$?? You can't because there is not $A\in \emptyset$. So since you cannot give an example of an $A\in \emptyset$ such that $x\notin A$, must mean that $x\in A$ for all $A\in \emptyset$.Also, please note that many authors leave $\bigcap \emptyset$ undefined. Why? Because there is not set that contains all sets.

 

[mahmoud2011]

Well, can you give me a $A\in \emptyset$ such that $x\notin A$?? You can't because there is not $A\in \emptyset$. So since you cannot give an example of an $A\in \emptyset$ such that $x\notin A$, must mean that $x\in A$ for all $A\in \emptyset$.


Also, please note that many authors leave $\bigcap \emptyset$ undefined. Why? Because there is not set that contains all sets.

Thanks
Now I used the Axiom of schema to prove that $\bigcap S$ exists for all S except
$S = \emptyset$, will that proof will be right

 

[mahmoud2011]

I forgot , Must I take a course in logic ?

 

[xxxx0xxxx]

Mahmoud,

In ZF,
$$\bigcap \emptyset = \emptyset$$

since the set of all sets does not exist in ZF, and
$$\forall B (B \in \emptyset \Rightarrow x \in B)$$
is vacuously true.

In NBG, however,
$$\bigcap \emptyset = \{ x|x=x \}$$

 

[micromass]

Mahmoud,

In ZF,
$$\bigcap \emptyset = \emptyset$$

This is false. In ZF, the intersection is undefined. It is not the empty set.

 

[xxxx0xxxx]

This is false. In ZF, the intersection is undefined. It is not the empty set.

Micromass, quit making blanket statements, this is shown in many texts on ZF

 

[micromass]

Micromass, quit making blanket statements, this is shown in many texts on ZF

Name one.

 

[micromass]

While I'm waiting for you to name some of the texts, let me look at my own texts:

- "Introduction to set theory" by Hrbacek and Jech. Page 15, exercise 4.6: states that the intersection $\bigcap S$ is defined if $S\neq \emptyset$.
- "Set Theory" by Jech states the same thing at the bottom of page 8
- "Set Theory: an introduction to independence proofs" by Kunen states it at page 13. It als states that $\bigcap \emptyset$ "should be" the set of all set, which does not exist. Thus we leave it undefined.
- "Lectures in Logic and set theory" by Tourlakis is a bit slower and takes until page 154. Proposition III.6.14.

 

[xxxx0xxxx]

While I'm waiting for you to name some of the texts, let me look at my own texts:

- "Introduction to set theory" by Hrbacek and Jech. Page 15, exercise 4.6: states that the intersection $\bigcap S$ is defined if $S\neq \emptyset$.
- "Set Theory" by Jech states the same thing at the bottom of page 8
- "Set Theory: an introduction to independence proofs" by Kunen states it at page 13. It als states that $\bigcap \emptyset$ "should be" the set of all set, which does not exist. Thus we leave it undefined.
- "Lectures in Logic and set theory" by Tourlakis is a bit slower and takes until page 154. Proposition III.6.14.

How about this:

The set of all set's does not exist in ZF: $$\neg \exists A \forall x (x \in A)$$

Thus $$\{ x | x=x \} = \emptyset$$

Suppose $$\bigcap \emptyset \not = \emptyset$$

Then $$\exists x (x \in \bigcap \emptyset)$$

But since, vacuously, $$\bigcap \emptyset = \{x | \forall B (B \in \emptyset \Rightarrow x \in B ) \} = \{ x |x = x \}$$
$$\exists x (x \in \emptyset)$$

which is a contradiction,

qed.

 

[micromass]

Thus $$\{ x | x=x \} = \emptyset$$

No, this is not correct. The left-hand side is not defined in ZF. That there is not set of sets, does not mean that the set of sets is empty. I don't know where you get such a thing?

Check your separation axiom, which states that set-builder notation must have the form

$$\{x\in A~\vert~P(x)\}$$

The set of sets cannot be written as such, and thus does not exist.

I suggest you take a book on ZF set theory and study it.

 

[micromass]

this is shown in many texts on ZF

I'm still waiting for one such text.

 

[xxxx0xxxx]

No, this is not correct. The left-hand side is not defined in ZF. That there is not set of sets, does not mean that the set of sets is empty. I don't know where you get such a thing?

Check your separation axiom, which states that set-builder notation must have the form

$$\{x\in A~\vert~P(x)\}$$

The set of sets cannot be written as such, and thus does not exist.

I suggest you take a book on ZF set theory and study it.

I beg to differ, $$P(x) = (x=x)$$ is a perfectly valid wff.

Since, it is a valid wff, it can occur in set builder notation: but it is a contradiction in ZF, and not a contradiction in NBG.

Since it is a contradiction in ZF it must empty, but in NBG it is the universal set.

 

[micromass]

I beg to differ, $$P(x) = (x=x)$$ is a perfectly valid wff.

True, it is a perfectly valid wff.

Since, it is a valid wff, it can occur in set builder notation:

In ZF, it can occur in set builder notation. But $\{x~\vert~x=x\}$ is not a valid set builder notation. Check the separation axiom.

Since it is a contradiction in ZF it must empty, but in NBG it is the universal set.

In NBG, it is indeed the universal class, but in ZF it does not exist. Does not exist is not the same as empty. Saying that

$$\{x~\vert~x=x\}=\emptyset$$

is fundamentally wrong. Since it would mean that an element of the left-hand side is an element of the right-hand side. But every set is an element of the left-hand site. That is, for every x it holds that x=x. But no set is an element of $\emptyset$.
The only problem is that the left-hand side is undefined. That is: it isn't a set. So it can not equal the empty set.

Also, I ask again to list one of the "many books" that you're getting your information from.

 

[xxxx0xxxx]

True, it is a perfectly valid wff.



In ZF, it can occur in set builder notation. But $\{x~\vert~x=x\}$ is not a valid set builder notation. Check the separation axiom.



In NBG, it is indeed the universal class, but in ZF it does not exist. Does not exist is not the same as empty. Saying that

$$\{x~\vert~x=x\}=\emptyset$$

is fundamentally wrong. Since it would mean that an element of the left-hand side is an element of the right-hand side. But every set is an element of the left-hand site. That is, for every x it holds that x=x. But no set is an element of $\emptyset$.
The only problem is that the left-hand side is undefined. That is: it isn't a set. So it can not equal the empty set.

Also, I ask again to list one of the "many books" that you're getting your information from.

In ZF, we deal with sets not "undefined," either a set exists, or it doesn't.

Is "undefined" a set? I think not.

if {x|x=x} is "undefined," then what is it?

it is either empty or it is not, and since it cannot contain any elements (otherwise it would be the universal set), it must be empty, not "undefined."

if we allow {x|x=x} to be Undefined, then:

$$\exists C \forall x(x \in C \Leftrightarrow x \in "Undefined" \wedge \phi (x))$$

Tell what does this separation axiom mean in ZF?

It doesn't mean anything, because then you are accepting another object into the theory that is neither set nor $$\emptyset$$.

 

[micromass]

In ZF, we deal with sets not "undefined," either a set exists, or it doesn't.

Is "undefined" a set? I think not.

if {x|x=x} is "undefined," then what is it?

it is either empty or it is not, and since it cannot contain any elements (otherwise it would be the universal set), it must be empty, not "undefined."

if we allow {x|x=x} to be Undefined, then:

$$\exists C \forall x(x \in C \Leftrightarrow x \in "Undefined" \wedge \phi (x))$$

Tell what does this separation axiom mean in ZF?

It doesn't mean anything, because then you are accepting another object into the theory that is neither set nor $$\emptyset$$.

You seem to have quite some misunderstandings here. The thing is that the notation is $\{x~\vert~x=x\}$ is undefined. The notation refers to something that does not exist!

Indeed, the set of all sets does not exist. That is:

$$\exists x:~\forall y:~y\in x$$

is a false statement in ZF. There is no such x, such an x does not exist.

And since such an x does not exist, this means that the notation $\{x~\vert~x=x\}$ does not refer to a valid set. Thus the notation is undefined.

You seem to think that "undefined" = "empty set". This is not true at all.

It's a bit like the notation $\frac{1}{0}$. There does not exist an x such that $0x=1$. Therefore the notation is undefined. And the quotient does not exist.

Please read a book on elementary logic.

 

[xxxx0xxxx]

You seem to have quite some misunderstandings here. The thing is that the notation is $\{x~\vert~x=x\}$ is undefined. The notation refers to something that does not exist!

Indeed, the set of all sets does not exist. That is:

$$\exists x:~\forall y:~y\in x$$

is a false statement in ZF. There is no such x, such an x does not exist.

And since such an x does not exist, this means that the notation $\{x~\vert~x=x\}$ does not refer to a valid set. Thus the notation is undefined.

You seem to think that "undefined" = "empty set". This is not true at all.

It's a bit like the notation $\frac{1}{0}$. There does not exist an x such that $0x=1$. Therefore the notation is undefined. And the quotient does not exist.

Please read a book on elementary logic.

Ah... Grasshopper, to introduce a new object in the theory of ZF is taboo. "Undefined" is not a constant of the theory of ZF, amongst whose constants, "undefined" may not be found.

Since "undefined" is not a constant of the object language, it cannot appear in any wff of the object language unless by definition. Therefore "undefined" must be a set (whose definition is "undefined") and the separation axiom requires C to be the set of objects which are "undefined" and satisfy Phi(x).

 

[micromass]

Ah... Grasshopper, to introduce a new object in the theory of ZF is taboo. "Undefined" is not a constant of the theory of ZF, amongst whose constants, "undefined" may not be found.

Since "undefined" is not a constant of the object language, it cannot appear in any wff of the object language unless by definition. Therefore "undefined" must be a set (whose definition is "undefined") and the separation axiom requires C to be the set of objects which are "undefined" and satisfy Phi(x).

You seem to be understanding me. I said that the notation

$$\{x~\vert~x=x\}$$

is undefined. This is ok, since this notation is an object of the metatheory. So "undefined" is not an element of the object language.

I simply say that

$$\exists x:~\forall y:~y\in x$$

is false in ZF. Any problems with that??

Again: please cite your sources. You say that many texts support you, so please give me one. You seem to be unable to do so.

 

[xxxx0xxxx]

You seem to be understanding me. I said that the notation

$$\{x~\vert~x=x\}$$

is undefined. This is ok, since this notation is an object of the metatheory. So "undefined" is not an element of the object language.

I simply say that

$$\exists x:~\forall y:~y\in x$$

is false in ZF. Any problems with that??

Again: please cite your sources. You say that many texts support you, so please give me one. You seem to be unable to do so.

I need cite no sources except the axioms of ZF and the object language, the definitions you have quoted specifically exclude
$$S= \emptyset$$ from the definition of $$\bigcap S$$ which is a perfectly correct definition for all cases except $$S= \emptyset$$

This in no way means $$\bigcap \emptyset = \emptyset$$ is false. Their definitions have already excluded $$\emptyset$$ from consideration (and unwittingly added a new constant to the object language). Or in other words, it is the definition that is undefined for $$S= \emptyset$$ not $$\bigcap \emptyset$$

$$\bigcap \emptyset = \emptyset$$ under the following definition:

$$\bigcap S = \{x | \forall B(B \in S \Rightarrow x \in B) \}$$

which does not exclude the case: $$S = \emptyset$$

 

[micromass]

$$\bigcap S = \{x | \forall B(B \in S \Rightarrow x \in B) \}$$

This is not a valid set builder notation. Review the axiom of separation and try again.

Please provide:

1) A formal proof that $\{x~\vert~x=x\}$ exists.

2) One of your "many texts".

 

[xxxx0xxxx]

This is not a valid set builder notation. Review the axiom of separation and try again.

Please provide:

1) A formal proof that $\{x~\vert~x=x\}$ exists.

2) One of your "many texts".

Oh my goodness

$$\forall B (B \in S \Rightarrow x \in Bf)$$ is a wff in which x is free and is vacuously true for $$S= \emptyset$$.

Since,

$$\{ x| x=x \} = \emptyset$$, it exists because the empty set exists.

You are asking for a proof of:

$$\{ x| x=x \} = \emptyset$$

which follows directly from the non-existence of the universal set and the axiom of separation:

$$\exists C \forall x (x \in C \Leftrightarrow x \in \{ x | x=x \} \wedge x \not \in x)$$

A contradiction,

which reduces to russell's paradox:

$$\exists C \forall x (x \in C \Leftrightarrow x \not \in x)$$

if


$$\{ x | x=x \} \not = \emptyset$$

Now I have work to do...

tata

 

[micromass]

$$\exists C \forall x (x \in C \Leftrightarrow x \in \{ x | x=x \} \wedge x \not \in x)$$

Not a valid use of seperation. Why not?? Because you did not yet shown $\{x~\vert~x=x\}$ to be a set. And the axiom of separation requires it to be a set.Please provide sources.

 

[xxxx0xxxx]

Not a valid use of seperation. Why not?? Because you did not yet shown $\{x~\vert~x=x\}$ to be a set. And the axiom of separation requires it to be a set.


Please provide sources.

My bad,

$$\{x|x=x\} = w \Leftrightarrow [ \forall x(x \in w \Leftrightarrow x=x) \wedge w \mbox{ is a set}] \vee [\neg \exists B \forall x(x \in B \Leftrightarrow x=x) \wedge w = \emptyset ]$$

Which covers all possible cases (of course), and uses a previously posted and perfectly valid definition of set in the object language that I know you hate ;) :

$$w \mbox{ is a set} \Leftrightarrow \exists x (x \in w \vee w = \emptyset)$$

And trust me, this definition really does deliver:

$$\{x |x=x \} = \emptyset$$ using the non-existence theorem of the universal set.

That you can work out for yourself since you have the full set of definitions that eliminate completely to object language, since now I really have to be going...


tata

 

[micromass]

Oh, I'm sorry then. You seem to be making up your own definitions and axioms at this state. Well, I'm sure that in your world, you are entirely correct!

 

[xxxx0xxxx]

Oh, I'm sorry then. You seem to be making up your own definitions and axioms at this state. Well, I'm sure that in your world, you are entirely correct!

Don't give up micromass you're almost there. It took me awhile to get it right.

That definition is a example of "definition by abstraction," which is fundamental to defining operations and proving the theorems of setbuilder theory.

It states what must be if $$\{ x| x=x \}$$ exists, and what must be if $$\{ x|x=x \}$$ doesn't exist,

it makes w unique if the set exists, empty if the set does not exist; w is fully derivable from the axiom of abstraction, which is incorporated in the definiens of the {...|...} operator.

Note that this definition is not immune to Russell's paradox, for that one still needs separation. If it exists by separation, then it exists by abstraction, but not the other way around (the paradox intervenes)

Because the set of all sets doesn't exist, this definition makes the set {x | x=x } empty, (requiring "w is a set" makes { ...|...} a set).

This covers all the bases for consistency, eliminability, non-creativity, and uniqueness.

 

[micromass]

It took me a couple of years to get it right.

Sadly, it will take you some more time

Let me explain the confusion that we have. You are accepting Suppes as a source for ZF set theory. I failed to see this before, because I never really payed attention to Suppes (since he's no mathematician, which I think is obvious from his book). And in Suppes, we indeed prove that $\{x~\vert~x=x\}=\emptyset$ is a valid statement. And it's correct too! In the sense that this follows immediately from the axioms and definitions that he uses.

The problem however is that Suppes unfortunetaly does something which is highly unstandard. The problem is (I think) in his separation axiom which is quite different from the standard separation axiom. I'm not saying that Suppes is wrong, but I am saying that no single mathematician does it the way Suppes indicates. We simply accept other axioms!

The axioms which I am referring to can be found in any book on set theory. For example, the bible of set theory: "Set theory" by Jech. Or staff.science.uva.nl/~vervoort/AST/ast.ps which does requires you to be able to read ps-files.
These two books are written by professional set theorists. And those people certainly do know what they're talking about. And according to the axioms and definitions that they use, we do not have that $\{x~\vert~x=x\}=\emptyset$.

So perhaps we should say that we are both right since we are working with different axiom systems...

 

[micromass]

OK, it is DEFINITION SCHEMA 11 on p34 that is nonstandard. All the authors (except Suppes) define

$$y\in \{x~\vert~\varphi(x)\}~\leftrightarrow~\varphi(y)$$

Thus like THEOREM SCHEMA 47 but with an equivalence this time.
You can check all the books, everybody will do it this way. Suppes does it another way and this was the cause for all the confusion.

 

[xxxx0xxxx]

OK, it is DEFINITION SCHEMA 11 on p34 that is nonstandard. All the authors (except Suppes) define

$$y\in \{x~\vert~\varphi(x)\}~\leftrightarrow~\varphi(y)$$

Thus like THEOREM SCHEMA 47 but with an equivalence this time.
You can check all the books, everybody will do it this way. Suppes does it another way and this was the cause for all the confusion.

It's not a case of standard or non-standard, Suppes defines what happens when the set doesn't exist, uniquely, in terms of the object language. The "standard" definition leaves out non-existence, and cannot account for it.

 

[micromass]

It's not a case of standard or non-standard, Suppes defines what happens when the set doesn't exist, uniquely, in terms of the object language. The "standard" definition leaves out non-existence, and cannot account for it.

This discussion will lead nowhere. I urge you to read some of the available literature to see what I'm talking about. Suppes is highly nonstandard in many things. He is not wrong, but it is nonstandard. Please, rent a good book on ZF set theory and read it. You will understand my point-of-view then.

 

[xxxx0xxxx]

This discussion will lead nowhere. I urge you to read some of the available literature to see what I'm talking about. Suppes is highly nonstandard in many things. He is not wrong, but it is nonstandard. Please, rent a good book on ZF set theory and read it. You will understand my point-of-view then.

I disagree on the following point.

I overlooked that you are referring to: y∈{x | φ(x)} ↔ φ(y) as though it is a definition, when it is actually not.

y∈{x | φ(x)} ↔ φ(y) is a theorem of naive set theory that restates the axiom of abstraction using the {...|...} operator.

In ZF, this theorem admits Russell's paradox, and cannot be a theorem of ZF.

The correct theorem in ZF is:

$$y \in \{x | \phi (x) \} \Rightarrow \phi (y)$$

The equivalence cannot be justified from the separation axiom, and must be an implication to avoid the paradox.

And as you may probably have already guessed, and operation definition for {x | φ(x)} is required in order to eliminate {x | φ(x)} entirely to wff's in the object language:

$$\{x | \phi (x) \} = w \Leftrightarrow [ \forall x (x \in w \Leftrightarrow \phi (x)) \wedge w \mbox{ is a set} ] \vee [\neg \exists B \forall x (x \in B \Leftrightarrow \phi (x)) \wedge w = \emptyset]$$

This definition along with

$$\{ x | \phi (x) \} \not = \emptyset$$

proves the implication, but the converse


$$\phi (y) \Rightarrow y \in \{x | \phi (x) \}$$

cannot be shown to follow from separation.

 

[micromass]

I disagree on the following point.

I overlooked that you are referring to: y∈{x | φ(x)} ↔ φ(y) as though it is a definition, when it is actually not.

y∈{x | φ(x)} ↔ φ(y) is a theorem of naive set theory that restates the axiom of abstraction using the {...|...} operator.

In ZF, this theorem admits Russell's paradox, and cannot be a theorem of ZF.

The correct theorem in ZF is:

$$y \in \{x | \phi (x) \} \Rightarrow \phi (y)$$

The equivalence cannot be justified from the separation axiom, and must be an implication to avoid the paradox.

And as you may probably have already guessed, and operation definition for {x | φ(x)} is required in order to eliminate {x | φ(x)} entirely to wff's in the object language:

$$\{x | \phi (x) \} = w \Leftrightarrow [ \forall x (x \in w \Leftrightarrow \phi (x)) \wedge w \mbox{ is a set} ] \vee [\neg \exists B \forall x (x \in B \Leftrightarrow \phi (x)) \wedge w = \emptyset]$$

This definition along with

$$\{ x | \phi (x) \} \not = \emptyset$$

proves the implication, but the converse


$$\phi (y) \Rightarrow y \in \{x | \phi (x) \}$$

cannot be shown to follow from separation.

Yes, I understand you completely. But I think you fail to understand me. You're taking the point-of-view of Suppes. This is not standard terminology!
Please, read another book on set theory and you will see what I'm talking about.

I can make suggestion to you that will increase your understanding and point-of-views. But only you can follow them

Good luck on your research!