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Why can't the tower number be $aleph_0$?

Does this make sense $aleph_0+aleph_1+aleph_2$?An increasing ${omega}_1$-sequence of sets of size $2^{aleph_0}$.$mathbb{N}$ with uncountably many infinite subsetsProving that for infinite $kappa$, $|[kappa]^lambda|=kappa^lambda$Proof that aleph null is the smallest transfinite number?The cardinality of the set of all linear order types over $omega$ is $2^{aleph_0}+aleph_1$ in ZF+AD?Partitioning an infinite cardinal numberConstruction of a Ramsey ultrafilterTower number is a regular uncountable cardinalDoubts about tower number.

2

$begingroup$

A set $A$ is said to be almost contained in a set $B$ if $Asetminus B$ is finite.
A sequence $(A_alpha)_{alpha<lambda}$ of infinite subsets of $mathbb N$ will be called a tower if for every $alpha<beta<lambda$, $A_beta$ is almost contained in $A_alpha$.
A tower is said to have a continuation if exists some infinite subset of $mathbb N$ that is almost contained in every element of the tower.
The tower number is defined to be the minimal cardinality of the set of towers that don't have a continuation.

My question is: why can't the tower number be $aleph_0$? Or equivalently, why does every tower of countable length necessarily have a continuation?

set-theory

share|cite|improve this question

edited Mar 10 at 21:38

Uri George Peterzil

asked Mar 10 at 21:22

Uri George PeterzilUri George Peterzil

10610

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you need to be a little more specific with your definitions here; for instance, every tower has a continuation because I can just take the continuing set to be $mathbb{N}$; presumably you want the continuing set to not show up in the sequence under consideration?
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is it. It must not appear in the sequence previously. Will now edit
  $endgroup$
  – Uri George Peterzil
  Mar 10 at 21:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wait, the set $mathbb N$ doesn't work as it is reverse inclusion. So no, any set.
  $endgroup$
  – Uri George Peterzil
  Mar 10 at 21:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The way you have defined almost containment makes $mathbb{N}$ almost contained in anything, as $Bsetminusmathbb{N}=emptyset$.
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, corrected my definition.
  $endgroup$
  – Uri George Peterzil
  Mar 10 at 21:38
  

  
  
  
  

add a comment | 

2

$begingroup$

A set $A$ is said to be almost contained in a set $B$ if $Asetminus B$ is finite.
A sequence $(A_alpha)_{alpha<lambda}$ of infinite subsets of $mathbb N$ will be called a tower if for every $alpha<beta<lambda$, $A_beta$ is almost contained in $A_alpha$.
A tower is said to have a continuation if exists some infinite subset of $mathbb N$ that is almost contained in every element of the tower.
The tower number is defined to be the minimal cardinality of the set of towers that don't have a continuation.

My question is: why can't the tower number be $aleph_0$? Or equivalently, why does every tower of countable length necessarily have a continuation?

set-theory

share|cite|improve this question

edited Mar 10 at 21:38

Uri George Peterzil

asked Mar 10 at 21:22

Uri George PeterzilUri George Peterzil

10610

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you need to be a little more specific with your definitions here; for instance, every tower has a continuation because I can just take the continuing set to be $mathbb{N}$; presumably you want the continuing set to not show up in the sequence under consideration?
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is it. It must not appear in the sequence previously. Will now edit
  $endgroup$
  – Uri George Peterzil
  Mar 10 at 21:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wait, the set $mathbb N$ doesn't work as it is reverse inclusion. So no, any set.
  $endgroup$
  – Uri George Peterzil
  Mar 10 at 21:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The way you have defined almost containment makes $mathbb{N}$ almost contained in anything, as $Bsetminusmathbb{N}=emptyset$.
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, corrected my definition.
  $endgroup$
  – Uri George Peterzil
  Mar 10 at 21:38
  

  
  
  
  

add a comment | 

$begingroup$

A set $A$ is said to be almost contained in a set $B$ if $Asetminus B$ is finite.
A sequence $(A_alpha)_{alpha<lambda}$ of infinite subsets of $mathbb N$ will be called a tower if for every $alpha<beta<lambda$, $A_beta$ is almost contained in $A_alpha$.
A tower is said to have a continuation if exists some infinite subset of $mathbb N$ that is almost contained in every element of the tower.
The tower number is defined to be the minimal cardinality of the set of towers that don't have a continuation.

My question is: why can't the tower number be $aleph_0$? Or equivalently, why does every tower of countable length necessarily have a continuation?

set-theory

share|cite|improve this question

edited Mar 10 at 21:38

Uri George Peterzil

asked Mar 10 at 21:22

Uri George PeterzilUri George Peterzil

10610

$endgroup$

share|cite|improve this question

edited Mar 10 at 21:38

Uri George Peterzil

asked Mar 10 at 21:22

Uri George PeterzilUri George Peterzil

10610

share|cite|improve this question

edited Mar 10 at 21:38

Uri George Peterzil

asked Mar 10 at 21:22

Uri George PeterzilUri George Peterzil

10610

asked Mar 10 at 21:22

Uri George PeterzilUri George Peterzil

10610

asked Mar 10 at 21:22

Uri George PeterzilUri George Peterzil

10610

1 Answer
1

active

oldest

votes

5

$begingroup$

This is a simple diagonalization argument. Suppose we have a countable tower. Taking a cofinal subsequence, we may assume it has length $omega$ and write it as $(A_n)_{n<omega}$. We may moreover assume that the $A_n$ are actually literally contained in each other instead of almost contained in each other, by replacing $A_n$ with $A_0capdotscap A_n$.

It's now easy to construct a set $B$ which is almost contained in each $A_n$: just pick one element from each $A_n$ to be in $B$. Picking these elements one by one, we can arrange that they are all distinct (since each $A_n$ is infinite), so that the resulting set $B$ is infinite. For each $n$, all but possibly the first $n$ elements we put in $B$ must be in $A_n$ (since they are in $A_m$ for some $mgeq n$), so $B$ is almost contained in $A_n$.

share|cite|improve this answer

answered Mar 10 at 21:36

Eric WofseyEric Wofsey

189k14216347

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does the edit to the definition of almost containment change your construction of $B$ at all? To be perfectly frank, I have a hard time following your answer due to the lack of justification for reducing to a sequence of length $omega$ of literally contained sets.
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I wrote this answer using the correct definition; it's a very commonly used definition in set theory.
  $endgroup$
  – Eric Wofsey
  Mar 10 at 21:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cool. I am not familiar with this at all, so I was just curious. Can you still provide more context on why your reduction is okay?
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is irrelevant whether $B$ is in the original sequence; what matters is that $B$ is almost contained in every term of the original sequence. That will be true because $B$ is almost contained in every term of a cofinal subsequence of the original sequence.
  $endgroup$
  – Eric Wofsey
  Mar 10 at 21:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The asker added that remark because they were confused by your questions and didn't realize the issue was just that they had miswritten the definition of "almost contain".
  $endgroup$
  – Eric Wofsey
  Mar 10 at 21:55
  

  
  
  
  

 | 
show 4 more comments

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5

$begingroup$

This is a simple diagonalization argument. Suppose we have a countable tower. Taking a cofinal subsequence, we may assume it has length $omega$ and write it as $(A_n)_{n<omega}$. We may moreover assume that the $A_n$ are actually literally contained in each other instead of almost contained in each other, by replacing $A_n$ with $A_0capdotscap A_n$.

It's now easy to construct a set $B$ which is almost contained in each $A_n$: just pick one element from each $A_n$ to be in $B$. Picking these elements one by one, we can arrange that they are all distinct (since each $A_n$ is infinite), so that the resulting set $B$ is infinite. For each $n$, all but possibly the first $n$ elements we put in $B$ must be in $A_n$ (since they are in $A_m$ for some $mgeq n$), so $B$ is almost contained in $A_n$.

share|cite|improve this answer

answered Mar 10 at 21:36

Eric WofseyEric Wofsey

189k14216347

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does the edit to the definition of almost containment change your construction of $B$ at all? To be perfectly frank, I have a hard time following your answer due to the lack of justification for reducing to a sequence of length $omega$ of literally contained sets.
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I wrote this answer using the correct definition; it's a very commonly used definition in set theory.
  $endgroup$
  – Eric Wofsey
  Mar 10 at 21:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cool. I am not familiar with this at all, so I was just curious. Can you still provide more context on why your reduction is okay?
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is irrelevant whether $B$ is in the original sequence; what matters is that $B$ is almost contained in every term of the original sequence. That will be true because $B$ is almost contained in every term of a cofinal subsequence of the original sequence.
  $endgroup$
  – Eric Wofsey
  Mar 10 at 21:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The asker added that remark because they were confused by your questions and didn't realize the issue was just that they had miswritten the definition of "almost contain".
  $endgroup$
  – Eric Wofsey
  Mar 10 at 21:55
  

  
  
  
  

 | 
show 4 more comments

5

$begingroup$

This is a simple diagonalization argument. Suppose we have a countable tower. Taking a cofinal subsequence, we may assume it has length $omega$ and write it as $(A_n)_{n<omega}$. We may moreover assume that the $A_n$ are actually literally contained in each other instead of almost contained in each other, by replacing $A_n$ with $A_0capdotscap A_n$.

It's now easy to construct a set $B$ which is almost contained in each $A_n$: just pick one element from each $A_n$ to be in $B$. Picking these elements one by one, we can arrange that they are all distinct (since each $A_n$ is infinite), so that the resulting set $B$ is infinite. For each $n$, all but possibly the first $n$ elements we put in $B$ must be in $A_n$ (since they are in $A_m$ for some $mgeq n$), so $B$ is almost contained in $A_n$.

share|cite|improve this answer

answered Mar 10 at 21:36

Eric WofseyEric Wofsey

189k14216347

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does the edit to the definition of almost containment change your construction of $B$ at all? To be perfectly frank, I have a hard time following your answer due to the lack of justification for reducing to a sequence of length $omega$ of literally contained sets.
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I wrote this answer using the correct definition; it's a very commonly used definition in set theory.
  $endgroup$
  – Eric Wofsey
  Mar 10 at 21:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cool. I am not familiar with this at all, so I was just curious. Can you still provide more context on why your reduction is okay?
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is irrelevant whether $B$ is in the original sequence; what matters is that $B$ is almost contained in every term of the original sequence. That will be true because $B$ is almost contained in every term of a cofinal subsequence of the original sequence.
  $endgroup$
  – Eric Wofsey
  Mar 10 at 21:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The asker added that remark because they were confused by your questions and didn't realize the issue was just that they had miswritten the definition of "almost contain".
  $endgroup$
  – Eric Wofsey
  Mar 10 at 21:55
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

This is a simple diagonalization argument. Suppose we have a countable tower. Taking a cofinal subsequence, we may assume it has length $omega$ and write it as $(A_n)_{n<omega}$. We may moreover assume that the $A_n$ are actually literally contained in each other instead of almost contained in each other, by replacing $A_n$ with $A_0capdotscap A_n$.

It's now easy to construct a set $B$ which is almost contained in each $A_n$: just pick one element from each $A_n$ to be in $B$. Picking these elements one by one, we can arrange that they are all distinct (since each $A_n$ is infinite), so that the resulting set $B$ is infinite. For each $n$, all but possibly the first $n$ elements we put in $B$ must be in $A_n$ (since they are in $A_m$ for some $mgeq n$), so $B$ is almost contained in $A_n$.

share|cite|improve this answer

answered Mar 10 at 21:36

Eric WofseyEric Wofsey

189k14216347

$endgroup$

share|cite|improve this answer

answered Mar 10 at 21:36

Eric WofseyEric Wofsey

189k14216347

answered Mar 10 at 21:36

Eric WofseyEric Wofsey

189k14216347

answered Mar 10 at 21:36

Eric WofseyEric Wofsey

189k14216347