How can we show that the intersection of a series of closed unbounded sets is closed unbounded? [duplicate]
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This question already has an answer here:
Club filter of $kappa$ is $kappa$-complete
1 answer
I am currently stuck on an exercise in set-theory and would like to get some help:
If $S$ is a stationary subset of a regular uncountable cardinal $κ$, and a subset $C$ of $κ$ is an S-cub if it is unbounded in $κ$ and $sup(x) ∈ C$
holds for every $x ⊆ C$ with $sup(x) ∈ S$, then how can we prove that for every sequence $(C_α | α < λ)$ of
S-cubs with λ < κ, $bigcap_{α<λ} C_α$ is an S-cub?
Thanks in advance.
set-theory
edited Jan 2 at 22:23
Davide Giraudo
127k16152266
asked Jan 2 at 21:50
P. DeckerP. Decker
112
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marked as duplicate by amWhy, Holo, Alessandro Codenotti, Cesareo, José Carlos Santos Jan 3 at 1:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Club filter of $kappa$ is $kappa$-complete
1 answer
I am currently stuck on an exercise in set-theory and would like to get some help:
If $S$ is a stationary subset of a regular uncountable cardinal $κ$, and a subset $C$ of $κ$ is an S-cub if it is unbounded in $κ$ and $sup(x) ∈ C$
holds for every $x ⊆ C$ with $sup(x) ∈ S$, then how can we prove that for every sequence $(C_α | α < λ)$ of
S-cubs with λ < κ, $bigcap_{α<λ} C_α$ is an S-cub?
Thanks in advance.
set-theory
edited Jan 2 at 22:23
Davide Giraudo
127k16152266
asked Jan 2 at 21:50
P. DeckerP. Decker
112
$endgroup$
marked as duplicate by amWhy, Holo, Alessandro Codenotti, Cesareo, José Carlos Santos Jan 3 at 1:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
It's not a duplicate of that question, since this question is about the $S$-clubs. The $C_alpha$ may not be club at all, but only $S$-club.
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– JDH
Jan 2 at 22:44
$begingroup$
Why was it closed as a duplicate? It isn't a duplicate.
$endgroup$
– JDH
Jan 3 at 3:14
add a comment |
$begingroup$
This question already has an answer here:
Club filter of $kappa$ is $kappa$-complete
1 answer
I am currently stuck on an exercise in set-theory and would like to get some help:
If $S$ is a stationary subset of a regular uncountable cardinal $κ$, and a subset $C$ of $κ$ is an S-cub if it is unbounded in $κ$ and $sup(x) ∈ C$
holds for every $x ⊆ C$ with $sup(x) ∈ S$, then how can we prove that for every sequence $(C_α | α < λ)$ of
S-cubs with λ < κ, $bigcap_{α<λ} C_α$ is an S-cub?
Thanks in advance.
set-theory
edited Jan 2 at 22:23
Davide Giraudo
127k16152266
asked Jan 2 at 21:50
P. DeckerP. Decker
112
$endgroup$
This question already has an answer here:
Club filter of $kappa$ is $kappa$-complete
1 answer
I am currently stuck on an exercise in set-theory and would like to get some help:
If $S$ is a stationary subset of a regular uncountable cardinal $κ$, and a subset $C$ of $κ$ is an S-cub if it is unbounded in $κ$ and $sup(x) ∈ C$
holds for every $x ⊆ C$ with $sup(x) ∈ S$, then how can we prove that for every sequence $(C_α | α < λ)$ of
S-cubs with λ < κ, $bigcap_{α<λ} C_α$ is an S-cub?
Thanks in advance.
This question already has an answer here:
Club filter of $kappa$ is $kappa$-complete
1 answer
set-theory
set-theory
edited Jan 2 at 22:23
Davide Giraudo
127k16152266
asked Jan 2 at 21:50
P. DeckerP. Decker
112
edited Jan 2 at 22:23
Davide Giraudo
127k16152266
asked Jan 2 at 21:50
P. DeckerP. Decker
112
edited Jan 2 at 22:23
Davide Giraudo
127k16152266
edited Jan 2 at 22:23
Davide Giraudo
127k16152266
edited Jan 2 at 22:23
Davide Giraudo
127k16152266
127k16152266
asked Jan 2 at 21:50
P. DeckerP. Decker
112
asked Jan 2 at 21:50
P. DeckerP. Decker
112
asked Jan 2 at 21:50
P. DeckerP. Decker
112
112
marked as duplicate by amWhy, Holo, Alessandro Codenotti, Cesareo, José Carlos Santos Jan 3 at 1:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by amWhy, Holo, Alessandro Codenotti, Cesareo, José Carlos Santos Jan 3 at 1:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
It's not a duplicate of that question, since this question is about the $S$-clubs. The $C_alpha$ may not be club at all, but only $S$-club.
$endgroup$
– JDH
Jan 2 at 22:44
$begingroup$
Why was it closed as a duplicate? It isn't a duplicate.
$endgroup$
– JDH
Jan 3 at 3:14
add a comment |
1
$begingroup$
It's not a duplicate of that question, since this question is about the $S$-clubs. The $C_alpha$ may not be club at all, but only $S$-club.
$endgroup$
– JDH
Jan 2 at 22:44
$begingroup$
Why was it closed as a duplicate? It isn't a duplicate.
$endgroup$
– JDH
Jan 3 at 3:14
1
1
$begingroup$
It's not a duplicate of that question, since this question is about the $S$-clubs. The $C_alpha$ may not be club at all, but only $S$-club.
$endgroup$
– JDH
Jan 2 at 22:44
$begingroup$
It's not a duplicate of that question, since this question is about the $S$-clubs. The $C_alpha$ may not be club at all, but only $S$-club.
$endgroup$
– JDH
Jan 2 at 22:44
$begingroup$
Why was it closed as a duplicate? It isn't a duplicate.
$endgroup$
– JDH
Jan 3 at 3:14
$begingroup$
Why was it closed as a duplicate? It isn't a duplicate.
$endgroup$
– JDH
Jan 3 at 3:14
add a comment |
1 Answer
1
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It is clear that we satisfy the closure part, since if a bounded set $x$ with $sup(x)in S$ is contained in the intersection $bigcap_alpha C_alpha$, then $x$ is contained in each $C_alpha$, and so $sup(x)$ is in each $C_alpha$ and hence in the intersection $bigcap_alpha C_alpha$, as desired.
So the only difficult part is to show that $bigcap_alpha C_alpha$ is unbounded in $kappa$. For this, let $bar C_alpha$ be the closure of $C_alpha$, that is, $C_alpha$ with all of its limit points. So this is a club set in $kappa$, and since the intersection of fewer than $kappa$ many clubs is club, it follows that $bigcap_alphabar C_alpha$ is closed and unbounded in $kappa$. Thus, since $S$ is stationary, there are unboundedly many $betain S$ that are limits of $bigcap_alpha bar C_alpha$, and these $beta$'s must all be in every $C_alpha$ since each $C_alpha$ is $S$-closed.
answered Jan 2 at 22:05
JDHJDH
32.6k680145
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is clear that we satisfy the closure part, since if a bounded set $x$ with $sup(x)in S$ is contained in the intersection $bigcap_alpha C_alpha$, then $x$ is contained in each $C_alpha$, and so $sup(x)$ is in each $C_alpha$ and hence in the intersection $bigcap_alpha C_alpha$, as desired.
So the only difficult part is to show that $bigcap_alpha C_alpha$ is unbounded in $kappa$. For this, let $bar C_alpha$ be the closure of $C_alpha$, that is, $C_alpha$ with all of its limit points. So this is a club set in $kappa$, and since the intersection of fewer than $kappa$ many clubs is club, it follows that $bigcap_alphabar C_alpha$ is closed and unbounded in $kappa$. Thus, since $S$ is stationary, there are unboundedly many $betain S$ that are limits of $bigcap_alpha bar C_alpha$, and these $beta$'s must all be in every $C_alpha$ since each $C_alpha$ is $S$-closed.
answered Jan 2 at 22:05
JDHJDH
32.6k680145
$endgroup$
add a comment |
$begingroup$
It is clear that we satisfy the closure part, since if a bounded set $x$ with $sup(x)in S$ is contained in the intersection $bigcap_alpha C_alpha$, then $x$ is contained in each $C_alpha$, and so $sup(x)$ is in each $C_alpha$ and hence in the intersection $bigcap_alpha C_alpha$, as desired.
So the only difficult part is to show that $bigcap_alpha C_alpha$ is unbounded in $kappa$. For this, let $bar C_alpha$ be the closure of $C_alpha$, that is, $C_alpha$ with all of its limit points. So this is a club set in $kappa$, and since the intersection of fewer than $kappa$ many clubs is club, it follows that $bigcap_alphabar C_alpha$ is closed and unbounded in $kappa$. Thus, since $S$ is stationary, there are unboundedly many $betain S$ that are limits of $bigcap_alpha bar C_alpha$, and these $beta$'s must all be in every $C_alpha$ since each $C_alpha$ is $S$-closed.
answered Jan 2 at 22:05
JDHJDH
32.6k680145
$endgroup$
add a comment |
$begingroup$
It is clear that we satisfy the closure part, since if a bounded set $x$ with $sup(x)in S$ is contained in the intersection $bigcap_alpha C_alpha$, then $x$ is contained in each $C_alpha$, and so $sup(x)$ is in each $C_alpha$ and hence in the intersection $bigcap_alpha C_alpha$, as desired.
So the only difficult part is to show that $bigcap_alpha C_alpha$ is unbounded in $kappa$. For this, let $bar C_alpha$ be the closure of $C_alpha$, that is, $C_alpha$ with all of its limit points. So this is a club set in $kappa$, and since the intersection of fewer than $kappa$ many clubs is club, it follows that $bigcap_alphabar C_alpha$ is closed and unbounded in $kappa$. Thus, since $S$ is stationary, there are unboundedly many $betain S$ that are limits of $bigcap_alpha bar C_alpha$, and these $beta$'s must all be in every $C_alpha$ since each $C_alpha$ is $S$-closed.
answered Jan 2 at 22:05
JDHJDH
32.6k680145
$endgroup$
It is clear that we satisfy the closure part, since if a bounded set $x$ with $sup(x)in S$ is contained in the intersection $bigcap_alpha C_alpha$, then $x$ is contained in each $C_alpha$, and so $sup(x)$ is in each $C_alpha$ and hence in the intersection $bigcap_alpha C_alpha$, as desired.
So the only difficult part is to show that $bigcap_alpha C_alpha$ is unbounded in $kappa$. For this, let $bar C_alpha$ be the closure of $C_alpha$, that is, $C_alpha$ with all of its limit points. So this is a club set in $kappa$, and since the intersection of fewer than $kappa$ many clubs is club, it follows that $bigcap_alphabar C_alpha$ is closed and unbounded in $kappa$. Thus, since $S$ is stationary, there are unboundedly many $betain S$ that are limits of $bigcap_alpha bar C_alpha$, and these $beta$'s must all be in every $C_alpha$ since each $C_alpha$ is $S$-closed.
answered Jan 2 at 22:05
JDHJDH
32.6k680145
edited Jan 2 at 22:11
answered Jan 2 at 22:05
JDHJDH
32.6k680145
answered Jan 2 at 22:05
JDHJDH
32.6k680145
answered Jan 2 at 22:05
JDHJDH
32.6k680145
32.6k680145
add a comment |
add a comment |
1
$begingroup$
It's not a duplicate of that question, since this question is about the $S$-clubs. The $C_alpha$ may not be club at all, but only $S$-club.
$endgroup$
– JDH
Jan 2 at 22:44
$begingroup$
Why was it closed as a duplicate? It isn't a duplicate.
$endgroup$
– JDH
Jan 3 at 3:14