“Weak” Ramsey conditions for cardinals
$begingroup$
Ok, so these questions just popped into my head and I can't seem to figure it out:
Ramsey's theorem tells us that for any $n,rinomega$ and any $f:[omega]^{n}rightarrow r$, exists an infinite set $Ssubseteqomega$ for which $f| [S]^{n}$ is constant.
A cardinal $kappa$ is said to be Ramsey if for every $f:[kappa]^{<omega}rightarrow {0,1}$ exists some set $Ssubseteqkappa$ of cardinality $kappa$ such that $f|[S]^{<omega}$ is constant.
ZFC can not prove the existence of a Ramsey cardinal (as it is quite large), but the amazing thing is that the difference in conditions seems to be minor, yet to the question: "How much must we enlarge our cardinal $omega$ to satisfy the improved condition", we must answer: "Unprovably much".
So, several questions arise on some conditions in-between. In order to ask them compactly, I will define the following:
Definition: A cardinal $lambda$ will be called K-Ramsey for some $Ksubseteqomega$ if for every $f:[lambda]^{K}rightarrow{0,1}$ (meaning the set of all subsets of $lambda$ with a cardinality in the set K), exists some $Ssubseteqlambda$ of cardinality $lambda$ which is homogeneous for f.
For the questions:
Which cardinal $kappa$ that have been proved to exist in ZFC:
Is K-Ramsey for $K={m,n}$, with $m>ngeq2$?
Is K-Ramsey for every set K of cardinality 2?
Is K-Ramsey for all finite sets $Ksubseteqomega$ that include every natural number up to their maximum (meaning every set of the form ${1,2,...,n}$)?
Is K-Ramsey for every finite set $Ksubseteqomega$?
For every $f:[kappa]^{<omega}rightarrow{0,1}$ exists some infinite set $Ksubseteqomega$ and some $Ssubseteqkappa$ of cardinality $kappa$, such that the set $[S]^{K}$ is monochromatic?
Does a cardinal being K-Ramsey for some infinite set $Ksubseteqomega$ imply it is Ramsey?
I am pretty sure a few of those finite questions can be answered using the pigeonhole principle countably many more times, but am not sure how to continue. If you have any leads or ideas or perhaps these results are well-known, please share them with me.
set-theory cardinals ramsey-theory large-cardinals
$endgroup$
add a comment |
$begingroup$
Ok, so these questions just popped into my head and I can't seem to figure it out:
Ramsey's theorem tells us that for any $n,rinomega$ and any $f:[omega]^{n}rightarrow r$, exists an infinite set $Ssubseteqomega$ for which $f| [S]^{n}$ is constant.
A cardinal $kappa$ is said to be Ramsey if for every $f:[kappa]^{<omega}rightarrow {0,1}$ exists some set $Ssubseteqkappa$ of cardinality $kappa$ such that $f|[S]^{<omega}$ is constant.
ZFC can not prove the existence of a Ramsey cardinal (as it is quite large), but the amazing thing is that the difference in conditions seems to be minor, yet to the question: "How much must we enlarge our cardinal $omega$ to satisfy the improved condition", we must answer: "Unprovably much".
So, several questions arise on some conditions in-between. In order to ask them compactly, I will define the following:
Definition: A cardinal $lambda$ will be called K-Ramsey for some $Ksubseteqomega$ if for every $f:[lambda]^{K}rightarrow{0,1}$ (meaning the set of all subsets of $lambda$ with a cardinality in the set K), exists some $Ssubseteqlambda$ of cardinality $lambda$ which is homogeneous for f.
For the questions:
Which cardinal $kappa$ that have been proved to exist in ZFC:
Is K-Ramsey for $K={m,n}$, with $m>ngeq2$?
Is K-Ramsey for every set K of cardinality 2?
Is K-Ramsey for all finite sets $Ksubseteqomega$ that include every natural number up to their maximum (meaning every set of the form ${1,2,...,n}$)?
Is K-Ramsey for every finite set $Ksubseteqomega$?
For every $f:[kappa]^{<omega}rightarrow{0,1}$ exists some infinite set $Ksubseteqomega$ and some $Ssubseteqkappa$ of cardinality $kappa$, such that the set $[S]^{K}$ is monochromatic?
Does a cardinal being K-Ramsey for some infinite set $Ksubseteqomega$ imply it is Ramsey?
I am pretty sure a few of those finite questions can be answered using the pigeonhole principle countably many more times, but am not sure how to continue. If you have any leads or ideas or perhaps these results are well-known, please share them with me.
set-theory cardinals ramsey-theory large-cardinals
$endgroup$
add a comment |
$begingroup$
Ok, so these questions just popped into my head and I can't seem to figure it out:
Ramsey's theorem tells us that for any $n,rinomega$ and any $f:[omega]^{n}rightarrow r$, exists an infinite set $Ssubseteqomega$ for which $f| [S]^{n}$ is constant.
A cardinal $kappa$ is said to be Ramsey if for every $f:[kappa]^{<omega}rightarrow {0,1}$ exists some set $Ssubseteqkappa$ of cardinality $kappa$ such that $f|[S]^{<omega}$ is constant.
ZFC can not prove the existence of a Ramsey cardinal (as it is quite large), but the amazing thing is that the difference in conditions seems to be minor, yet to the question: "How much must we enlarge our cardinal $omega$ to satisfy the improved condition", we must answer: "Unprovably much".
So, several questions arise on some conditions in-between. In order to ask them compactly, I will define the following:
Definition: A cardinal $lambda$ will be called K-Ramsey for some $Ksubseteqomega$ if for every $f:[lambda]^{K}rightarrow{0,1}$ (meaning the set of all subsets of $lambda$ with a cardinality in the set K), exists some $Ssubseteqlambda$ of cardinality $lambda$ which is homogeneous for f.
For the questions:
Which cardinal $kappa$ that have been proved to exist in ZFC:
Is K-Ramsey for $K={m,n}$, with $m>ngeq2$?
Is K-Ramsey for every set K of cardinality 2?
Is K-Ramsey for all finite sets $Ksubseteqomega$ that include every natural number up to their maximum (meaning every set of the form ${1,2,...,n}$)?
Is K-Ramsey for every finite set $Ksubseteqomega$?
For every $f:[kappa]^{<omega}rightarrow{0,1}$ exists some infinite set $Ksubseteqomega$ and some $Ssubseteqkappa$ of cardinality $kappa$, such that the set $[S]^{K}$ is monochromatic?
Does a cardinal being K-Ramsey for some infinite set $Ksubseteqomega$ imply it is Ramsey?
I am pretty sure a few of those finite questions can be answered using the pigeonhole principle countably many more times, but am not sure how to continue. If you have any leads or ideas or perhaps these results are well-known, please share them with me.
set-theory cardinals ramsey-theory large-cardinals
$endgroup$
Ok, so these questions just popped into my head and I can't seem to figure it out:
Ramsey's theorem tells us that for any $n,rinomega$ and any $f:[omega]^{n}rightarrow r$, exists an infinite set $Ssubseteqomega$ for which $f| [S]^{n}$ is constant.
A cardinal $kappa$ is said to be Ramsey if for every $f:[kappa]^{<omega}rightarrow {0,1}$ exists some set $Ssubseteqkappa$ of cardinality $kappa$ such that $f|[S]^{<omega}$ is constant.
ZFC can not prove the existence of a Ramsey cardinal (as it is quite large), but the amazing thing is that the difference in conditions seems to be minor, yet to the question: "How much must we enlarge our cardinal $omega$ to satisfy the improved condition", we must answer: "Unprovably much".
So, several questions arise on some conditions in-between. In order to ask them compactly, I will define the following:
Definition: A cardinal $lambda$ will be called K-Ramsey for some $Ksubseteqomega$ if for every $f:[lambda]^{K}rightarrow{0,1}$ (meaning the set of all subsets of $lambda$ with a cardinality in the set K), exists some $Ssubseteqlambda$ of cardinality $lambda$ which is homogeneous for f.
For the questions:
Which cardinal $kappa$ that have been proved to exist in ZFC:
Is K-Ramsey for $K={m,n}$, with $m>ngeq2$?
Is K-Ramsey for every set K of cardinality 2?
Is K-Ramsey for all finite sets $Ksubseteqomega$ that include every natural number up to their maximum (meaning every set of the form ${1,2,...,n}$)?
Is K-Ramsey for every finite set $Ksubseteqomega$?
For every $f:[kappa]^{<omega}rightarrow{0,1}$ exists some infinite set $Ksubseteqomega$ and some $Ssubseteqkappa$ of cardinality $kappa$, such that the set $[S]^{K}$ is monochromatic?
Does a cardinal being K-Ramsey for some infinite set $Ksubseteqomega$ imply it is Ramsey?
I am pretty sure a few of those finite questions can be answered using the pigeonhole principle countably many more times, but am not sure how to continue. If you have any leads or ideas or perhaps these results are well-known, please share them with me.
set-theory cardinals ramsey-theory large-cardinals
set-theory cardinals ramsey-theory large-cardinals
edited Dec 12 '18 at 18:45
Jean Marie
29.8k42051
29.8k42051
asked Dec 12 '18 at 18:34
Uri George PeterzilUri George Peterzil
949
949
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2 Answers
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$begingroup$
Even for the simplest nontrivial situation $K={2}$ - let alone $2$-element $K$s - we're forced to leave ZFC:
The statement "There is a cardinal $kappa$ such that every two-coloring of pairs $c:[kappa]^2rightarrow 2$ has a homogeneous set of size $kappa$" is independent of ZFC.
In fact, a cardinal satisfying Ramsey's theorem for pairs (= your "${2}$-Ramsey") is called weakly compact, and this is a significant large cardinal property. So the pigeonhole principle is actually extremely limited in the uncountable setting! (Or perhaps more accurately, "simple" combinatorial facts about $omega$ secretly depend on more than just the pigeonhole principle.)
$endgroup$
add a comment |
$begingroup$
A couple of results, using the "arrow notation": For any set $A$ and any (finite or infinite) cardinals $b,c,d,$ the notation $$Ato (b)^c_d$$ means that for every $f:[A]^cto d$ there exists $Sin [A]^b$ such that $|f(x):xin S|=1.$ (I.e. $S$ is homogeneous for $f$.)
For infinite cardinal ordinal $a$ we have $neg (,2^ato (a^+)^2_2,).$
For uncountable cardinal ordinal $a,$ if $ato (a)^2_2,$ we can show readily that $a$ is regular, and we can use 1. to show that $a$ is a strong limit cardinal, so $a$ is strongly-inaccessible. Such an $a$ is called weakly compact, and is "rather large", much larger than the least strongly-inaccessible.
In particular, for 1., for $|a|=omega:$ Let $<_W$ be a well-order on $Bbb R$ and let $<$ be the arithmetic order. Define $f: [Bbb R]^2to {0,1}$ where $f({x,y})=1$ iff $<_W$ and $<$ agree on ${x,y}.$ And let $x<^*y$ iff $y<x. $ Now if $S$ is homogenous for $f,$ then $<$ or $<^*$ is a well-order on $S.$ But since $Ssubset Bbb R,$ this implies $S$ is countable.
For 1. in general, let $<_L$ be the lexicographic order on $^{{0,1}}a:$ For functions $g,h$ from $a$ to ${0,1}$ with $gne h,$ let $x=min {yin a: g(y)ne h(y)}.$ Let $g<_L h$ iff $g(x)=0$ and $h(x)=1.$ And let $<_W$ be a well-order on $^{{0,1}}a.$.... Now define $f:[^{{0,1}}a]^2to {0,1}$ by letting $f({g,h})=1$ iff $<_L$ and $<_W$ agree on ${g,h}.$ A theorem of "Erdos & company" is that if $S$ is homogeneous for $f$ then $|S|leq a.$
Paul Erdos is an author or co-author of over 1500 published papers, and the number of other co-authors on them is in the thousands.
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2 Answers
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2 Answers
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$begingroup$
Even for the simplest nontrivial situation $K={2}$ - let alone $2$-element $K$s - we're forced to leave ZFC:
The statement "There is a cardinal $kappa$ such that every two-coloring of pairs $c:[kappa]^2rightarrow 2$ has a homogeneous set of size $kappa$" is independent of ZFC.
In fact, a cardinal satisfying Ramsey's theorem for pairs (= your "${2}$-Ramsey") is called weakly compact, and this is a significant large cardinal property. So the pigeonhole principle is actually extremely limited in the uncountable setting! (Or perhaps more accurately, "simple" combinatorial facts about $omega$ secretly depend on more than just the pigeonhole principle.)
$endgroup$
add a comment |
$begingroup$
Even for the simplest nontrivial situation $K={2}$ - let alone $2$-element $K$s - we're forced to leave ZFC:
The statement "There is a cardinal $kappa$ such that every two-coloring of pairs $c:[kappa]^2rightarrow 2$ has a homogeneous set of size $kappa$" is independent of ZFC.
In fact, a cardinal satisfying Ramsey's theorem for pairs (= your "${2}$-Ramsey") is called weakly compact, and this is a significant large cardinal property. So the pigeonhole principle is actually extremely limited in the uncountable setting! (Or perhaps more accurately, "simple" combinatorial facts about $omega$ secretly depend on more than just the pigeonhole principle.)
$endgroup$
add a comment |
$begingroup$
Even for the simplest nontrivial situation $K={2}$ - let alone $2$-element $K$s - we're forced to leave ZFC:
The statement "There is a cardinal $kappa$ such that every two-coloring of pairs $c:[kappa]^2rightarrow 2$ has a homogeneous set of size $kappa$" is independent of ZFC.
In fact, a cardinal satisfying Ramsey's theorem for pairs (= your "${2}$-Ramsey") is called weakly compact, and this is a significant large cardinal property. So the pigeonhole principle is actually extremely limited in the uncountable setting! (Or perhaps more accurately, "simple" combinatorial facts about $omega$ secretly depend on more than just the pigeonhole principle.)
$endgroup$
Even for the simplest nontrivial situation $K={2}$ - let alone $2$-element $K$s - we're forced to leave ZFC:
The statement "There is a cardinal $kappa$ such that every two-coloring of pairs $c:[kappa]^2rightarrow 2$ has a homogeneous set of size $kappa$" is independent of ZFC.
In fact, a cardinal satisfying Ramsey's theorem for pairs (= your "${2}$-Ramsey") is called weakly compact, and this is a significant large cardinal property. So the pigeonhole principle is actually extremely limited in the uncountable setting! (Or perhaps more accurately, "simple" combinatorial facts about $omega$ secretly depend on more than just the pigeonhole principle.)
answered Dec 12 '18 at 19:35
Noah SchweberNoah Schweber
124k10150287
124k10150287
add a comment |
add a comment |
$begingroup$
A couple of results, using the "arrow notation": For any set $A$ and any (finite or infinite) cardinals $b,c,d,$ the notation $$Ato (b)^c_d$$ means that for every $f:[A]^cto d$ there exists $Sin [A]^b$ such that $|f(x):xin S|=1.$ (I.e. $S$ is homogeneous for $f$.)
For infinite cardinal ordinal $a$ we have $neg (,2^ato (a^+)^2_2,).$
For uncountable cardinal ordinal $a,$ if $ato (a)^2_2,$ we can show readily that $a$ is regular, and we can use 1. to show that $a$ is a strong limit cardinal, so $a$ is strongly-inaccessible. Such an $a$ is called weakly compact, and is "rather large", much larger than the least strongly-inaccessible.
In particular, for 1., for $|a|=omega:$ Let $<_W$ be a well-order on $Bbb R$ and let $<$ be the arithmetic order. Define $f: [Bbb R]^2to {0,1}$ where $f({x,y})=1$ iff $<_W$ and $<$ agree on ${x,y}.$ And let $x<^*y$ iff $y<x. $ Now if $S$ is homogenous for $f,$ then $<$ or $<^*$ is a well-order on $S.$ But since $Ssubset Bbb R,$ this implies $S$ is countable.
For 1. in general, let $<_L$ be the lexicographic order on $^{{0,1}}a:$ For functions $g,h$ from $a$ to ${0,1}$ with $gne h,$ let $x=min {yin a: g(y)ne h(y)}.$ Let $g<_L h$ iff $g(x)=0$ and $h(x)=1.$ And let $<_W$ be a well-order on $^{{0,1}}a.$.... Now define $f:[^{{0,1}}a]^2to {0,1}$ by letting $f({g,h})=1$ iff $<_L$ and $<_W$ agree on ${g,h}.$ A theorem of "Erdos & company" is that if $S$ is homogeneous for $f$ then $|S|leq a.$
Paul Erdos is an author or co-author of over 1500 published papers, and the number of other co-authors on them is in the thousands.
$endgroup$
add a comment |
$begingroup$
A couple of results, using the "arrow notation": For any set $A$ and any (finite or infinite) cardinals $b,c,d,$ the notation $$Ato (b)^c_d$$ means that for every $f:[A]^cto d$ there exists $Sin [A]^b$ such that $|f(x):xin S|=1.$ (I.e. $S$ is homogeneous for $f$.)
For infinite cardinal ordinal $a$ we have $neg (,2^ato (a^+)^2_2,).$
For uncountable cardinal ordinal $a,$ if $ato (a)^2_2,$ we can show readily that $a$ is regular, and we can use 1. to show that $a$ is a strong limit cardinal, so $a$ is strongly-inaccessible. Such an $a$ is called weakly compact, and is "rather large", much larger than the least strongly-inaccessible.
In particular, for 1., for $|a|=omega:$ Let $<_W$ be a well-order on $Bbb R$ and let $<$ be the arithmetic order. Define $f: [Bbb R]^2to {0,1}$ where $f({x,y})=1$ iff $<_W$ and $<$ agree on ${x,y}.$ And let $x<^*y$ iff $y<x. $ Now if $S$ is homogenous for $f,$ then $<$ or $<^*$ is a well-order on $S.$ But since $Ssubset Bbb R,$ this implies $S$ is countable.
For 1. in general, let $<_L$ be the lexicographic order on $^{{0,1}}a:$ For functions $g,h$ from $a$ to ${0,1}$ with $gne h,$ let $x=min {yin a: g(y)ne h(y)}.$ Let $g<_L h$ iff $g(x)=0$ and $h(x)=1.$ And let $<_W$ be a well-order on $^{{0,1}}a.$.... Now define $f:[^{{0,1}}a]^2to {0,1}$ by letting $f({g,h})=1$ iff $<_L$ and $<_W$ agree on ${g,h}.$ A theorem of "Erdos & company" is that if $S$ is homogeneous for $f$ then $|S|leq a.$
Paul Erdos is an author or co-author of over 1500 published papers, and the number of other co-authors on them is in the thousands.
$endgroup$
add a comment |
$begingroup$
A couple of results, using the "arrow notation": For any set $A$ and any (finite or infinite) cardinals $b,c,d,$ the notation $$Ato (b)^c_d$$ means that for every $f:[A]^cto d$ there exists $Sin [A]^b$ such that $|f(x):xin S|=1.$ (I.e. $S$ is homogeneous for $f$.)
For infinite cardinal ordinal $a$ we have $neg (,2^ato (a^+)^2_2,).$
For uncountable cardinal ordinal $a,$ if $ato (a)^2_2,$ we can show readily that $a$ is regular, and we can use 1. to show that $a$ is a strong limit cardinal, so $a$ is strongly-inaccessible. Such an $a$ is called weakly compact, and is "rather large", much larger than the least strongly-inaccessible.
In particular, for 1., for $|a|=omega:$ Let $<_W$ be a well-order on $Bbb R$ and let $<$ be the arithmetic order. Define $f: [Bbb R]^2to {0,1}$ where $f({x,y})=1$ iff $<_W$ and $<$ agree on ${x,y}.$ And let $x<^*y$ iff $y<x. $ Now if $S$ is homogenous for $f,$ then $<$ or $<^*$ is a well-order on $S.$ But since $Ssubset Bbb R,$ this implies $S$ is countable.
For 1. in general, let $<_L$ be the lexicographic order on $^{{0,1}}a:$ For functions $g,h$ from $a$ to ${0,1}$ with $gne h,$ let $x=min {yin a: g(y)ne h(y)}.$ Let $g<_L h$ iff $g(x)=0$ and $h(x)=1.$ And let $<_W$ be a well-order on $^{{0,1}}a.$.... Now define $f:[^{{0,1}}a]^2to {0,1}$ by letting $f({g,h})=1$ iff $<_L$ and $<_W$ agree on ${g,h}.$ A theorem of "Erdos & company" is that if $S$ is homogeneous for $f$ then $|S|leq a.$
Paul Erdos is an author or co-author of over 1500 published papers, and the number of other co-authors on them is in the thousands.
$endgroup$
A couple of results, using the "arrow notation": For any set $A$ and any (finite or infinite) cardinals $b,c,d,$ the notation $$Ato (b)^c_d$$ means that for every $f:[A]^cto d$ there exists $Sin [A]^b$ such that $|f(x):xin S|=1.$ (I.e. $S$ is homogeneous for $f$.)
For infinite cardinal ordinal $a$ we have $neg (,2^ato (a^+)^2_2,).$
For uncountable cardinal ordinal $a,$ if $ato (a)^2_2,$ we can show readily that $a$ is regular, and we can use 1. to show that $a$ is a strong limit cardinal, so $a$ is strongly-inaccessible. Such an $a$ is called weakly compact, and is "rather large", much larger than the least strongly-inaccessible.
In particular, for 1., for $|a|=omega:$ Let $<_W$ be a well-order on $Bbb R$ and let $<$ be the arithmetic order. Define $f: [Bbb R]^2to {0,1}$ where $f({x,y})=1$ iff $<_W$ and $<$ agree on ${x,y}.$ And let $x<^*y$ iff $y<x. $ Now if $S$ is homogenous for $f,$ then $<$ or $<^*$ is a well-order on $S.$ But since $Ssubset Bbb R,$ this implies $S$ is countable.
For 1. in general, let $<_L$ be the lexicographic order on $^{{0,1}}a:$ For functions $g,h$ from $a$ to ${0,1}$ with $gne h,$ let $x=min {yin a: g(y)ne h(y)}.$ Let $g<_L h$ iff $g(x)=0$ and $h(x)=1.$ And let $<_W$ be a well-order on $^{{0,1}}a.$.... Now define $f:[^{{0,1}}a]^2to {0,1}$ by letting $f({g,h})=1$ iff $<_L$ and $<_W$ agree on ${g,h}.$ A theorem of "Erdos & company" is that if $S$ is homogeneous for $f$ then $|S|leq a.$
Paul Erdos is an author or co-author of over 1500 published papers, and the number of other co-authors on them is in the thousands.
edited Dec 17 '18 at 9:37
answered Dec 17 '18 at 9:22
DanielWainfleetDanielWainfleet
35.1k31648
35.1k31648
add a comment |
add a comment |
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