PROOF THREE of the Finite Canonical Ramsey
Theorem: Mileti’s SECOND Proof
William Gasarch-U of MD
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
PROOF THREE: a-ary Case
MODIFY PROOF TWO by GETTING RID of Ra−1 .
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Bill’s Mantra
It is often easier to proof something harder.
Theorem: For all a, for all α ∈ N, for all k there exists n such
that for all COL : [n]
a → ω × [α] there exists a set H of size k
such that
1. There exists I ⊆ [a] such that H is I -homog with respect to
Π1 (COL).
2. H is homog with respect to Π2 (COL).
Definition: GERa (k, α) is the least n that works.
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
How to Use This
1. GER1 (k, α) EASY to bound (your HW)
2. Using modification of PROOF THREE can bound GERa (k, α)
using GERa−1 (−, −) WITHOUT using Ra−1 or any R at all!
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Proof in the Style of Ramsey
Given COL : [n]
a → ω × [α] define a sequence.
Stage a − 2 (∀1 ≤ i ≤ a − 2)[xi = i]. X = {x1 , . . . , xa−1 }.
Aa−1 = [n] − X .
Stage s: Have
X = {x1 , . . . , xs−1 },
X
→ ω × [α] × {homog, rain}, and As−1 .
COL0 : a−1
KEY: Will use GERa−1 (k 0 , 2α) for some k 0 later.
Let A0s = As−1 and xs be least element of As−1 .
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Construction
( s )
Form A0s , A1s , . . ., As a−2
Assume have AL−1
and COL0 (X1 , xs ), . . ., COL0 (XL−1 , xs ) defined.
s
Notation: We denote ALs by AL throughout.
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
THE REAL KEY DIFFERENCE
This is the KEY diff from PROOF TWO.
Before doing ANYTHING else we do the following: Let i be the
number that MAXIMIZES
{y ∈ AL−1 | Π2 (COL(XL , xs , y )) = i}.
We ONLY work with these, we KILL all of the others. Let
AL−1
= {y ∈ AL−1 | Π2 (COL(XL , xs , y )) = i}.
0
Note
|AL−1
0 | ≥ |AL−1 |/α.
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Case 1
Case 1: (∃c)[|{x ∈ AL−1
: COL(XL , xs , x) = c}| ≥
0
q
|AL−1
0 |.
COL0 (XL , xs ) = (c, (i, homog))
AL = {x ∈ AL−1
: COL(XL , xs , x) = (c, i)}
0
q
p
Note: |AL | ≥ |AL−1
|AL−1 |/α.
0 |≥
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Can Ramsey Proof
p
Case 2: (∀c)[|{x ∈ AL−1 : COL(XL , xs , x) = (c, i)}| < |AL−1 |.
Make all colors coming out of (XL , xs ) to the right different:
Let AL be the set of all x ∈ AL−1 such that x is the LEAST
number with the color COL(XL , xs , x).
Formally AL = {x ∈ AL−1 :
COL(XL , xs , x) ∈
/ {COL(XL , xs , y ) : xs < y < x ∧ y ∈ AL−1 }
}
Now have
(∀y , y 0 ∈ AL )[COL(XL , xs , y ) 6= COL(XL , xs , y 0 )].
Note: |AL | ≥
p
|AL−1 |/α.
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Want to make colors DIFF
Important Note and Convention: For the rest of Case 2 we only
X
care about Z ∈ a−1
such that COL0 (Z ) = (−, (i, rain)).
Want to make the following true
X
)(∀y , y 0 ∈ As )[COL(Z , y 0 ) 6= COL(X L , y )]
(∀Z ∈
a−1
Its OKAY if COL(Z , y ) = COL(X L , y ).
For each y ∈ AL we thin out AL so that:
X
0
0
L
I (∀Z ∈
a−1 )(∀y ∈ AL − {y })[COL(Z , y ) 6= COL(X , y )].
X
0
L 0
I (∀Z ∈
a−1 )(∀y ∈ AL − {y })[COL(Z , y ) 6= COL(X , y )].
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
More to do!
Use C for COL for space
T = AL (elements to process)
w h i l e T 6= ∅
y = l e a s t element of T .
T = T − {y } ( b u t y s t a y s i n AL )
0
X
I f (∃Z ∈ a−1
, y ∈ T )[C (XL , xs , y ) = C (Z , y 0 )] t h e n
T = T − {y 0 } AL = AL − {y 0 }
0
X
I f (∃Z ∈ a−1
y ∈ T )[C (XL , xs , y 0 ) = C (Z , y )] t h e n
T = T − {y 0 } AL = AL − {y 0 }
p
√
Note: At end |AL | ≥ |AL−1 / αs a−1 .
Note: At end
0
X
(∀Z ∈ a−1
, y ∈ AL ))[COL(XL , xs , y ) 6= COL(Z , y 0 )].
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
OKAY- What is COL0 (XL , xs )?
RECAP:
(∀y , y 0 ∈ AL )[COL(XL , xs , y ) 6= COL(XL , xs , y 0 )]
X
0
I (∀y , y 0 ∈ AL )(∀Z ∈
a−1 )[COL(XL , xs , x) 6= COL(Z , y )]
√
|A
|
f (s) TBD. Let t = |AL | ≥ √αsL−1
a−1
Case 2.1: X
t
(∃Z ∈ a−1
)[|{y : COL(XL , xs , y ) = COL(Z , y )}| ≥ f (s)
].
I
COL0 (XL , xs ) = COL(XL , xs )
AL = {y ∈ AL : COL(XL , xs , y ) = COL(Z , y )}
Note: This will be a color of the form (−, (i, rain)).
t
Note: |AL | ≥ f (s)
.
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
OKAY- What is COL0 (XL , xs )
Case 2.2: X
(∀Z ∈ a−1
)[|{y ∈ AL : COL(XL , xs , y ) = COL(Z , y )}| <
t
f (s) ].
COL0 (XL , xs ) = (`, (i, rain)) ` is least not-used-for-rain color.
X
AL = AL − {y : (∃Z ∈ a−1
[COL(XL , xs , y ) = COL(Z , y )].
Note: |AL | ≥ t −
s−1 t
a−1 f (s)
≥ t(1 −
William Gasarch-U of MD
s−1 1
a−1 f (s) )
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Picking f (s)
Case 1 yields |AL | ≥
t
f (s) .
s−1 1
Case 2 yields |AL | ≥ t(1 − a−1
f (s) )
s−1
Take f (s) = 1 + a−1
≤ s a /a!. Both cases yield:
p
p
|AL−1 | a!
t
|AL | ≥
≥ √ a−1 a ≥ c |AL−1 |
f (s)
s
αs
a!
2a−1 but that would not gain
where c = √αs
2a (could have used s
us much).
We later see how far we need to go.
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Whats Really Going on?
c=
√ a! 2a .
αs
We assume c < 1. (If c ≥ 1 then we ignore it.)
b0 = bp
= as−1
bL ≥ c bL−1
By Rec Lemma
L
bL ≥ c 2 b 1/2 .
In stage s do this for ≤ s a−1 times. Hence
as ≥ bs a−1 ≥ c 2 b 1/2
Where d =
s a−1
≥
(a!)2 1/2s a−1
s a−1
b
≥ dm1/2
4a
αs
(a!)2
.
s 4a α
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Bound on As
Let as = |As |.
a0 = n
as
1/2s
a−1
≥ das−1
.
By Rec Lemma
as ≥ d 2 n1/2
sa
We later see how far we need to go.
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
AVOID RAMSEY
We will run the construction until X has r elements— we
determine r later
Have X = {x
1 , x2 , . . . , xr },
X
0
COL : a−1 → ω × ([α] × {homog, rain}). We can apply
GERa−1 (k, 2α)
KEY: In PROOF TWO we applied Ramsey at this step to get
either all homog or all rain. Here we don’t need to since GER will
take care of that.
Get I -homog set wrt to Π1 ◦ COL0 that is also homog wrt
Π2 ◦ COL0 .
H = {z1 , z2 , . . . , zk }.
Cases depend on if Π2 ◦ COL0 homog color is (−, homog) or
(−, rain).
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Homog of color (−, homog)
Case 1: Π2 Color is (−, homog) (real colors).
Have
H
(∀Y ∈
, z1 , z2 ∈ H)[COL(Y , z1 ) = COL(Y , z2 )].
a−1
H is I -homog set where I ⊆ [a − 1] wrt Π1 ◦ COL0 .
COL(y1 , . . . , ya ) = COL(z1 , . . . , za ) iff
COL0 (y1 , . . . , ya−1 ) = COL0 (z1 , . . . , za−1 )(def of COL0 iff
(∀i ∈ I )[yi = zi ](def of I -homog)
So H is I -homog set.
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Homog of color rain
Case 2: Π2 Color is (−, rain).
Have
H
(∀Y ∈
, z1 , z2 )[COL(Y , z1 ) 6= COL(Y , z2 )].
a−1
H is I -homog set where I ⊆ [a − 1] wrt Π1 ◦ COL0 .
COL(y1 , . . . , ya ) = COL(z1 , . . . , za ) iff
ya = za ∧ COL0 (y1 , . . . , ya−1 ) = COL0 (z1 , . . . , za−1 ) (from const.
iff ya = za ∧ (∀i ∈ I )[yi = zi ] (def of I -homog)).
So H is I ∪ {a}-homog set.
Need r = GERa−1 (k, 2α).
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
Estimate n
LET r = GERa−1 (k, 2α).
NEED
ar ≥ d
2n
1/2r
a
≥1
r 8a
ra
ar ≥ (a!/αr 4a α)2 n1/2 ≥ 1
ra
n1/2 ≥
e 4a
where e =
r
n ≥ er 4a2
1
d2
=
α2
(a!)4
ra
If suffices to take
n = Γ2 (ear a )
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
GER1 (k, α) = αk 2 .
α2
GERa (k) ≤ Γ2 ( (a!)
4 × a × GERa−1 (k, 2α))
For simplicity lets not use the a. (The reader is challenged to get a
better bound using it.
GERa (k) ≤ Γ2 (α2 GERa−1 (k, 2α))
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
GER1 (k, α) = αk 2 .
GER1 (k, 2α) = 2αk 2 .
GER2 (k, α) ≤ Γ2 (α2 GER1 (k, 2α)) ≤ Γ2 (α2 2α2 k 2 ) = Γ2 (2α4 k 2 )
GER2 (k, 2α) ≤ Γ2 (2(2α)4 k 2 ) = Γ2 (25 α4 k 2 )
GER3 (k, α)) ≤ Γ2 (α2 Γ2 (25 α4 k 2 ) ≤ Γ4 (25 α6 k 2 )
GER3 (k, 2α)) ≤ Γ4 (211 α6 k 2 )
GER4 (k, α)) ≤ Γ2 (α2 Γ4 (211 α6 k 2 ) ≤ Γ6 (211 α8 k 2 )
GER4 (k, 2α)) ≤ Γ6 (219 α8 k 2 )
GER5 (k, α)) ≤ Γ2 (α2 Γ6 (219 α8 k 2 )) ≤ Γ8 (219 α10 k 2 )
GER5 (k, 2α)) ≤ Γ8 (229 α10 k 2 )
Can show
2
GERa (k, α) ≤ Γ2a−2 (2a+(a+1) α2a k 2 ))
In particular:
2
ERa (k) = GERa (k, 1) ≤ Γ2a−2 (2a+(a+1) k 2 )
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil
PROS and CONS
1. GOOD-Proof reminsicent of Ramsey Proof.
2. GOOD-Seemed to be able to avoid alot of cases.
3. BAD-Proof complicated(?).
2
4. GOOD: ERa (k) ≤ Γ2a−2 (2a+(a+1) k 2 ) BIG improvement!
William Gasarch-U of MD
PROOF THREE of the Finite Canonical Ramsey Theorem: Mil