Thread: Ambiguity on the Axiom of Union...

Hello friends,
Take a look at the image below:

The blue part is of Paul.R.Halmos naive set theory.
I looked for the formal statement of the axiom and i found 2 different statements on net!
The red and the green one!
Certainly they are different. In fact the red is a part of the green as you know.
I myself think the red is true and this also corresponds to Halmos assertion.
Because if the green were true, then the brown weren't so!(Since every element of B can be found in at least one of the sets contained in A!)

What do you think?!
(Sorry for different letters in different assertions-It's just a collection of copies and so i didn't change the letters!)

Regards.

[I'll use 'A' and 'E' for the universal and existential quantifiers, respectively, and 'e' for 'is a member of'.]

(1) EuAz(Ex(xeC & zex) -> zeu)

and

(2) EuAz(Ex(xeC & zex) <-> zeu)

With an instance of the axiom schema of separation, we can derive (2) from (1):

EuAz(Ex(xeC & zex) -> zeu).
So let Az(Ex(xeC & zex) -> zeu).
Separation gives us EvAz((zeu & Ex(xeC & zex)) <-> zev).
Let Az((zeu & Ex(xeC & zex)) <-> zev).
So Az(Ex(xeC & zex) <-> zev).
So, generalizing 'v' to 'u', we get EuAz(Ex(xeC & zex) <-> zeu).

The difference then is just a slight technical detail.

Thanks,
But as you yourself assert and also mentioned in the Halmos book(blue), WE CAN CREATE SUCH A SET USING THE AXIOM SCHEMA OF SPECIFICATION but i mean this set is not necessarily the same that the axiom asserts!

So i think my 1st assertion is true.
Right?

Originally Posted by Mathelogician

this set is not necessarily the same that the axiom asserts

Here is the most exact answer I can give:

Axiom (1) does not name a PARTICULAR set u. It asserts only that there is at least one u such that Az(Ex(xeC & zex) -> zeu). And the axiom alone does not preclude that there may be members of such a u other than the z such that Ex(xeC & zex). But then with the axiom schema of separation, we derive that there is at least one u such that Az(Ex(xeC & zex) <-> zeu) and thus that it IS precluded that there may be members of a u of THAT kind other than the z such that Ex(xeC & zex). Then from the axiom of extensionality, we derive that there is exactly one such u.

Axiom (2) asserts that there is at least one u such that Az(Ex(xeC & zex) -> zeu). And the axiom alone does preclude that there are members of such a u other than the z such that Ex(xeC & zex). Then, also, with the axiom of extentionsionality, we derive that there is a exactly one such u, a particular one that we may name as 'UC'.

So let S be the relevant instance of the axiom schema of separation. Then we have:

S |- (1) <-> (2).

And, yes, of course it is not the case that

|- (1) <-> (2)

as it is not the case that

|- (1) -> (2)

though it is the case that

|- (2) -> (1).

Originally Posted by MoeBlee

Here is the most exact answer I can give:

Axiom (1) does not name a PARTICULAR set u. It asserts only that there is at least one u such that Az(Ex(xeC & zex) -> zeu). And the axiom alone does not preclude that there may be members of such a u other than the z such that Ex(xeC & zex). But then with the axiom schema of separation, we derive that there is at least one u such that Az(Ex(xeC & zex) <-> zeu) and thus that it IS precluded that there may be members of a u of THAT kind other than the z such that Ex(xeC & zex). Then from the axiom of extensionality, we derive that there is exactly one such u.

Axiom (2) asserts that there is at least one u such that Az(Ex(xeC & zex) -> zeu). And the axiom alone does preclude that there are members of such a u other than the z such that Ex(xeC & zex). Then, also, with the axiom of extentionsionality, we derive that there is a exactly one such u, a particular one that we may name as 'UC'.

Therefore we see that axiom 1 does not preclude the existence of some elements in u such that they are not contained in any of the sets of C. And after using the axiom schema of specification we know there exists at least a set v(or any letter!)that all of its elements are contained in at least one of the sets of C. And in fact the set v (which derived by the axiom of specification) is certainly a subset any kind of set that the axiom of union asserts to exist!

Therefore, since we are searching for a formal sentence for the axiom itself (not the assertion derived by using the axiom schema of specification), then the proper is the 1st.

So let S be the relevant instance of the axiom schema of separation. Then we have:

S |- (1) <-> (2).

And, yes, of course it is not the case that

|- (1) <-> (2)

as it is not the case that

|- (1) -> (2)

though it is the case that

|- (2) -> (1).

What do you mean bye |- ??
More clear please...

Originally Posted by Mathelogician

axiom 1 does not preclude the existence of some elements in u such that they are not contained in any of the sets of C.

Right.

Originally Posted by Mathelogician

And after using the axiom schema of specification we know there exists at least a set v(or any letter!)that all of its elements are contained in at least one of the sets of C. And in fact the set v (which derived by the axiom of specification) is certainly a subset any kind of set that the axiom of union asserts to exist!

Right. But I don't know why you are so excited about it.

Originally Posted by Mathelogician

Therefore, since we are searching for a formal sentence for the axiom itself (not the assertion derived by using the axiom schema of specification), then the proper is the 1st.

I don't know what you mean by "proper" in this context. We are free to axiomatize any way we want.

(1) has the possible advantage that it assumes less than (2).

(2) has the possible advantage that it is self-contained in the sense that we don't need the axiom schema of separation to get the actual desired set.

In any case, with the axiom schema of separation, (1) and (2) are equivalent.

Originally Posted by Mathelogician

What do you mean bye |- ??

|-

is the standard symbol in mathematical logic for "proves".

Where G is a set of formulas, and P is a formula

G |- P

stands for

There is a proof of P from G.

So, to be precise, I should have written:

{S} |- (1) <-> (2)

When there is no G on the left side, then

|- P

means that P is a theorem of pure logic alone.

Right. But I don't know why you are so excited about it.

Oh i know... Because i want to insist on the fact that the axiom of union does not assert the existence of such a set...And because the purpose of this topic is certainly to find a formal sentence for the axiom of of union itself not any derived conclusion.

I don't know what you mean by "proper" in this context

I mean a suitable formal sentence for the axiom itself.

In a formal context, the axiom itself IS a formal sentence.

We are free to choose whatever formal sentences we wish to choose for axioms. Any given author is free to state whatever formal sentences he wants to state as the axioms. (1) happens to be weaker than (2), but they are equivalent given the axiom schema of separation. It's not a matter of what is or is not "proper".

You'll find that there are lots of examples of axiomatizations being given differently by different authors.

Any way thank you.
Good discussion.