A property of an ultrafilter

Clash Royale CLAN TAG#URR8PPP

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Let $mathcal U$ be a free ultrafilter on a set $X$. For $ninmathbb N$ let $mathcal F$ be a family of $n$-element subsets of $X$ such that $bigcupmathcal Finmathcal U$.

Question. Is there a set $Uinmathcal U$ and a subfamily $mathcal Esubsetmathcal F$ such that $bigcupmathcal Einmathcal U$ and $|Ucap E|=1$ for every $Einmathcal E$?

I do not know the answer even for $n=2$ and countable set $X$.

set-theory ultrafilters

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asked 2 hours ago

Taras Banakh

14.1k12882

add a comment | 

up vote
5
down vote

favorite

Let $mathcal U$ be a free ultrafilter on a set $X$. For $ninmathbb N$ let $mathcal F$ be a family of $n$-element subsets of $X$ such that $bigcupmathcal Finmathcal U$.

Question. Is there a set $Uinmathcal U$ and a subfamily $mathcal Esubsetmathcal F$ such that $bigcupmathcal Einmathcal U$ and $|Ucap E|=1$ for every $Einmathcal E$?

I do not know the answer even for $n=2$ and countable set $X$.

set-theory ultrafilters

share|cite|improve this question

asked 2 hours ago

Taras Banakh

14.1k12882

add a comment | 

up vote
5
down vote

favorite

up vote
5
down vote

favorite

Let $mathcal U$ be a free ultrafilter on a set $X$. For $ninmathbb N$ let $mathcal F$ be a family of $n$-element subsets of $X$ such that $bigcupmathcal Finmathcal U$.

Question. Is there a set $Uinmathcal U$ and a subfamily $mathcal Esubsetmathcal F$ such that $bigcupmathcal Einmathcal U$ and $|Ucap E|=1$ for every $Einmathcal E$?

I do not know the answer even for $n=2$ and countable set $X$.

set-theory ultrafilters

share|cite|improve this question

asked 2 hours ago

Taras Banakh

14.1k12882

Let $mathcal U$ be a free ultrafilter on a set $X$. For $ninmathbb N$ let $mathcal F$ be a family of $n$-element subsets of $X$ such that $bigcupmathcal Finmathcal U$.

Question. Is there a set $Uinmathcal U$ and a subfamily $mathcal Esubsetmathcal F$ such that $bigcupmathcal Einmathcal U$ and $|Ucap E|=1$ for every $Einmathcal E$?

I do not know the answer even for $n=2$ and countable set $X$.

set-theory ultrafilters

set-theory ultrafilters

share|cite|improve this question

asked 2 hours ago

Taras Banakh

14.1k12882

share|cite|improve this question

asked 2 hours ago

Taras Banakh

14.1k12882

share|cite|improve this question

share|cite|improve this question

asked 2 hours ago

Taras Banakh

14.1k12882

asked 2 hours ago

Taras Banakh

14.1k12882

asked 2 hours ago

Taras Banakh

14.1k12882

14.1k12882

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

Here's a try for $n=3$ which I think generalizes to any $n$.

Well-order $mathcalF$ as $F_alpha$. By discarding any $F_alpha$ which is contained in $bigcup_beta <alpha F_beta$, we can ensure that each $F_alpha$ contains at least one point which is not in any previous $F_beta$, without affecting $U = bigcup F_alpha$.

We can now inductively construct a set $V subset U$ with the property that $|V cap F_alpha|= 1$ or $2$ for all alpha. At each $alpha$ at least one element of $F_alpha$ is new and we can include it or not to ensure the condition for $F_alpha$.

Now $U setminus V$ has the same property, so since $U in mathcalU$, wlog we can assume $V in mathcalU$.

Let $mathcalE_1 subseteq mathcalF$ consist of those $F_alpha$ which intersect $V$ in exactly one point, and let $mathcalE_2$ consist of the other $F_alpha$, which all intersect $V$ in two points. If the union of $mathcalE_1$ belongs to $mathcalU$ then we are done. Otherwise the union of $mathcalE_2$ belongs to $mathcalU$ and this reduces us to the $n=2$ case.

The general reduction turns the problem for an arbitrary $n$ into the same problem for some smaller $n$.

share|cite|improve this answer

answered 20 mins ago

Nik Weaver

18.9k145117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So nice argument! Thank you for the help.
  – Taras Banakh
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome!
  – Nik Weaver
  35 secs ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

For $n=2$ and any $X$: for each $xin cup mathcal F$ fix any edge $(x,y)in mathcal F$ and draw an arrow from $x$ to $y$. Remove all other edges. Remaining edges still have the same union as $mathcalF$, but they form a directed graph with outdegrees at most 1. Such a graph has a proper 3-coloring (it suffices to prove this for finite subgraphs by compactness theorem, and for them the result is well known and easy: remove the vertex with minimal in-degree, it is at most 1, and proceed by induction). One of three colors works.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Fedor Petrov

45.5k5109214

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You can alternatively take a covering forest of the graph (which constructs $mathcalE$), and then you have a 2-coloring (fix 1 vertex of some color in each component, and elements of a single component have the same color iff they're at even distance), and then take the set of vertices of some fixed color.
  – YCor
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It's just a maximal subset of edges with containing no loop.
  – YCor
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the answers. What about the case $nge 3$? (Starting from $n=3$ I have some applications to the problem of normality of hyperballeans).
  – Taras Banakh
  27 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  My argument with covering forest does not work, I think. Instead, one consider a maximal non-covering subfamily of $mathcalE$, so it has the same union: then the corresponding graph has the property that no edge joins two vertices of valency $ge 2$. It follows that each of its component is a tree, with one vertex joined to all others.
  – YCor
  14 mins ago
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Here's a try for $n=3$ which I think generalizes to any $n$.

Well-order $mathcalF$ as $F_alpha$. By discarding any $F_alpha$ which is contained in $bigcup_beta <alpha F_beta$, we can ensure that each $F_alpha$ contains at least one point which is not in any previous $F_beta$, without affecting $U = bigcup F_alpha$.

We can now inductively construct a set $V subset U$ with the property that $|V cap F_alpha|= 1$ or $2$ for all alpha. At each $alpha$ at least one element of $F_alpha$ is new and we can include it or not to ensure the condition for $F_alpha$.

Now $U setminus V$ has the same property, so since $U in mathcalU$, wlog we can assume $V in mathcalU$.

Let $mathcalE_1 subseteq mathcalF$ consist of those $F_alpha$ which intersect $V$ in exactly one point, and let $mathcalE_2$ consist of the other $F_alpha$, which all intersect $V$ in two points. If the union of $mathcalE_1$ belongs to $mathcalU$ then we are done. Otherwise the union of $mathcalE_2$ belongs to $mathcalU$ and this reduces us to the $n=2$ case.

The general reduction turns the problem for an arbitrary $n$ into the same problem for some smaller $n$.

share|cite|improve this answer

answered 20 mins ago

Nik Weaver

18.9k145117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So nice argument! Thank you for the help.
  – Taras Banakh
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome!
  – Nik Weaver
  35 secs ago
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

Here's a try for $n=3$ which I think generalizes to any $n$.

Well-order $mathcalF$ as $F_alpha$. By discarding any $F_alpha$ which is contained in $bigcup_beta <alpha F_beta$, we can ensure that each $F_alpha$ contains at least one point which is not in any previous $F_beta$, without affecting $U = bigcup F_alpha$.

We can now inductively construct a set $V subset U$ with the property that $|V cap F_alpha|= 1$ or $2$ for all alpha. At each $alpha$ at least one element of $F_alpha$ is new and we can include it or not to ensure the condition for $F_alpha$.

Now $U setminus V$ has the same property, so since $U in mathcalU$, wlog we can assume $V in mathcalU$.

Let $mathcalE_1 subseteq mathcalF$ consist of those $F_alpha$ which intersect $V$ in exactly one point, and let $mathcalE_2$ consist of the other $F_alpha$, which all intersect $V$ in two points. If the union of $mathcalE_1$ belongs to $mathcalU$ then we are done. Otherwise the union of $mathcalE_2$ belongs to $mathcalU$ and this reduces us to the $n=2$ case.

The general reduction turns the problem for an arbitrary $n$ into the same problem for some smaller $n$.

share|cite|improve this answer

answered 20 mins ago

Nik Weaver

18.9k145117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So nice argument! Thank you for the help.
  – Taras Banakh
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome!
  – Nik Weaver
  35 secs ago
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Here's a try for $n=3$ which I think generalizes to any $n$.

Well-order $mathcalF$ as $F_alpha$. By discarding any $F_alpha$ which is contained in $bigcup_beta <alpha F_beta$, we can ensure that each $F_alpha$ contains at least one point which is not in any previous $F_beta$, without affecting $U = bigcup F_alpha$.

We can now inductively construct a set $V subset U$ with the property that $|V cap F_alpha|= 1$ or $2$ for all alpha. At each $alpha$ at least one element of $F_alpha$ is new and we can include it or not to ensure the condition for $F_alpha$.

Now $U setminus V$ has the same property, so since $U in mathcalU$, wlog we can assume $V in mathcalU$.

Let $mathcalE_1 subseteq mathcalF$ consist of those $F_alpha$ which intersect $V$ in exactly one point, and let $mathcalE_2$ consist of the other $F_alpha$, which all intersect $V$ in two points. If the union of $mathcalE_1$ belongs to $mathcalU$ then we are done. Otherwise the union of $mathcalE_2$ belongs to $mathcalU$ and this reduces us to the $n=2$ case.

The general reduction turns the problem for an arbitrary $n$ into the same problem for some smaller $n$.

share|cite|improve this answer

answered 20 mins ago

Nik Weaver

18.9k145117

Here's a try for $n=3$ which I think generalizes to any $n$.

Well-order $mathcalF$ as $F_alpha$. By discarding any $F_alpha$ which is contained in $bigcup_beta <alpha F_beta$, we can ensure that each $F_alpha$ contains at least one point which is not in any previous $F_beta$, without affecting $U = bigcup F_alpha$.

We can now inductively construct a set $V subset U$ with the property that $|V cap F_alpha|= 1$ or $2$ for all alpha. At each $alpha$ at least one element of $F_alpha$ is new and we can include it or not to ensure the condition for $F_alpha$.

Now $U setminus V$ has the same property, so since $U in mathcalU$, wlog we can assume $V in mathcalU$.

Let $mathcalE_1 subseteq mathcalF$ consist of those $F_alpha$ which intersect $V$ in exactly one point, and let $mathcalE_2$ consist of the other $F_alpha$, which all intersect $V$ in two points. If the union of $mathcalE_1$ belongs to $mathcalU$ then we are done. Otherwise the union of $mathcalE_2$ belongs to $mathcalU$ and this reduces us to the $n=2$ case.

The general reduction turns the problem for an arbitrary $n$ into the same problem for some smaller $n$.

share|cite|improve this answer

answered 20 mins ago

Nik Weaver

18.9k145117

share|cite|improve this answer

share|cite|improve this answer

answered 20 mins ago

Nik Weaver

18.9k145117

answered 20 mins ago

Nik Weaver

18.9k145117

answered 20 mins ago

Nik Weaver

18.9k145117

18.9k145117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So nice argument! Thank you for the help.
  – Taras Banakh
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome!
  – Nik Weaver
  35 secs ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So nice argument! Thank you for the help.
  – Taras Banakh
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are welcome!
  – Nik Weaver
  35 secs ago
  

  
  
  
  

So nice argument! Thank you for the help.
– Taras Banakh
11 mins ago

So nice argument! Thank you for the help.
– Taras Banakh
11 mins ago

You are welcome!
– Nik Weaver
35 secs ago

You are welcome!
– Nik Weaver
35 secs ago

add a comment | 

up vote
2
down vote

For $n=2$ and any $X$: for each $xin cup mathcal F$ fix any edge $(x,y)in mathcal F$ and draw an arrow from $x$ to $y$. Remove all other edges. Remaining edges still have the same union as $mathcalF$, but they form a directed graph with outdegrees at most 1. Such a graph has a proper 3-coloring (it suffices to prove this for finite subgraphs by compactness theorem, and for them the result is well known and easy: remove the vertex with minimal in-degree, it is at most 1, and proceed by induction). One of three colors works.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Fedor Petrov

45.5k5109214

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You can alternatively take a covering forest of the graph (which constructs $mathcalE$), and then you have a 2-coloring (fix 1 vertex of some color in each component, and elements of a single component have the same color iff they're at even distance), and then take the set of vertices of some fixed color.
  – YCor
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It's just a maximal subset of edges with containing no loop.
  – YCor
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the answers. What about the case $nge 3$? (Starting from $n=3$ I have some applications to the problem of normality of hyperballeans).
  – Taras Banakh
  27 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  My argument with covering forest does not work, I think. Instead, one consider a maximal non-covering subfamily of $mathcalE$, so it has the same union: then the corresponding graph has the property that no edge joins two vertices of valency $ge 2$. It follows that each of its component is a tree, with one vertex joined to all others.
  – YCor
  14 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

For $n=2$ and any $X$: for each $xin cup mathcal F$ fix any edge $(x,y)in mathcal F$ and draw an arrow from $x$ to $y$. Remove all other edges. Remaining edges still have the same union as $mathcalF$, but they form a directed graph with outdegrees at most 1. Such a graph has a proper 3-coloring (it suffices to prove this for finite subgraphs by compactness theorem, and for them the result is well known and easy: remove the vertex with minimal in-degree, it is at most 1, and proceed by induction). One of three colors works.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Fedor Petrov

45.5k5109214

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You can alternatively take a covering forest of the graph (which constructs $mathcalE$), and then you have a 2-coloring (fix 1 vertex of some color in each component, and elements of a single component have the same color iff they're at even distance), and then take the set of vertices of some fixed color.
  – YCor
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It's just a maximal subset of edges with containing no loop.
  – YCor
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the answers. What about the case $nge 3$? (Starting from $n=3$ I have some applications to the problem of normality of hyperballeans).
  – Taras Banakh
  27 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  My argument with covering forest does not work, I think. Instead, one consider a maximal non-covering subfamily of $mathcalE$, so it has the same union: then the corresponding graph has the property that no edge joins two vertices of valency $ge 2$. It follows that each of its component is a tree, with one vertex joined to all others.
  – YCor
  14 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

For $n=2$ and any $X$: for each $xin cup mathcal F$ fix any edge $(x,y)in mathcal F$ and draw an arrow from $x$ to $y$. Remove all other edges. Remaining edges still have the same union as $mathcalF$, but they form a directed graph with outdegrees at most 1. Such a graph has a proper 3-coloring (it suffices to prove this for finite subgraphs by compactness theorem, and for them the result is well known and easy: remove the vertex with minimal in-degree, it is at most 1, and proceed by induction). One of three colors works.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Fedor Petrov

45.5k5109214

For $n=2$ and any $X$: for each $xin cup mathcal F$ fix any edge $(x,y)in mathcal F$ and draw an arrow from $x$ to $y$. Remove all other edges. Remaining edges still have the same union as $mathcalF$, but they form a directed graph with outdegrees at most 1. Such a graph has a proper 3-coloring (it suffices to prove this for finite subgraphs by compactness theorem, and for them the result is well known and easy: remove the vertex with minimal in-degree, it is at most 1, and proceed by induction). One of three colors works.

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Fedor Petrov

45.5k5109214

share|cite|improve this answer

share|cite|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 1 hour ago

Fedor Petrov

45.5k5109214

answered 1 hour ago

Fedor Petrov

45.5k5109214

answered 1 hour ago

Fedor Petrov

45.5k5109214

45.5k5109214

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You can alternatively take a covering forest of the graph (which constructs $mathcalE$), and then you have a 2-coloring (fix 1 vertex of some color in each component, and elements of a single component have the same color iff they're at even distance), and then take the set of vertices of some fixed color.
  – YCor
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It's just a maximal subset of edges with containing no loop.
  – YCor
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the answers. What about the case $nge 3$? (Starting from $n=3$ I have some applications to the problem of normality of hyperballeans).
  – Taras Banakh
  27 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  My argument with covering forest does not work, I think. Instead, one consider a maximal non-covering subfamily of $mathcalE$, so it has the same union: then the corresponding graph has the property that no edge joins two vertices of valency $ge 2$. It follows that each of its component is a tree, with one vertex joined to all others.
  – YCor
  14 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You can alternatively take a covering forest of the graph (which constructs $mathcalE$), and then you have a 2-coloring (fix 1 vertex of some color in each component, and elements of a single component have the same color iff they're at even distance), and then take the set of vertices of some fixed color.
  – YCor
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  It's just a maximal subset of edges with containing no loop.
  – YCor
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the answers. What about the case $nge 3$? (Starting from $n=3$ I have some applications to the problem of normality of hyperballeans).
  – Taras Banakh
  27 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  My argument with covering forest does not work, I think. Instead, one consider a maximal non-covering subfamily of $mathcalE$, so it has the same union: then the corresponding graph has the property that no edge joins two vertices of valency $ge 2$. It follows that each of its component is a tree, with one vertex joined to all others.
  – YCor
  14 mins ago
  

  
  
  
  

2

2

You can alternatively take a covering forest of the graph (which constructs $mathcalE$), and then you have a 2-coloring (fix 1 vertex of some color in each component, and elements of a single component have the same color iff they're at even distance), and then take the set of vertices of some fixed color.
– YCor
1 hour ago

You can alternatively take a covering forest of the graph (which constructs $mathcalE$), and then you have a 2-coloring (fix 1 vertex of some color in each component, and elements of a single component have the same color iff they're at even distance), and then take the set of vertices of some fixed color.
– YCor
1 hour ago

2

2

It's just a maximal subset of edges with containing no loop.
– YCor
1 hour ago

It's just a maximal subset of edges with containing no loop.
– YCor
1 hour ago

Thank you for the answers. What about the case $nge 3$? (Starting from $n=3$ I have some applications to the problem of normality of hyperballeans).
– Taras Banakh
27 mins ago

Thank you for the answers. What about the case $nge 3$? (Starting from $n=3$ I have some applications to the problem of normality of hyperballeans).
– Taras Banakh
27 mins ago

1

1

My argument with covering forest does not work, I think. Instead, one consider a maximal non-covering subfamily of $mathcalE$, so it has the same union: then the corresponding graph has the property that no edge joins two vertices of valency $ge 2$. It follows that each of its component is a tree, with one vertex joined to all others.
– YCor
14 mins ago

My argument with covering forest does not work, I think. Instead, one consider a maximal non-covering subfamily of $mathcalE$, so it has the same union: then the corresponding graph has the property that no edge joins two vertices of valency $ge 2$. It follows that each of its component is a tree, with one vertex joined to all others.
– YCor
14 mins ago

add a comment | 

 

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Name Email

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