EXACT DICKSON ORDINALS
MARKUS MICHELBRINK, ANTON SETZER
Abstract.
We argue classicaly. We identify a natural number with the set of its predecessors.
Definition 0.1. Let n ∈ N, ~x, ~y ∈ Nn and M, N ⊆ Nn .
~x ≤ ~y :⇔ ∀i < n.xi ≤ yi
~x < ~y :⇔ ~x ≤ ~y ∧ ~x 6= ~y
M ∗ := {~y ∈ N|∀~x ∈ M.~x 6≤ ~y}
N ⊑ M :⇔ N ∗ ⊆ M ∗
N ⊏ M :⇔ N ⊑ M ∧ N 6= M
Proposition 0.1.
metric.
1. The relation ≤ is reflexive, transitive and antisym-
2. The relation ⊑ is reflexive and transitive.
⊣
Proof. Simple verification.
n
Proposition 0.2. Every non empty set M ⊆ N has an <-minimal element.
⊣
Proof. Natural induction.
Proposition 0.3.
N ∗ ⊆ M ∗ ⇔ ∀~x ∈ M.∃~y ∈ N.~y ≤ ~x.
Proof. ‘⇒’: Let ~x ∈ M . We have ~x ≤ ~x. Therefore ~x ∈ ∁M ∗ ⊆ ∁N ∗ . It
follows ∃~y ∈ N.~y ≤ ~x. ‘⇐’: Let ~x ∈ ∁M ∗ . It follows that there is an ~x ∈ M with
~x ≤ ~z. Further there is a ~
y ∈ N with ~y ≤ ~x ≤ ~z. It follows ~z ∈ ∁N ∗ .
⊣
Definition 0.2.
Incompn := {M ⊆ Nn |∀~x, ~y ∈ M.~x 6< ~y ∧ ~y 6< ~x}
Proposition 0.4. The relation ⊑ is reflexive, transitive and antisymmetric
on Incompn .
Proof. Antisymmetric: Let M, N ∈ Incompn with M ⊑ N and N ⊑ M . By
Definition we have M ∗ = N ∗ . Therefore also ∁M ∗ = ∁N ∗ . Let ~x ∈ M . We have
~x ∈ M ⊆ ∁M ∗ = ∁N ∗ .
1
2
MARKUS MICHELBRINK, ANTON SETZER
Therefore there is a ~y ∈ N with ~y ≤ ~x. Analog it follows that there is a ~z ∈ M
with ~z ≤ ~
y ≤ x. Since M ∈ Incompn we have
~x = ~z = ~y ∈ N.
This shows M ⊆ N . That N ⊆ M follows analog.
⊣
Definition 0.3. Let M ⊆ Nn .
I(M ) := M \ {~x ∈ M |∃~y ∈ M.~y < ~x}.
Proposition 0.5. 1. I(M ) ∈ Incompn
2. I(M ) ⊆ M
3. I(I(M )) ⊆ I(M ).
Proof. Ad 3.:
~x ∈ I(M ) ⇒ ~x ∈ M ∧ ∀~y ∈ M.~y 6< ~x
⇒ ~x ∈ I(M ) ∧ ∀~y ∈ I(M ).~y 6< ~x (since I(M ) ⊆ M )
⇒ ~x ∈ I(I(M )).
⊣
Remark. Note that
M ⊆ N 6⇒ I(M ) ⊆ I(N )
e.g. M = {h1, 1i}, N = {h0, 0i, h1, 1i}.
Proposition 0.6.
M ∗ = I(M )∗
Proof. ‘⊆’: Immediate since I(M ) ⊆ M .
‘⊇’: Let ~y ∈ I(M )∗ and ~x ∈ M . We must show ~x 6≤ ~y. Assume ~x ≤ ~y. Since
∀~x ∈ I(M ).~x 6≤ ~y we have ~x 6∈ I(M ), i.e. ∃~z ∈ M.~z < ~x. Let ~z be minimal with
this property. It follows ~z ∈ I(M ) and ~z < ~y in contradiction to ~y ∈ I(M ). ⊣
Proposition 0.7. Every M ∈ Incompn is finite.
Proof. We show by induction on n that for every infinite sequence (~xi )i∈N ∈
(Nn )N there are i, j ∈ N with i < j and ~xi ≤ ~xj .
The claim follows immediate for n = 0. Let (~xi )i∈N ∈ (Nn+1 )N . We can build
a subsequence (~yi )i∈N ∈ (Nn+1 )N with yi 0 ≤ yi+1 0 for all i ∈ N. By Induction
Hypothesis we have i, j ∈ N with i < j and y~′ i ≤ y~′ j where y~′ is ~y without the
first component. It follows ~yi ≤ ~yj .
⊣
We are going to define ord M ∈ ω n for non empty M ∈ Incompn . We give the
definition only for n = 3. It is simple to extend our work to higher dimensions.
EXACT DICKSON ORDINALS
3
Definition 0.4. Let M ∈ Incomp3 . We define
M̄20 := M20 := {i ∈ N|{i} × N × N ⊆ M ∗ }
M̄21 := M21 := {i ∈ N|N × {i} × N ⊆ M ∗ }
M̄22 := M22 := {i ∈ N|N × N × {i} ⊆ M ∗ }
M̄10,1 := {hi, ji ∈ N2 |{i} × {j} × N ⊆ M ∗ }
M̄10,2 := {hi, ji ∈ N2 |{i} × N × {j} ⊆ M ∗ }
M̄11,2 := {hi, ji ∈ N2 |N × {i} × {j} ⊆ M ∗ }
M10,1 := {hi, ji ∈ M̄10,1 |i 6∈ M̄20 ∧ j 6∈ M̄21 }
M10,2 := {hi, ji ∈ M̄10,2 |i 6∈ M̄20 ∧ j 6∈ M̄22 }
M11,2 := {hi, ji ∈ M̄11,2 |i 6∈ M̄21 ∧ j 6∈ M̄22 }
M̄0 := {hi, j, ki ∈ N3 |{i} × {j} × {k} ⊆ M ∗ }
M0 := {hi, j, ki ∈ M̄0 |hi, ji 6∈ M̄10,1 ∧ hi, ki 6∈ M̄10,2 ∧ hj, ki 6∈ M̄11,2 }.
We are going to show that the “non bar” sets in this definition are finite and
that the functions giving the number of elements of these sets are computable
from M . We start by showing that the membership relation for the sets in the
Definition above is decidable.
Proposition 0.8. Let M ∈ Incomp3 non empty.
1.
2.
3.
4.
5.
6.
{i} × N × N ⊆ M ∗ ⇔ i < min(pr0 M )
N × {i} × N ⊆ M ∗ ⇔ i < min(pr1 M )
N × N × {i} ⊆ M ∗ ⇔ i < min(pr2 M )
{i} × {j} × N ⊆ M ∗ ⇔ ∀~x ∈ M.i < x0 ∨ j < x1
{i} × N × {j} ⊆ M ∗ ⇔ ∀~x ∈ M.i < x0 ∨ j < x2
N × {i} × {j} ⊆ M ∗ ⇔ ∀~x ∈ M.i < x1 ∨ j < x2 .
Proof. Ad 1.:
{i} × N × N ⊆ M ∗ ⇔ ∀j, k ∈ N.hi, j, ki ∈ M ∗
⇔ ∀j, k ∈ N.∀~x ∈ M.~x 6≤ hi, j, ki
⇔ ∀j, k ∈ N.∀~x ∈ M.i < x0 ∨ j < x1 ∨ k < x2
⇔ ∀~x ∈ M.i < x0
⇔ i < min(pr0 M ).
Ad 2.,3.: Analog.
Ad 4.:
{i} × {j} × N ⊆ M ∗ ⇔ ∀k ∈ N.hi, j, ki ∈ M ∗
⇔ ∀k ∈ N.∀~x ∈ M.~x 6≤ hi, j, ki
⇔ ∀k ∈ N.∀~x ∈ M.i < x0 ∨ j < x1 ∨ k < x2
⇔ ∀~x ∈ M.i < x0 ∨ j < x1
Ad 5.,6.: Analog.
⊣
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MARKUS MICHELBRINK, ANTON SETZER
Corollary 0.9. The membership relation for the sets of Definition 0.4 is
decidable.
Next we show that we only need to test a finite set of tuples to count the
elements of the “non bar” sets in Definition 0.4.
Proposition 0.10. Let M ∈ Incomp3 non empty.
1. M20 = min(pr0 M ), M21 = min(pr1 M ), M22 = min(pr2 M )
2. M10,1 = M10,1 ∩ (max(pr0 M ) × max(pr1 M ))
M10,2 = M10,2 ∩ (max(pr0 M ) × max(pr2 M ))
M11,2 = M11,2 ∩ (max(pr1 M ) × max(pr2 M ))
3. M0 = M0 ∩ (max(pr0 M ) × max(pr1 M ) × max(pr2 M )).
Proof. Ad 1.: This follows immediatly from the previous Proposition.
Ad 2.: Let hi, ji ∈ M10,1 . We must prove i < max(pr0 M ) and j < max(pr1 M ).
We have i < x0 or j < x1 for ~x ∈ M by the previous Proposition. Since i 6∈ M̄20
exists a ~y ∈ M with y0 ≤ i. It follows
j < y1 ≤ max(pr1 M ).
Analog follows i < max(pr0 M ) and the claims for the other sets.
Ad 3.: We simply repeat the argument from ad 2. Let hi, j, ki ∈ M10,1 . We must
prove i < max(pr0 M ), j < max(pr1 M ) and k < max(pr2 M ). We have i < x0 ,
j < x1 or k < x2 for ~x ∈ M by the previous Proposition. Since hi, ji 6∈ M̄10,1
exists a ~y ∈ M with y0 ≤ i and y1 ≤ j. It follows
k < y2 ≤ max(pr2 M ).
⊣
Corollary 0.11. The functions giving the number of elements of the sets of
Definition 0.4 are computable.
Definition 0.5. Let M ∈ Incomp3 non empty.
ord M := ω 2 · (|M20 | + |M21 | + |M22 |) + ω · (|M10,1 + |M10,2 | + |M11,2 |) + |M0 |
where | · | denotes the cardinality function.
We are going to prove that
ord M = sup{ord N + 1|N ∈ Incomp3 ∧ N ⊏ M }.
We will frequently use the following fact:
Proposition 0.12. Let M ∈ Incomp3 non empty. The set M ∗ is the union
of the following sets
{hi, j, ki ∈ N3 | i ∈ M20 },
{hi, j, ki ∈ N3 | j ∈ M21 },
{hi, j, ki ∈ N3 | k ∈ M22 },
{hi, j, ki ∈ N3 | hi, ji ∈ M10,1 },
{hi, j, ki ∈ N3 | hi, ki ∈ M10,2 },
{hi, j, ki ∈ N3 | hj, ki ∈ M11,2 },
and M0 .
EXACT DICKSON ORDINALS
5
Further we have
M̄10,1 = M10,1 ∪ {hi, ji ∈ N2 |i ∈ M20 } ∪ {hi, ji ∈ N2 |j ∈ M21 }
etc.
Proof. Simple verification.
To shorten our statements we introduce some new notations.
⊣
Definition 0.6. Let M, N ∈ Incomp3 non empty.
N2 E M2 :⇔ N20 ⊆ M20 ∧ N21 ⊆ M20 ∧ N22 ⊆ M22
N1 E M1 :⇔ N10,1 ⊆ M10,1 ∧ N10,2 ⊆ M10,2 ∧ N11,2 ⊆ M21,2
N0 E M0 :⇔ N0 ⊆ M0
N2 = M2 :⇔ N20 = M20 ∧ N21 = M20 ∧ N22 = M22
N1 = M1 :⇔ N10,1 = M10,1 ∧ N10,2 = M10,2 ∧ N11,2 = M21,2
Ni Mi :⇔ Ni E Mi ∧ Ni 6= Mi for i = 0, 1, 2.
Proposition 0.13. Let M, N ∈ Incomp3 non empty.
N ⊑ M ⇒ N2 E M2 ∧(N2 = M2 ⇒ N1 E M1 )∧(N2 = M2 ∧N1 = M1 ⇒ N0 E M0 )
Proof. Let N ⊑ M . We have ∀~x ∈ M ∃~y ∈ N.~y ≤ ~x by Proposition 0.3.
Further we have
(1)
min(pri N ) ≤ min(pri M )
for i = 0, 1, 2: Let ~x ∈ M with xi = min(pri M ). There is a ~y ∈ N with ~y ≤ ~x.
It follows
min(pri N ) ≤ yi ≤ xi = min(pri M ).
This proves already N2 E M2 . Further we have
(2)
{i} × {j} × N ⊆ N ∗ ⇒ {i} × {j} × N ⊆ M ∗
for i, j ∈ N: We have {i} × {j} × N ⊆ N ∗ if and only if ∀~y ∈ N.i < y0 ∨ j < y1 .
Let ~x ∈ M . There is a ~
y ∈ N with ~y ≤ ~x. It follows
i < y0 ≤ x0 ∨ j < y1 ≤ x1 .
Therefore {i} × {j} × N ⊆ M ∗ . This means
N̄10,1 ⊆ M̄10,1
by Definitions. Analog follows N̄10,2 ⊆ M̄10,2 and N̄11,2 ⊆ M̄11,2 .
Let N2 = M2 and hi, ji ∈ N10,1 . We have hi, ji ∈ N̄10,1 ⊆ M̄10,1 , i 6∈ N̄20 = M̄20
and j 6∈ N̄21 = M̄21 . Ergo hi, ji ∈ M10,1 which proves
N10,1 ⊆ M10,1 .
Analog follows N10,2 ⊆ M10,2 and N11,2 ⊆ M11,2 and therefore
N2 = M 2 ⇒ N1 E M 1 .
Finally let N2 = M2 , N1 = M1 . We have
N̄10,1 = M̄10,1
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MARKUS MICHELBRINK, ANTON SETZER
since
N̄10,1 = N10,1 ∪ {hi, ji ∈ N2 |i ∈ N20 } ∪ {hi, ji ∈ N2 |j ∈ N21 }
= M10,1 ∪ {hi, ji ∈ N2 |i ∈ M20 } ∪ {hi, ji ∈ N2 |j ∈ M21 }
= M̄10,1
Analog N̄10,2 = M̄10,2 and N̄11,2 = M̄11,2 . Therefore
N0 = N ∗ \ {hi, j, ki ∈ N3 |hi, ji ∈ N̄10,1 ∨ hi, ki ∈ N̄10,2 ∨ hj, ki ∈ N̄11,2 }
⊆ M ∗ \ {hi, j, ki ∈ N3 |hi, ji ∈ M̄10,1 ∨ hi, ki ∈ M̄10,2 ∨ hj, ki ∈ M̄11,2 }
= M0 .
⊣
Corollary 0.14. Let M, N ∈ Incomp3 non empty.
N ⊑ M ⇒ ord N ≤ ord M.
Proposition 0.15. Let M, N ∈ Incomp3 non empty.
N ⊏ M ⇒ N2 M 2 ∨ N1 M 1 ∨ N0 M 0 .
Proof. Let N2 ⋪ M2 and N1 ⋪ M1 . By the previous Proposition this means
N2 = M2 , N1 = M1 and N0 E M0 . Since N ⊏ M there is an hi, j, ki ∈ M ∗ \ N ∗ .
It follows hi, j, ki ∈ M0 \ N0 .
⊣
Corollary 0.16. Let M, N ∈ Incomp3 non empty.
N ⊏ M ⇒ ord N < ord M.
Proposition 0.17. Let M ∈ Incomp3 non empty. Then
ord M = sup{ord N + 1|N ∈ Incomp3 ∧ N ⊏ M }.
Proof. With respect to the previous Corollary it remains to find sets M (m) ⊏
M in Incomp3 with
ord M = sup{ord M (m) + 1|m ∈ N}.
We go through the different cases. Let ord M = ω 2 · α2 + ω · α1 + α0 and
ord M (m) = ω 2 · α(m)2 + ω · α(m)1 + α(m)0 .
First Case: M0 6= ∅ i.e., α0 = α′0 + 1 for some α′0 .
Let i0 := max(pr0 M0 ), j0 := max(pr1 {hi, j, ki ∈ M0 |i0 = i}), k0 := max(pr2 {hi, j, ki ∈
M0 |i = i0 ∧ j = j0 }). Let M (m) = I(M ∪ {hi0 , j0 , k0 i}) for m ∈ N. We show
(3)
M ∗ (m) = M ∗ \ {hi0 , j0 , k0 i}
‘⊆’: Let ~x ∈ M (m)∗ . By Proposition 0.6 we have ~x ∈ (M ∪ {hi0 , j0 , k0 i})∗ . This
proves ∀~y ∈ M.~y 6≤ ~x and ~x 6= hi0 , j0 , k0 i i.e., ~x ∈ M \ {hi0 , j0 , k0 i}.
‘⊇’: Let ~x ∈ M ∗ and ~x 6= hi0 , j0 , k0 i. Let ~y ∈ M . Since ~x ∈ M ∗ we have
y 6≤ ~x. Assume hi0 , j0 , k0 i ≤ ~x. Then we have hi0 , j0 , k0 i < ~x. Further we have
~
~x ∈ M0 since from hx0 , x1 i ∈ M̄10,1 follows hi0 , j0 i ∈ M̄10,1 etc. But now we got a
contradiction since i0 , j0 and k0 were choosen maximal with this property.
This proves equation 3. Since the sets giving the coefficients α(m)2 and α(m)1
remain unchanged and we just took one element out of M0 we have
ord M = ord M (m) + 1
EXACT DICKSON ORDINALS
7
for m ∈ N.
Second Case: M0 = ∅, M10,1 6= ∅ i.e., α0 = 0 and α1 = α′1 + 1 for some α′1 .
Let i0 := max(pr0 M10,1 ), j0 := max(pr1 {hi, ji ∈ M10,1 |i0 = i}), k0 = max(pr2 M ).
Let M (m) = I(M ∪ {hi0 , j0 , k0 + mi}) for m ∈ N. We show
M (m)02 = M20 .
(4)
Since hi0 , j0 i ∈ M10,1 we have i0 6∈ M20 and therefore i0 6≤ min(pr0 M ). Ergo
i ∈ M (m)02 ⇔ i < min(pr0 M (m)) = min(pr0 M ) ⇔ i ∈ M20 .
This proves the equation. Analog follows
M (m)12 = M21
(5)
and by using min(pr2 M ) ≤ k0 we can prove in the same way
M (m)22 = M22 .
(6)
Next we show
M̄ (m)10,1 = M̄10,1 \ {hi0 , j0 i}.
We calculate
hi, ji ∈ M̄ (m)0,1
x ∈ M (m).i < x0 ∨ j < x1
1 ⇔ ∀~
⇔ (∀~x ∈ M.i < x0 ∨ j < x1 ) ∧ (i < i0 ∨ j < j0 )
⇔ (∀~x ∈ M.i < x0 ∨ j < x1 ) ∧ (i 6= i0 ∨ j 6= j0 )
(7)
⇔ hi, ji ∈ M̄10,1 \ {hi0 , j0 i}
where the direction from left to right in 7 follows immediate. For the other
direction assume (∀~x ∈ M.i < x0 ∨ j < x0 ) ∧ (i 6= i0 ∨ j 6= j0 ), i0 ≤ i and j0 ≤ j.
Since ∀~x ∈ M.i < x0 ∨ j < x0 we have hi, ji ∈ M̄10,1 . Since min(pr0 M ) ≤ i0 ≤ i
and min(pr1 M ) ≤ j0 ≤ j we have i 6∈ M20 and j 6∈ M21 . Therefore hi, ji ∈ M10,1 .
But this contradicts i0 resp. j0 maximal with this property. We can conclude
0,1
M (m)0,1
1 = M1 \ {hi0 , j0 i}.
(8)
An analogous calculation as above gives
0,2
M̄ (m)0,2
1 = M̄1
where we use
(∀~x ∈ M.i < x0 ∨ k < x2 ) ∧ (i < i0 ∨ k < k0 ) ⇔ (∀~x ∈ M.i < x0 ∨ k < x2 )
instead of 7. The direction from left to right follows again immediate. Since
i0 6∈ M̄20 there is an ~x ∈ M with ~x ≤ i0 . Since k0 = max(pr2 M ) we also have
x2 ≤ k0 . Assume hi0 , k0 i ≤ hi, ki. Then we have
hx0 , x2 i ≤ hi0 , k0 i ≤ hi, ki
and therefore
(9)
∃~x ∈ M.x0 ≤ i ∧ x2 ≤ k.
With the previous calculations we can conclude
(10)
0,2
M (m)0,2
1 = M1
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MARKUS MICHELBRINK, ANTON SETZER
and analogous follows
1,2
M (m)1,2
1 = M1 .
(11)
Finally we understand
|m| ≤ |M (m)0 |.
(12)
Therefore we show hi0 , j0 , li ∈ M0 (m) for k0 ≤ l < k0 + m. First let us see
hi0 , j0 , li ∈ M̄ (m)0 : Since hi0 , j0 i ∈ M10,1 we have ~x 6≤ hi0 , j0 , li for all ~x ∈ M
and for ~x = hi0 , j0 , k0 + mi we have hi0 , j0 , li < hi0 , j0 , k0 + mi. Therefore
hi0 , j0 , li ∈ M̄ (m)0 .
On the other side we have hi0 , j0 i 6∈ M̄ (m)0,1
since hi0 , j0 , k0 + mi 6∈ M (m)∗ .
1
The existential statement 9 implies hi0 , k0 i 6∈ M̄ (m)0,2
and hj0 , k0 i 6∈ M̄ (m)1,2
1
1
can be shown analog. Altogether this gives hi0 , j0 , li ∈ M (m)0 . Summarising
the equations 4,5,6,8,10,11 and 12 we receive
ord M = sup{ord M (m) + 1|m ∈ N}.
M10,1
The cases M0 =
= ∅, M10,2 6= ∅ and M0 = M10,1 = M10,2 = ∅, M11,2 6= ∅
follow analog.
Third case: M0 = M10,1 = M10,2 = M11,2 = ∅ and M20 6= ∅.
Actually in this case the set M is a singleton i.e., it contains only one element.
However we do not use this information and proceed in a way which also applies
to higher dimensions.
Let i0 := max(pr0 M20 ), j0 := max(pr1 M ), k0 := max(pr2 M ). Let M (m) =
I(M ∪ {hi0 , j0 , k0 + mi}). We have
(13)
M (m)02 = M20 \ {i0 }
as well as
(14)
M (m)12 = M21
and M (m)22 = M22
and
(15)
|m| ≤ |M10,1 (m)|.
The equations 13,14 and 15 prove
ord M = sup{ord M (m) + 1|m ∈ N}.
The remaining cases follow analog.
⊣
Remark. For dimensions above 3 we lose some of our geometrical intuition.
Already the case n = 3 is hard to visualise. However the geometrical intuition
which guided our proof applies as well to higher dimensions. Note how the different cases are treated in a completely similiar way. We can proceed in the same
way for n > 3 .