Constant functions in set-theory
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I need some help with an exercise in set theory, which is about certain constant functions.
Let $S$ be a stationary subset of a regular uncountable cardinal $lambda$. Given an ordinal $alpha$, let $c_alpha^lambda$ denote the constant function with domain $lambda$ and range ${alpha}$.
Letting $psi,varphi$ range over all ordinal-valued functions with domain $lambda$, define $$varphi<_Spsimbox{ if and only if }{deltain Smid varphi(delta)≥psi(delta)}mbox{ is non-stationary}.$$ The relation $<_S$ is well-founded, so we can use it to define a rank $|cdot|_S$ by recursion as
$$ |psi|_S=bigcup{|varphi|_S+1mid varphi<_Spsi}. $$
How can we prove that, for all $alphain{rm Ord}$, $|c_alpha^lambda|_S gealpha$ holds?
How can we determine the value of $|c_alpha^lambda|_S$ for all $alpha<lambda$?
Can we prove that $|c_lambda^lambda|_S >lambda$?
set-theory
asked Jan 8 at 13:36
N. LevelingN. Leveling
52
$endgroup$
|
show 9 more comments
$begingroup$
I need some help with an exercise in set theory, which is about certain constant functions.
Let $S$ be a stationary subset of a regular uncountable cardinal $lambda$. Given an ordinal $alpha$, let $c_alpha^lambda$ denote the constant function with domain $lambda$ and range ${alpha}$.
Letting $psi,varphi$ range over all ordinal-valued functions with domain $lambda$, define $$varphi<_Spsimbox{ if and only if }{deltain Smid varphi(delta)≥psi(delta)}mbox{ is non-stationary}.$$ The relation $<_S$ is well-founded, so we can use it to define a rank $|cdot|_S$ by recursion as
$$ |psi|_S=bigcup{|varphi|_S+1mid varphi<_Spsi}. $$
How can we prove that, for all $alphain{rm Ord}$, $|c_alpha^lambda|_S gealpha$ holds?
How can we determine the value of $|c_alpha^lambda|_S$ for all $alpha<lambda$?
Can we prove that $|c_lambda^lambda|_S >lambda$?
set-theory
asked Jan 8 at 13:36
N. LevelingN. Leveling
52
$endgroup$
$begingroup$
What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 8 at 13:39
1
$begingroup$
You might want to provide a definition of $|f|_S$ as well.
$endgroup$
– Asaf Karagila♦
Jan 8 at 13:44
$begingroup$
What precisely is difficult here?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 14:47
$begingroup$
$||Ψ||_S= bigcup { ||φ||_S +1 | φ<_S Ψ}$ is the definition of the S-rank, with S being stationary and $φ<_S ψ $ iff $ {δ∈S|φ(δ)> ψ(δ)} ∈ NS_λ$ (NS is the Non-stationary ideal on λ). I just can't seem to figure out how I can determine the rank of the constant function, and thus don't know how to prove anything about it.
$endgroup$
– N. Leveling
Jan 8 at 16:02
$begingroup$
Do you see that $|c_alpha^lambda|_S$ is an ordinal for all $alpha$, and can you prove that $|c_alpha^lambda|_S>|c_beta^lambda|_S$ whenever $alpha>beta$?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 18:27
|
show 9 more comments
$begingroup$
I need some help with an exercise in set theory, which is about certain constant functions.
Let $S$ be a stationary subset of a regular uncountable cardinal $lambda$. Given an ordinal $alpha$, let $c_alpha^lambda$ denote the constant function with domain $lambda$ and range ${alpha}$.
Letting $psi,varphi$ range over all ordinal-valued functions with domain $lambda$, define $$varphi<_Spsimbox{ if and only if }{deltain Smid varphi(delta)≥psi(delta)}mbox{ is non-stationary}.$$ The relation $<_S$ is well-founded, so we can use it to define a rank $|cdot|_S$ by recursion as
$$ |psi|_S=bigcup{|varphi|_S+1mid varphi<_Spsi}. $$
How can we prove that, for all $alphain{rm Ord}$, $|c_alpha^lambda|_S gealpha$ holds?
How can we determine the value of $|c_alpha^lambda|_S$ for all $alpha<lambda$?
Can we prove that $|c_lambda^lambda|_S >lambda$?
set-theory
asked Jan 8 at 13:36
N. LevelingN. Leveling
52
$endgroup$
I need some help with an exercise in set theory, which is about certain constant functions.
Let $S$ be a stationary subset of a regular uncountable cardinal $lambda$. Given an ordinal $alpha$, let $c_alpha^lambda$ denote the constant function with domain $lambda$ and range ${alpha}$.
Letting $psi,varphi$ range over all ordinal-valued functions with domain $lambda$, define $$varphi<_Spsimbox{ if and only if }{deltain Smid varphi(delta)≥psi(delta)}mbox{ is non-stationary}.$$ The relation $<_S$ is well-founded, so we can use it to define a rank $|cdot|_S$ by recursion as
$$ |psi|_S=bigcup{|varphi|_S+1mid varphi<_Spsi}. $$
How can we prove that, for all $alphain{rm Ord}$, $|c_alpha^lambda|_S gealpha$ holds?
How can we determine the value of $|c_alpha^lambda|_S$ for all $alpha<lambda$?
Can we prove that $|c_lambda^lambda|_S >lambda$?
set-theory
set-theory
asked Jan 8 at 13:36
N. LevelingN. Leveling
52
asked Jan 8 at 13:36
N. LevelingN. Leveling
52
edited Jan 8 at 19:50
N. Leveling
asked Jan 8 at 13:36
N. LevelingN. Leveling
52
asked Jan 8 at 13:36
N. LevelingN. Leveling
52
asked Jan 8 at 13:36
N. LevelingN. Leveling
52
52
$begingroup$
What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 8 at 13:39
1
$begingroup$
You might want to provide a definition of $|f|_S$ as well.
$endgroup$
– Asaf Karagila♦
Jan 8 at 13:44
$begingroup$
What precisely is difficult here?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 14:47
$begingroup$
$||Ψ||_S= bigcup { ||φ||_S +1 | φ<_S Ψ}$ is the definition of the S-rank, with S being stationary and $φ<_S ψ $ iff $ {δ∈S|φ(δ)> ψ(δ)} ∈ NS_λ$ (NS is the Non-stationary ideal on λ). I just can't seem to figure out how I can determine the rank of the constant function, and thus don't know how to prove anything about it.
$endgroup$
– N. Leveling
Jan 8 at 16:02
$begingroup$
Do you see that $|c_alpha^lambda|_S$ is an ordinal for all $alpha$, and can you prove that $|c_alpha^lambda|_S>|c_beta^lambda|_S$ whenever $alpha>beta$?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 18:27
|
show 9 more comments
$begingroup$
What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 8 at 13:39
1
$begingroup$
You might want to provide a definition of $|f|_S$ as well.
$endgroup$
– Asaf Karagila♦
Jan 8 at 13:44
$begingroup$
What precisely is difficult here?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 14:47
$begingroup$
$||Ψ||_S= bigcup { ||φ||_S +1 | φ<_S Ψ}$ is the definition of the S-rank, with S being stationary and $φ<_S ψ $ iff $ {δ∈S|φ(δ)> ψ(δ)} ∈ NS_λ$ (NS is the Non-stationary ideal on λ). I just can't seem to figure out how I can determine the rank of the constant function, and thus don't know how to prove anything about it.
$endgroup$
– N. Leveling
Jan 8 at 16:02
$begingroup$
Do you see that $|c_alpha^lambda|_S$ is an ordinal for all $alpha$, and can you prove that $|c_alpha^lambda|_S>|c_beta^lambda|_S$ whenever $alpha>beta$?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 18:27
$begingroup$
What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 8 at 13:39
$begingroup$
What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 8 at 13:39
1
1
$begingroup$
You might want to provide a definition of $|f|_S$ as well.
$endgroup$
– Asaf Karagila♦
Jan 8 at 13:44
$begingroup$
You might want to provide a definition of $|f|_S$ as well.
$endgroup$
– Asaf Karagila♦
Jan 8 at 13:44
$begingroup$
What precisely is difficult here?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 14:47
$begingroup$
What precisely is difficult here?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 14:47
$begingroup$
$||Ψ||_S= bigcup { ||φ||_S +1 | φ<_S Ψ}$ is the definition of the S-rank, with S being stationary and $φ<_S ψ $ iff $ {δ∈S|φ(δ)> ψ(δ)} ∈ NS_λ$ (NS is the Non-stationary ideal on λ). I just can't seem to figure out how I can determine the rank of the constant function, and thus don't know how to prove anything about it.
$endgroup$
– N. Leveling
Jan 8 at 16:02
$begingroup$
$||Ψ||_S= bigcup { ||φ||_S +1 | φ<_S Ψ}$ is the definition of the S-rank, with S being stationary and $φ<_S ψ $ iff $ {δ∈S|φ(δ)> ψ(δ)} ∈ NS_λ$ (NS is the Non-stationary ideal on λ). I just can't seem to figure out how I can determine the rank of the constant function, and thus don't know how to prove anything about it.
$endgroup$
– N. Leveling
Jan 8 at 16:02
$begingroup$
Do you see that $|c_alpha^lambda|_S$ is an ordinal for all $alpha$, and can you prove that $|c_alpha^lambda|_S>|c_beta^lambda|_S$ whenever $alpha>beta$?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 18:27
$begingroup$
Do you see that $|c_alpha^lambda|_S$ is an ordinal for all $alpha$, and can you prove that $|c_alpha^lambda|_S>|c_beta^lambda|_S$ whenever $alpha>beta$?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 18:27
|
show 9 more comments
1 Answer
1
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oldest
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$begingroup$
Note that $alphamapsto|c_alpha^lambda|_S$ is strictly increasing (trivially): After all, $${deltain Smid c_beta^lambda(delta)ge c_alpha^lambda(delta)}={deltain Smidbetage alpha}=emptyset$$ if $beta<alpha$. This immediately gives that $|c_alpha^lambda|_Sgealpha$ for all $alpha$.
Suppose now that $f<c_alpha^lambda$. This means that ${deltain Smid f(delta)ge alpha}$ is non-stationary, or, what is the same, $f(delta)<alpha$ for almost every $deltain S$. If, in addition, $alpha<lambda$, then in fact $f(delta)<delta$ for almost every $deltain S$. Use Fodor's lemma to conclude that $f$ coincides with some $c_beta^lambda$ for some $beta<alpha$ ("coincides" in the sense of $=_S$, where $f=_S g$ implies in particular that $|f|_S=|g|_S$). This should give you that $|c_alpha^lambda|_S=alpha$ for all $alpha<lambda$.
Finally, check that the identity map is above all $c_alpha^lambda$, $alpha<lambda$, and below $c_lambda^lambda$.
answered Jan 8 at 19:31
Andrés E. CaicedoAndrés E. Caicedo
65.8k8160252
$endgroup$
$begingroup$
Thank you very much! Just one more question: Why is the identity map below $c_λ^λ$? I don't understand this point
$endgroup$
– N. Leveling
Jan 8 at 19:55
$begingroup$
Use the definitions, it is just as immediate as the other properties we discussed in the comments above.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 19:57
add a comment |
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$begingroup$
Note that $alphamapsto|c_alpha^lambda|_S$ is strictly increasing (trivially): After all, $${deltain Smid c_beta^lambda(delta)ge c_alpha^lambda(delta)}={deltain Smidbetage alpha}=emptyset$$ if $beta<alpha$. This immediately gives that $|c_alpha^lambda|_Sgealpha$ for all $alpha$.
Suppose now that $f<c_alpha^lambda$. This means that ${deltain Smid f(delta)ge alpha}$ is non-stationary, or, what is the same, $f(delta)<alpha$ for almost every $deltain S$. If, in addition, $alpha<lambda$, then in fact $f(delta)<delta$ for almost every $deltain S$. Use Fodor's lemma to conclude that $f$ coincides with some $c_beta^lambda$ for some $beta<alpha$ ("coincides" in the sense of $=_S$, where $f=_S g$ implies in particular that $|f|_S=|g|_S$). This should give you that $|c_alpha^lambda|_S=alpha$ for all $alpha<lambda$.
Finally, check that the identity map is above all $c_alpha^lambda$, $alpha<lambda$, and below $c_lambda^lambda$.
answered Jan 8 at 19:31
Andrés E. CaicedoAndrés E. Caicedo
65.8k8160252
$endgroup$
$begingroup$
Thank you very much! Just one more question: Why is the identity map below $c_λ^λ$? I don't understand this point
$endgroup$
– N. Leveling
Jan 8 at 19:55
$begingroup$
Use the definitions, it is just as immediate as the other properties we discussed in the comments above.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 19:57
add a comment |
$begingroup$
Note that $alphamapsto|c_alpha^lambda|_S$ is strictly increasing (trivially): After all, $${deltain Smid c_beta^lambda(delta)ge c_alpha^lambda(delta)}={deltain Smidbetage alpha}=emptyset$$ if $beta<alpha$. This immediately gives that $|c_alpha^lambda|_Sgealpha$ for all $alpha$.
Suppose now that $f<c_alpha^lambda$. This means that ${deltain Smid f(delta)ge alpha}$ is non-stationary, or, what is the same, $f(delta)<alpha$ for almost every $deltain S$. If, in addition, $alpha<lambda$, then in fact $f(delta)<delta$ for almost every $deltain S$. Use Fodor's lemma to conclude that $f$ coincides with some $c_beta^lambda$ for some $beta<alpha$ ("coincides" in the sense of $=_S$, where $f=_S g$ implies in particular that $|f|_S=|g|_S$). This should give you that $|c_alpha^lambda|_S=alpha$ for all $alpha<lambda$.
Finally, check that the identity map is above all $c_alpha^lambda$, $alpha<lambda$, and below $c_lambda^lambda$.
answered Jan 8 at 19:31
Andrés E. CaicedoAndrés E. Caicedo
65.8k8160252
$endgroup$
$begingroup$
Thank you very much! Just one more question: Why is the identity map below $c_λ^λ$? I don't understand this point
$endgroup$
– N. Leveling
Jan 8 at 19:55
$begingroup$
Use the definitions, it is just as immediate as the other properties we discussed in the comments above.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 19:57
add a comment |
$begingroup$
Note that $alphamapsto|c_alpha^lambda|_S$ is strictly increasing (trivially): After all, $${deltain Smid c_beta^lambda(delta)ge c_alpha^lambda(delta)}={deltain Smidbetage alpha}=emptyset$$ if $beta<alpha$. This immediately gives that $|c_alpha^lambda|_Sgealpha$ for all $alpha$.
Suppose now that $f<c_alpha^lambda$. This means that ${deltain Smid f(delta)ge alpha}$ is non-stationary, or, what is the same, $f(delta)<alpha$ for almost every $deltain S$. If, in addition, $alpha<lambda$, then in fact $f(delta)<delta$ for almost every $deltain S$. Use Fodor's lemma to conclude that $f$ coincides with some $c_beta^lambda$ for some $beta<alpha$ ("coincides" in the sense of $=_S$, where $f=_S g$ implies in particular that $|f|_S=|g|_S$). This should give you that $|c_alpha^lambda|_S=alpha$ for all $alpha<lambda$.
Finally, check that the identity map is above all $c_alpha^lambda$, $alpha<lambda$, and below $c_lambda^lambda$.
answered Jan 8 at 19:31
Andrés E. CaicedoAndrés E. Caicedo
65.8k8160252
$endgroup$
Note that $alphamapsto|c_alpha^lambda|_S$ is strictly increasing (trivially): After all, $${deltain Smid c_beta^lambda(delta)ge c_alpha^lambda(delta)}={deltain Smidbetage alpha}=emptyset$$ if $beta<alpha$. This immediately gives that $|c_alpha^lambda|_Sgealpha$ for all $alpha$.
Suppose now that $f<c_alpha^lambda$. This means that ${deltain Smid f(delta)ge alpha}$ is non-stationary, or, what is the same, $f(delta)<alpha$ for almost every $deltain S$. If, in addition, $alpha<lambda$, then in fact $f(delta)<delta$ for almost every $deltain S$. Use Fodor's lemma to conclude that $f$ coincides with some $c_beta^lambda$ for some $beta<alpha$ ("coincides" in the sense of $=_S$, where $f=_S g$ implies in particular that $|f|_S=|g|_S$). This should give you that $|c_alpha^lambda|_S=alpha$ for all $alpha<lambda$.
Finally, check that the identity map is above all $c_alpha^lambda$, $alpha<lambda$, and below $c_lambda^lambda$.
answered Jan 8 at 19:31
Andrés E. CaicedoAndrés E. Caicedo
65.8k8160252
answered Jan 8 at 19:31
Andrés E. CaicedoAndrés E. Caicedo
65.8k8160252
answered Jan 8 at 19:31
Andrés E. CaicedoAndrés E. Caicedo
65.8k8160252
answered Jan 8 at 19:31
Andrés E. CaicedoAndrés E. Caicedo
65.8k8160252
65.8k8160252
$begingroup$
Thank you very much! Just one more question: Why is the identity map below $c_λ^λ$? I don't understand this point
$endgroup$
– N. Leveling
Jan 8 at 19:55
$begingroup$
Use the definitions, it is just as immediate as the other properties we discussed in the comments above.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 19:57
add a comment |
$begingroup$
Thank you very much! Just one more question: Why is the identity map below $c_λ^λ$? I don't understand this point
$endgroup$
– N. Leveling
Jan 8 at 19:55
$begingroup$
Use the definitions, it is just as immediate as the other properties we discussed in the comments above.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 19:57
$begingroup$
Thank you very much! Just one more question: Why is the identity map below $c_λ^λ$? I don't understand this point
$endgroup$
– N. Leveling
Jan 8 at 19:55
$begingroup$
Thank you very much! Just one more question: Why is the identity map below $c_λ^λ$? I don't understand this point
$endgroup$
– N. Leveling
Jan 8 at 19:55
$begingroup$
Use the definitions, it is just as immediate as the other properties we discussed in the comments above.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 19:57
$begingroup$
Use the definitions, it is just as immediate as the other properties we discussed in the comments above.
$endgroup$
– Andrés E. Caicedo
Jan 8 at 19:57
add a comment |
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$begingroup$
What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 8 at 13:39
1
$begingroup$
You might want to provide a definition of $|f|_S$ as well.
$endgroup$
– Asaf Karagila♦
Jan 8 at 13:44
$begingroup$
What precisely is difficult here?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 14:47
$begingroup$
$||Ψ||_S= bigcup { ||φ||_S +1 | φ<_S Ψ}$ is the definition of the S-rank, with S being stationary and $φ<_S ψ $ iff $ {δ∈S|φ(δ)> ψ(δ)} ∈ NS_λ$ (NS is the Non-stationary ideal on λ). I just can't seem to figure out how I can determine the rank of the constant function, and thus don't know how to prove anything about it.
$endgroup$
– N. Leveling
Jan 8 at 16:02
$begingroup$
Do you see that $|c_alpha^lambda|_S$ is an ordinal for all $alpha$, and can you prove that $|c_alpha^lambda|_S>|c_beta^lambda|_S$ whenever $alpha>beta$?
$endgroup$
– Andrés E. Caicedo
Jan 8 at 18:27