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A normal filter including the tail sets is $kappa$-complete

0

$begingroup$

I need to show that

if $cal F$ is a normal filter on a regular uncountable cardinal $kappa$, and if $cal F$ contains all tail sets, i.e. all
$$ C_alpha=beta : alpha<beta<kappa$$
Then $cal F$ is $kappa$-complete

My thought

Let $X$ be a collection of less than $kappa$ elements of $cal F$, i.e. for some $lambda<kappa$:
$$ X = X_alpha, alpha<lambda$$
With $X_alphain cal F$ for all $alpha$.

We need to show that $bigcap_alpha<lambda X_alpha in cal F$. Let $Y$ be the collection $X$ adding the tail set with $alphageqlambda$:
$$Y = X_alpha: alpha<lambda cup C_alpha : alpha geq lambda$$

Lets define $Y_alpha$ as
$$Y_alpha = left{ beginarrayll
X_alpha&text if alpha<lambda\
C_alpha&text if alphageqlambda
endarrayright.$$

For all $alpha$, $Y_alphaincal F$, then because $cal F$ is normal,
$$triangle_alpha<kappa Y_alpha in cal F$$
But by definition of the diagonal intersection $triangle_alpha<kappa Y_alpha = beta<kappa : beta in bigcap_alpha<betaY_alpha$ we have $triangle_alpha<kappaY_alpha subseteq bigcap_alpha<lambda X_alpha$. Therefore, because $cal F$ is a filter
$$ bigcap_alpha<lambda X_alphain cal F$$

My issue is with the statement
$$triangle_alpha<kappaY_alpha subseteq bigcap_alpha<lambda X_alpha$$

Let $beta in triangle_alpha<kappaY_alpha = beta<kappa : beta in bigcap_alpha<betaY_alpha$

Clearly if $lambda < beta < kappa$, then $betainbigcap_alpha<betaY_alpha$ and $betainbigcap_alpha<lambdaY_alpha=bigcap_alpha<lambdaX_alpha$

But if $beta < lambda$, how can I prove that $betainbigcap_alpha<lambdaX_alpha$ ?

set-theory filters

share|cite|improve this question

asked Mar 19 at 10:41

Thomas LesgourguesThomas Lesgourgues

1,285220

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Intersect $X_alpha$ with $C_lambda$ in your definition of $Y_alpha.$ That is still in $mathcal F$ of course, and it takes care of the last part for you.
  $endgroup$
  – spaceisdarkgreen
  Mar 19 at 11:19
  

  
  
  
  

add a comment | 

0

$begingroup$

I need to show that

if $cal F$ is a normal filter on a regular uncountable cardinal $kappa$, and if $cal F$ contains all tail sets, i.e. all
$$ C_alpha=beta : alpha<beta<kappa$$
Then $cal F$ is $kappa$-complete

My thought

Let $X$ be a collection of less than $kappa$ elements of $cal F$, i.e. for some $lambda<kappa$:
$$ X = X_alpha, alpha<lambda$$
With $X_alphain cal F$ for all $alpha$.

We need to show that $bigcap_alpha<lambda X_alpha in cal F$. Let $Y$ be the collection $X$ adding the tail set with $alphageqlambda$:
$$Y = X_alpha: alpha<lambda cup C_alpha : alpha geq lambda$$

Lets define $Y_alpha$ as
$$Y_alpha = left{ beginarrayll
X_alpha&text if alpha<lambda\
C_alpha&text if alphageqlambda
endarrayright.$$

For all $alpha$, $Y_alphaincal F$, then because $cal F$ is normal,
$$triangle_alpha<kappa Y_alpha in cal F$$
But by definition of the diagonal intersection $triangle_alpha<kappa Y_alpha = beta<kappa : beta in bigcap_alpha<betaY_alpha$ we have $triangle_alpha<kappaY_alpha subseteq bigcap_alpha<lambda X_alpha$. Therefore, because $cal F$ is a filter
$$ bigcap_alpha<lambda X_alphain cal F$$

My issue is with the statement
$$triangle_alpha<kappaY_alpha subseteq bigcap_alpha<lambda X_alpha$$

Let $beta in triangle_alpha<kappaY_alpha = beta<kappa : beta in bigcap_alpha<betaY_alpha$

Clearly if $lambda < beta < kappa$, then $betainbigcap_alpha<betaY_alpha$ and $betainbigcap_alpha<lambdaY_alpha=bigcap_alpha<lambdaX_alpha$

But if $beta < lambda$, how can I prove that $betainbigcap_alpha<lambdaX_alpha$ ?

set-theory filters

share|cite|improve this question

asked Mar 19 at 10:41

Thomas LesgourguesThomas Lesgourgues

1,285220

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Intersect $X_alpha$ with $C_lambda$ in your definition of $Y_alpha.$ That is still in $mathcal F$ of course, and it takes care of the last part for you.
  $endgroup$
  – spaceisdarkgreen
  Mar 19 at 11:19
  

  
  
  
  

add a comment | 

$begingroup$

I need to show that

if $cal F$ is a normal filter on a regular uncountable cardinal $kappa$, and if $cal F$ contains all tail sets, i.e. all
$$ C_alpha=beta : alpha<beta<kappa$$
Then $cal F$ is $kappa$-complete

My thought

Let $X$ be a collection of less than $kappa$ elements of $cal F$, i.e. for some $lambda<kappa$:
$$ X = X_alpha, alpha<lambda$$
With $X_alphain cal F$ for all $alpha$.

We need to show that $bigcap_alpha<lambda X_alpha in cal F$. Let $Y$ be the collection $X$ adding the tail set with $alphageqlambda$:
$$Y = X_alpha: alpha<lambda cup C_alpha : alpha geq lambda$$

Lets define $Y_alpha$ as
$$Y_alpha = left{ beginarrayll
X_alpha&text if alpha<lambda\
C_alpha&text if alphageqlambda
endarrayright.$$

For all $alpha$, $Y_alphaincal F$, then because $cal F$ is normal,
$$triangle_alpha<kappa Y_alpha in cal F$$
But by definition of the diagonal intersection $triangle_alpha<kappa Y_alpha = beta<kappa : beta in bigcap_alpha<betaY_alpha$ we have $triangle_alpha<kappaY_alpha subseteq bigcap_alpha<lambda X_alpha$. Therefore, because $cal F$ is a filter
$$ bigcap_alpha<lambda X_alphain cal F$$

My issue is with the statement
$$triangle_alpha<kappaY_alpha subseteq bigcap_alpha<lambda X_alpha$$

Let $beta in triangle_alpha<kappaY_alpha = beta<kappa : beta in bigcap_alpha<betaY_alpha$

Clearly if $lambda < beta < kappa$, then $betainbigcap_alpha<betaY_alpha$ and $betainbigcap_alpha<lambdaY_alpha=bigcap_alpha<lambdaX_alpha$

But if $beta < lambda$, how can I prove that $betainbigcap_alpha<lambdaX_alpha$ ?

set-theory filters

share|cite|improve this question

asked Mar 19 at 10:41

Thomas LesgourguesThomas Lesgourgues

1,285220

$endgroup$

share|cite|improve this question

asked Mar 19 at 10:41

Thomas LesgourguesThomas Lesgourgues

1,285220

share|cite|improve this question

asked Mar 19 at 10:41

Thomas LesgourguesThomas Lesgourgues

1,285220

asked Mar 19 at 10:41

Thomas LesgourguesThomas Lesgourgues

1,285220

asked Mar 19 at 10:41

Thomas LesgourguesThomas Lesgourgues

1,285220