Axiomatic set theory

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Axiomatic set theory

• Jouko Väänänen

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Axiom of Choice AC
x
y
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Another formulation AC
x
f
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AC? AC
Choose elements using AC
Make them disjoint
Return to the original sets
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AC ? AC
x
f
AC gives f
Then take the range of f
You get the set required to exist in AC
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Well-ordering principle
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WO ?AC
Well-order the union
The choice set y is the set of points z in the
union such that in the particular unique (!) set
that z belongs to, it is lt-smallest.
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A useful fact

• If X is a set, there is no one-one class function
  fOn ? X.
• Proof. Otherwise, by the Axiom of Subset
  Selection, Yf(On)?X is a set. Thus we can form
  the function f-1Y ? On. By the Axiom of
  Replacement, f-1(Y) is a set of ordinals. Let ?
  be an ordinal greater than all ordinals in
  f-1(Y). Then f(?) is in Y, so ?f-1(f(?)) is in
  f-1(Y), a contradiction.

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AC ? WO
We need a function which picks an element like
this
From this non-empty set
The point is AC gives such a function
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Zorns Lemma
A maximal element
.... an upper bound
Every chain has an ....
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Intuitive proof of ZL
Starting from any element we keep picking larger
and larger elements until we hit a maximal
element. If we do not hit any such, we have
anyway built a chain. By assumption, this chain
has an upper bound. So we have to hit a maximal
element.
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WO ?Zorns Lemma
P is first well ordered
g(?) is chosen to be the least (in the w.o.)
strict upper bound, if any
g(?)
g(?) ?lt?
Note that we use recursion on ordinals
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An application of AC

• We show that there is a non-measurable set of
  real numbers.
• The collection of measurable sets of real numbers
  have the following properties
• ? and all intervals a,b are measurable.
• The complement of a measurable set is measurable.
• The union of countably many measurable sets is
  always measurable.
• If A is measurable, then Arxrx?A is
  measurable, what ever r is.

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Lebesgue-measure

• The Lebesgue-measure m(A) of a measurable set A
  is defined so that the following hold
• m(?)0, m(a,b)b-a
• If the measurable sets An are disjoint, then
  m(?nAn)?n m(An)
• If A is measurable, then m(Ar)m(A) for all
  reals r.

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A non-measurable set

• Define on 0,1 x?y iff x-y is rational.
• This is an equivalence relation.
• By AC there is a set A which has exactly one
  element from each ?-equivalence class.
• We show that A is non-measurable.
• Assume otherwise. Let rm(A).

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We derive a contradiction

• 0,1? ?q(Aq) ?-1,2, where q ranges over
  Q?0,1.
• 1?m(?q(Aq))?q m(Aq)?q r?3.
• If r0, we cannot have 1 ? ?q r
• If rgt0, we cannot have ?q r?3.
• We have arrived at a contradiction.
• So A is not measurable. QED

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Banach-Tarski Paradox

• Another application of AC
• The unit sphere in the three dimensional
  Euclidean space can be divided into five pieces
  so that by reassembling the pieces in space we
  get two unit spheres of the same size as the
  original.
• The trick the pieces are non-measurable, so we
  cannot say that e.g. volume has been doubled. The
  pieces (or at least some of them) simply do not
  have volume.

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Ordinal addition
?
?
??
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Ordinal addition formally

• A(? x 0)?(? x 1)
• (?,i)ltA(?,j) iff (iltj) v (ij ?lt?)

?
1
?
0
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Non-commutativity
1 ? ? ? 1
?
1
has last element
does not have last element
?
1
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Associativity
?
?
?
??
??
(??)?
?(??)
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Cancellation laws??
? ? ? ? ? ? ? ?
Yes
? ? ? ? ? ? ? ?
No
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Ordinal sum
?0
?1
?2
?0?1?2
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Ordinal sum
?0
?1
??
...
...
??lt???
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Example
1
1
1
...
...
?nlt?1 ?
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Ordinal sum
??lt???
??
...
...
...
?
?