“Forcing” satisfying $kappa^+$-chain condition
$begingroup$
Lemma 15.4 of Jech: Let $kappa$ be a regular cardinal such that $2^{< kappa} = kappa$.
Let $S$ be an arbitrary set and let $|C| leq kappa$.
Let $P$ be the set of all functions $p$ whose domains are subsets of $S$ of size $< kappa$ with values in $C$.
Let $p < q$ iff $p supset q$.
Then $P$ satisfies the $kappa^+$-chain condition.
Proof: Let $W subset P$ be an antichain.
We construct sequences $A_0 subset A_1 subset ... subset A_{alpha} subset ... (alpha < kappa)$ of subsets of $S$ and $W_0 subset ... subset W_{alpha} subset ... (alpha < kappa)$ of subsets of $W$.
If $alpha$ is a limit ordinal, let $W_{alpha} = bigcup_{beta < alpha} W_{beta}$ and $A_{alpha} = bigcup_{beta < alpha} A_{beta}$.
Given $A_{alpha}$ and $W_{alpha}$, we choose for each $p in P$ with dom$(p) subset A_{alpha}$ some $q in W$ (if there is one) such that $p = q|_{A_{alpha}}$.
Then we let $W_{alpha+1} = W_{alpha} cup {text{the chosen $q$'s}}$ and $A_{alpha+1} = bigcup {text{dom}(q) : q in W_{alpha+1}}$.
Finally, $A = bigcup_{alpha < kappa} A_{alpha}$. (The proof continues)
My problem: Who is $A_0$? How do I guarantee that $A_1$ and $W_1$ are not empty?
set-theory proof-explanation
asked Nov 28 '18 at 6:13
Bias of PrieneBias of Priene
284112
$endgroup$
|
show 2 more comments
$begingroup$
Lemma 15.4 of Jech: Let $kappa$ be a regular cardinal such that $2^{< kappa} = kappa$.
Let $S$ be an arbitrary set and let $|C| leq kappa$.
Let $P$ be the set of all functions $p$ whose domains are subsets of $S$ of size $< kappa$ with values in $C$.
Let $p < q$ iff $p supset q$.
Then $P$ satisfies the $kappa^+$-chain condition.
Proof: Let $W subset P$ be an antichain.
We construct sequences $A_0 subset A_1 subset ... subset A_{alpha} subset ... (alpha < kappa)$ of subsets of $S$ and $W_0 subset ... subset W_{alpha} subset ... (alpha < kappa)$ of subsets of $W$.
If $alpha$ is a limit ordinal, let $W_{alpha} = bigcup_{beta < alpha} W_{beta}$ and $A_{alpha} = bigcup_{beta < alpha} A_{beta}$.
Given $A_{alpha}$ and $W_{alpha}$, we choose for each $p in P$ with dom$(p) subset A_{alpha}$ some $q in W$ (if there is one) such that $p = q|_{A_{alpha}}$.
Then we let $W_{alpha+1} = W_{alpha} cup {text{the chosen $q$'s}}$ and $A_{alpha+1} = bigcup {text{dom}(q) : q in W_{alpha+1}}$.
Finally, $A = bigcup_{alpha < kappa} A_{alpha}$. (The proof continues)
My problem: Who is $A_0$? How do I guarantee that $A_1$ and $W_1$ are not empty?
set-theory proof-explanation
asked Nov 28 '18 at 6:13
Bias of PrieneBias of Priene
284112
$endgroup$
$begingroup$
Does Jech consider $0$ to be a limit ordinal?
$endgroup$
– bof
Nov 28 '18 at 8:08
$begingroup$
Supposing it does, how can I prove my second question?
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:06
$begingroup$
Why the quotes?
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 12:44
$begingroup$
I'm using the result in Theorem 15.17, actually, and there I have a Forcing, but only after asking I realized that Jech doesn't mention $P$ as being a Forcing in the Lemma itself.
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:50
1
$begingroup$
$A_0=cup _{b<0}A_b=emptyset.$ The empty function $p_0=emptyset$ belongs to $P$ and its domain is $A_0.$ Assuming $W$ has at least $2$ members, no $qin W$ is $p_0$ (else $W$ would not be an anti-chain because $p_0$ is compatible with all $pin P$). And any $qin W$ satisfies $p_0=q|_{A_0}$ so some $q in W$ belongs to $W_1$, and its domain is not the empty set so $A_1$ is not empty either.... I think you were concerned as to whether every $A_a$ might end up being empty.
$endgroup$
– DanielWainfleet
Nov 28 '18 at 15:38
|
show 2 more comments
$begingroup$
Lemma 15.4 of Jech: Let $kappa$ be a regular cardinal such that $2^{< kappa} = kappa$.
Let $S$ be an arbitrary set and let $|C| leq kappa$.
Let $P$ be the set of all functions $p$ whose domains are subsets of $S$ of size $< kappa$ with values in $C$.
Let $p < q$ iff $p supset q$.
Then $P$ satisfies the $kappa^+$-chain condition.
Proof: Let $W subset P$ be an antichain.
We construct sequences $A_0 subset A_1 subset ... subset A_{alpha} subset ... (alpha < kappa)$ of subsets of $S$ and $W_0 subset ... subset W_{alpha} subset ... (alpha < kappa)$ of subsets of $W$.
If $alpha$ is a limit ordinal, let $W_{alpha} = bigcup_{beta < alpha} W_{beta}$ and $A_{alpha} = bigcup_{beta < alpha} A_{beta}$.
Given $A_{alpha}$ and $W_{alpha}$, we choose for each $p in P$ with dom$(p) subset A_{alpha}$ some $q in W$ (if there is one) such that $p = q|_{A_{alpha}}$.
Then we let $W_{alpha+1} = W_{alpha} cup {text{the chosen $q$'s}}$ and $A_{alpha+1} = bigcup {text{dom}(q) : q in W_{alpha+1}}$.
Finally, $A = bigcup_{alpha < kappa} A_{alpha}$. (The proof continues)
My problem: Who is $A_0$? How do I guarantee that $A_1$ and $W_1$ are not empty?
set-theory proof-explanation
asked Nov 28 '18 at 6:13
Bias of PrieneBias of Priene
284112
$endgroup$
Lemma 15.4 of Jech: Let $kappa$ be a regular cardinal such that $2^{< kappa} = kappa$.
Let $S$ be an arbitrary set and let $|C| leq kappa$.
Let $P$ be the set of all functions $p$ whose domains are subsets of $S$ of size $< kappa$ with values in $C$.
Let $p < q$ iff $p supset q$.
Then $P$ satisfies the $kappa^+$-chain condition.
Proof: Let $W subset P$ be an antichain.
We construct sequences $A_0 subset A_1 subset ... subset A_{alpha} subset ... (alpha < kappa)$ of subsets of $S$ and $W_0 subset ... subset W_{alpha} subset ... (alpha < kappa)$ of subsets of $W$.
If $alpha$ is a limit ordinal, let $W_{alpha} = bigcup_{beta < alpha} W_{beta}$ and $A_{alpha} = bigcup_{beta < alpha} A_{beta}$.
Given $A_{alpha}$ and $W_{alpha}$, we choose for each $p in P$ with dom$(p) subset A_{alpha}$ some $q in W$ (if there is one) such that $p = q|_{A_{alpha}}$.
Then we let $W_{alpha+1} = W_{alpha} cup {text{the chosen $q$'s}}$ and $A_{alpha+1} = bigcup {text{dom}(q) : q in W_{alpha+1}}$.
Finally, $A = bigcup_{alpha < kappa} A_{alpha}$. (The proof continues)
My problem: Who is $A_0$? How do I guarantee that $A_1$ and $W_1$ are not empty?
set-theory proof-explanation
set-theory proof-explanation
asked Nov 28 '18 at 6:13
Bias of PrieneBias of Priene
284112
asked Nov 28 '18 at 6:13
Bias of PrieneBias of Priene
284112
edited Nov 28 '18 at 6:23
Bias of Priene
asked Nov 28 '18 at 6:13
Bias of PrieneBias of Priene
284112
asked Nov 28 '18 at 6:13
Bias of PrieneBias of Priene
284112
asked Nov 28 '18 at 6:13
Bias of PrieneBias of Priene
284112
284112
$begingroup$
Does Jech consider $0$ to be a limit ordinal?
$endgroup$
– bof
Nov 28 '18 at 8:08
$begingroup$
Supposing it does, how can I prove my second question?
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:06
$begingroup$
Why the quotes?
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 12:44
$begingroup$
I'm using the result in Theorem 15.17, actually, and there I have a Forcing, but only after asking I realized that Jech doesn't mention $P$ as being a Forcing in the Lemma itself.
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:50
1
$begingroup$
$A_0=cup _{b<0}A_b=emptyset.$ The empty function $p_0=emptyset$ belongs to $P$ and its domain is $A_0.$ Assuming $W$ has at least $2$ members, no $qin W$ is $p_0$ (else $W$ would not be an anti-chain because $p_0$ is compatible with all $pin P$). And any $qin W$ satisfies $p_0=q|_{A_0}$ so some $q in W$ belongs to $W_1$, and its domain is not the empty set so $A_1$ is not empty either.... I think you were concerned as to whether every $A_a$ might end up being empty.
$endgroup$
– DanielWainfleet
Nov 28 '18 at 15:38
|
show 2 more comments
$begingroup$
Does Jech consider $0$ to be a limit ordinal?
$endgroup$
– bof
Nov 28 '18 at 8:08
$begingroup$
Supposing it does, how can I prove my second question?
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:06
$begingroup$
Why the quotes?
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 12:44
$begingroup$
I'm using the result in Theorem 15.17, actually, and there I have a Forcing, but only after asking I realized that Jech doesn't mention $P$ as being a Forcing in the Lemma itself.
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:50
1
$begingroup$
$A_0=cup _{b<0}A_b=emptyset.$ The empty function $p_0=emptyset$ belongs to $P$ and its domain is $A_0.$ Assuming $W$ has at least $2$ members, no $qin W$ is $p_0$ (else $W$ would not be an anti-chain because $p_0$ is compatible with all $pin P$). And any $qin W$ satisfies $p_0=q|_{A_0}$ so some $q in W$ belongs to $W_1$, and its domain is not the empty set so $A_1$ is not empty either.... I think you were concerned as to whether every $A_a$ might end up being empty.
$endgroup$
– DanielWainfleet
Nov 28 '18 at 15:38
$begingroup$
Does Jech consider $0$ to be a limit ordinal?
$endgroup$
– bof
Nov 28 '18 at 8:08
$begingroup$
Does Jech consider $0$ to be a limit ordinal?
$endgroup$
– bof
Nov 28 '18 at 8:08
$begingroup$
Supposing it does, how can I prove my second question?
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:06
$begingroup$
Supposing it does, how can I prove my second question?
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:06
$begingroup$
Why the quotes?
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 12:44
$begingroup$
Why the quotes?
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 12:44
$begingroup$
I'm using the result in Theorem 15.17, actually, and there I have a Forcing, but only after asking I realized that Jech doesn't mention $P$ as being a Forcing in the Lemma itself.
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:50
$begingroup$
I'm using the result in Theorem 15.17, actually, and there I have a Forcing, but only after asking I realized that Jech doesn't mention $P$ as being a Forcing in the Lemma itself.
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:50
1
1
$begingroup$
$A_0=cup _{b<0}A_b=emptyset.$ The empty function $p_0=emptyset$ belongs to $P$ and its domain is $A_0.$ Assuming $W$ has at least $2$ members, no $qin W$ is $p_0$ (else $W$ would not be an anti-chain because $p_0$ is compatible with all $pin P$). And any $qin W$ satisfies $p_0=q|_{A_0}$ so some $q in W$ belongs to $W_1$, and its domain is not the empty set so $A_1$ is not empty either.... I think you were concerned as to whether every $A_a$ might end up being empty.
$endgroup$
– DanielWainfleet
Nov 28 '18 at 15:38
$begingroup$
$A_0=cup _{b<0}A_b=emptyset.$ The empty function $p_0=emptyset$ belongs to $P$ and its domain is $A_0.$ Assuming $W$ has at least $2$ members, no $qin W$ is $p_0$ (else $W$ would not be an anti-chain because $p_0$ is compatible with all $pin P$). And any $qin W$ satisfies $p_0=q|_{A_0}$ so some $q in W$ belongs to $W_1$, and its domain is not the empty set so $A_1$ is not empty either.... I think you were concerned as to whether every $A_a$ might end up being empty.
$endgroup$
– DanielWainfleet
Nov 28 '18 at 15:38
|
show 2 more comments
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$begingroup$
Does Jech consider $0$ to be a limit ordinal?
$endgroup$
– bof
Nov 28 '18 at 8:08
$begingroup$
Supposing it does, how can I prove my second question?
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:06
$begingroup$
Why the quotes?
$endgroup$
– Andrés E. Caicedo
Nov 28 '18 at 12:44
$begingroup$
I'm using the result in Theorem 15.17, actually, and there I have a Forcing, but only after asking I realized that Jech doesn't mention $P$ as being a Forcing in the Lemma itself.
$endgroup$
– Bias of Priene
Nov 28 '18 at 12:50
1
$begingroup$
$A_0=cup _{b<0}A_b=emptyset.$ The empty function $p_0=emptyset$ belongs to $P$ and its domain is $A_0.$ Assuming $W$ has at least $2$ members, no $qin W$ is $p_0$ (else $W$ would not be an anti-chain because $p_0$ is compatible with all $pin P$). And any $qin W$ satisfies $p_0=q|_{A_0}$ so some $q in W$ belongs to $W_1$, and its domain is not the empty set so $A_1$ is not empty either.... I think you were concerned as to whether every $A_a$ might end up being empty.
$endgroup$
– DanielWainfleet
Nov 28 '18 at 15:38