A question about intersection of system of sets

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 .
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 Quote by 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 .
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.

 Quote by micromass 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

A question about intersection of system of sets

I forgot , Must I take a course in logic ?
 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, $$\bigcap \emptyset = \{ x|x=x \}$$

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 Quote by xxxx0xxxx Mahmoud, In ZF, $$\bigcap \emptyset = \emptyset$$
This is false. In ZF, the intersection is undefined. It is not the empty set.

 Quote by micromass 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

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 Quote by xxxx0xxxx Micromass, quit making blanket statements, this is shown in many texts on ZF
Name one.
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

 Quote by 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.

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

qed.

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 Quote by xxxx0xxxx 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.

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 Quote by xxxx0xxxx this is shown in many texts on ZF
I'm still waiting for one such text.

 Quote by micromass 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.

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 Quote by xxxx0xxxx 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.

 Quote by micromass 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 $$\emptyset$$.

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 Quote by xxxx0xxxx 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.

 Quote by micromass 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.