PROFESSOR: Let's take a quick look
at the axioms of Zermelo-Frankel Set Theory With Choice.
So the axioms of ZFC define the standard theory
of sets, which is now accepted by most mathematicians
as a reliable and simple basis for developing and justifying
all of mathematics.
Among the axioms, maybe a simple want to understand
and really the motivation for this short video is twofold.
One is practice with writing predicate formulas,
and the other is to think a little bit more
about self application.
So one of the basic axioms of set theory
is called extensionality, which is
capturing the idea that a set is determined by its members.
So let's consider the assertion that two sets x and y
have the same elements, which we could write
as a predicate formula in set theory as for all
x, x is a member of y, if and only if x is a member of z.
Now we could use this is a definition of equality.
It's what we mean by y and z are equal.
But we don't really need to even introduce equality
as a basic part of the language and add axioms
about how it behaves.
There's one axiom that covers things adequately,
and that is that if two sets have the same members, then
they are members of the same sets.
So if all the members of x and y are the same, then x
and y are members of exactly the same thing, which
we could say this way, for every x,
y is an x, if and only if z is an x.
So that is one of the basic axioms of Set Theory, maybe
the starting one.
Another one is the Power Set axiom,
which simply says that every set has a power set.
How would you say that in the language of predicate set
theory?
Well, you'd say that for every x,
there is a p, which is going to be the power set effects, such
that for every set s, s is a subset of x, if and only
if s is a member of p.
Remember, we know how to express s as a subset of s
in the language of predicate calculus,
mentioning only membership.
So this is a good axiom that says, yes, there
is a set p consisting of precisely the subsets of x.
That set p called the powers set of x.
When you're trying to deal with the Russell's paradox
kind of issue, where you define a set of element
or a collection of sets that satisfies
some property, the safe conservative version of saying
that a set of elements that satisfy some property really
is a set, a collection of elements that satisfy
some property, really is a set, the comprehension axiom's
a simple version of an axiom that allows you to do that.
So basically, it says that if s is
a set and p of x is an arbitrary predicate of set theory, which
might in fact be one of these dangerous things
like x is not a member of x, nevertheless,
if you look at those elements in the set
S that satisfy P of x, that's a set.
In other words, the set of x and s, such that P of x is a set,
it means that any definable collection
of elements within a set also form a proper subset.
And the reason why this matters is,
remember, if I just talked about not the set of x
in a particular set s the satisfied P
of x, if I just talked about the collection of x's that
satisfied P of x, that's when I start getting into Russell's
paradox areas, when I declare that the set of x
such that P of x is a set for unrestricted P of x.
But all I get to do is put a bound on the elements
that x ranges over, that x is a member of some particular set.
Then it's safe to take all of those x's that satisfy P of x.
Now another particularly interesting axiom
of ZF which addresses this issue of self membership and self
reference is that the intuitive idea that the elements of a set
have to come before the set itself.
They have to be simpler than the set itself,
if you think about sort of building up
a set from successively simpler elements to more
complicated ones.
In particular, you can't have a set be a member of itself
because then it's not being built
from things that are simpler than it is
or that came before it.
In fact, you can't even have a set that's
a member of a member of itself.
All of that kind of indirect membership is forbidden.
Now, how do you say that is a nice axiom?
Well, there's a very elegant way to do it,
and that is to say that all sets are well
founded under membership, which means that you can't find
an infinite sequence of sets where each one has
the next one as a member.
Let's give a precise way to formulate that.
It's also good practice with the formulas of set theory.
Let me say that x is membership minimal, epsilon minimal, in y
means that x is a member of y, but there's no element of x
that's also in y.
In other words, x is built out of things that are not in y,
but x itself is in y.
So x kind of comes before any of the other elements in y.
It's built out of non-y stuff.
So to say this with a formula we could just say that x is in y,
and for every z, if it's in x, then it's not in y.
So that's the definition of x is membership minimal in y.
And then the basic axiom of ZF, called the Foundation Axiom,
simply says that every nonempty set has a membership
minimal element.
This is actually a kind of generalization
of the well ordering principle that
says that every nonempty set of non-negative integers
has a least element.
This is a direct analogy.
Just as the in principle for integers
implies that you can't have an infinite decreasing
sequence of non-negative integers,
the Foundation Axiom actually implies
that you can't have an infinite sequence of sets, each of which
is a member of the previous one.
Here is a formula that's asserting Foundation.
For every x, if x is not empty, that implies that there is a y,
such that y is membership minimal in x.
What is the Foundation got to do with membership?
Well, the Foundation Axiom actually
will very quickly let us conclude that no set is
a member of itself.
How does that work?
Well, suppose that you are interested in some set,
and you'd like to verify that the set can't
be a member of itself.
Well, let R be the set consisting of just this set S
that you're interested in.
R is the singleton S, its only element in S. Well,
R is not empty.
And by the Foundation Axiom, it must have a membership
minimal element.
Now suppose that S isn't S. We're
going to reach a contradiction.
The claim is that R has no membership minimal element,
and that violates the Foundation Axiom,
so you can't have S is a member of S. Why does this follow?
Well, look, R is supposed to have a membership
minimal element.
Well, R's only got one element.
So if it's got any membership element, it's got to be S.
But S this can't be membership minimal
because S is in R, which means that S
has an element in R in it.
So S is not R minimal.
And the Foundation Axiom then immediately
implies that you can't have S be a member of S. S is not
membership minimal in R. And this argument
extends in a nice way to member of a member and member
of a member, and we'll throw a feedback on one question
about that at you shortly.
So looking at the Foundation Axiom and the conclusion
that no set is a member of itself, what we can immediately
conclude is that, first of all, the collection of all sets
can't be a set because if the collection of all sets
was a set, then it would be a member of itself,
and that's forbidden by the S can't
be a member of S consequence of the Foundation Axiom.
The second thing it tells us is remember the set W
from Russell's paradox?
W was the collection of those sets which
are not members of themselves.
Well, now we've just figured out that this is all sets
because no set is a member of itself.
So the sets that are not members of themselves
is everything, and that's why W is not a set and not
a member of itself, which explains finally
how the Foundation Axiom resolves the Russell paradox.