A question about intersection of system of sets

1. Sep 9, 2011

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 .

Last edited: Sep 9, 2011
2. Sep 9, 2011

micromass

Staff Emeritus
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.

3. Sep 10, 2011

mahmoud2011

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

Last edited: Sep 10, 2011
4. Sep 10, 2011

mahmoud2011

I forgot , Must I take a course in logic ?

5. Sep 19, 2011

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

Last edited: Sep 19, 2011
6. Sep 19, 2011

micromass

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

7. Sep 19, 2011

xxxx0xxxx

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

8. Sep 19, 2011

micromass

Staff Emeritus
Name one.

9. Sep 19, 2011

micromass

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.

10. Sep 19, 2011

xxxx0xxxx

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.

11. Sep 19, 2011

micromass

Staff Emeritus
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.

12. Sep 19, 2011

micromass

Staff Emeritus
I'm still waiting for one such text.

13. Sep 19, 2011

xxxx0xxxx

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.

14. Sep 19, 2011

micromass

Staff Emeritus
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.

15. Sep 19, 2011

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 $$\emptyset$$.

16. Sep 19, 2011

micromass

Staff Emeritus
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.

17. Sep 19, 2011

xxxx0xxxx

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).

18. Sep 19, 2011

micromass

Staff Emeritus
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.

19. Sep 19, 2011

xxxx0xxxx

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: $$S = \emptyset$$

20. Sep 19, 2011

micromass

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

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