$alpha + beta = sup_{v<gamma} (alpha + beta_{v})$? [closed]
$begingroup$
On page 124 of Introduction to Set Theory, Jech claims that ordinal functions below are continuous in the second variable: If $gamma$ is a limit ordinal and $beta= sup_{v<gamma} beta_{v}$ then
$$alpha + beta = sup_{v<gamma} (alpha + beta_{v})$$
$$alpha . beta = sup_{v<gamma} (alpha . beta_{v})$$
$$alpha ^ beta = sup_{v<gamma} (alpha ^ {beta_{v}})$$
Then it is said that the results above are followed directly from the definitions!
Somebody has answered something over here which somehow asks the same question but the problem is question above which is directly taken from Jech book is stated in a slightly different way considering indexes and it has made some confusion for me since I am new to the topic. I would appreciate if you could clarify the exact arguments above.
logic set-theory proof-explanation ordinals
asked Jan 2 at 23:25
FreeMindFreeMind
9231133
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closed as off-topic by Cameron Buie, Lord Shark the Unknown, user91500, Lord_Farin, ancientmathematician Jan 5 at 14:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cameron Buie, Lord Shark the Unknown, user91500, Lord_Farin, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
On page 124 of Introduction to Set Theory, Jech claims that ordinal functions below are continuous in the second variable: If $gamma$ is a limit ordinal and $beta= sup_{v<gamma} beta_{v}$ then
$$alpha + beta = sup_{v<gamma} (alpha + beta_{v})$$
$$alpha . beta = sup_{v<gamma} (alpha . beta_{v})$$
$$alpha ^ beta = sup_{v<gamma} (alpha ^ {beta_{v}})$$
Then it is said that the results above are followed directly from the definitions!
Somebody has answered something over here which somehow asks the same question but the problem is question above which is directly taken from Jech book is stated in a slightly different way considering indexes and it has made some confusion for me since I am new to the topic. I would appreciate if you could clarify the exact arguments above.
logic set-theory proof-explanation ordinals
asked Jan 2 at 23:25
FreeMindFreeMind
9231133
$endgroup$
closed as off-topic by Cameron Buie, Lord Shark the Unknown, user91500, Lord_Farin, ancientmathematician Jan 5 at 14:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cameron Buie, Lord Shark the Unknown, user91500, Lord_Farin, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
What definitions are you given for the ordinal arithmetic operations? It will make a big difference.
$endgroup$
– Cameron Buie
Jan 2 at 23:32
$begingroup$
I have already told you the reference book I cannot copy the whole book on SE just to make it clear for you!
$endgroup$
– FreeMind
Jan 3 at 15:05
1
$begingroup$
The whole book isn't necessary; only (a description of) the definitions in question will be needed. Without them, the only people who can help you are those who have access to the text and choose to avail themselves of it.
$endgroup$
– Cameron Buie
Jan 3 at 18:10
add a comment |
$begingroup$
On page 124 of Introduction to Set Theory, Jech claims that ordinal functions below are continuous in the second variable: If $gamma$ is a limit ordinal and $beta= sup_{v<gamma} beta_{v}$ then
$$alpha + beta = sup_{v<gamma} (alpha + beta_{v})$$
$$alpha . beta = sup_{v<gamma} (alpha . beta_{v})$$
$$alpha ^ beta = sup_{v<gamma} (alpha ^ {beta_{v}})$$
Then it is said that the results above are followed directly from the definitions!
Somebody has answered something over here which somehow asks the same question but the problem is question above which is directly taken from Jech book is stated in a slightly different way considering indexes and it has made some confusion for me since I am new to the topic. I would appreciate if you could clarify the exact arguments above.
logic set-theory proof-explanation ordinals
asked Jan 2 at 23:25
FreeMindFreeMind
9231133
$endgroup$
On page 124 of Introduction to Set Theory, Jech claims that ordinal functions below are continuous in the second variable: If $gamma$ is a limit ordinal and $beta= sup_{v<gamma} beta_{v}$ then
$$alpha + beta = sup_{v<gamma} (alpha + beta_{v})$$
$$alpha . beta = sup_{v<gamma} (alpha . beta_{v})$$
$$alpha ^ beta = sup_{v<gamma} (alpha ^ {beta_{v}})$$
Then it is said that the results above are followed directly from the definitions!
Somebody has answered something over here which somehow asks the same question but the problem is question above which is directly taken from Jech book is stated in a slightly different way considering indexes and it has made some confusion for me since I am new to the topic. I would appreciate if you could clarify the exact arguments above.
logic set-theory proof-explanation ordinals
logic set-theory proof-explanation ordinals
asked Jan 2 at 23:25
FreeMindFreeMind
9231133
asked Jan 2 at 23:25
FreeMindFreeMind
9231133
asked Jan 2 at 23:25
FreeMindFreeMind
9231133
asked Jan 2 at 23:25
FreeMindFreeMind
9231133
asked Jan 2 at 23:25
FreeMindFreeMind
9231133
9231133
closed as off-topic by Cameron Buie, Lord Shark the Unknown, user91500, Lord_Farin, ancientmathematician Jan 5 at 14:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cameron Buie, Lord Shark the Unknown, user91500, Lord_Farin, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Cameron Buie, Lord Shark the Unknown, user91500, Lord_Farin, ancientmathematician Jan 5 at 14:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cameron Buie, Lord Shark the Unknown, user91500, Lord_Farin, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
What definitions are you given for the ordinal arithmetic operations? It will make a big difference.
$endgroup$
– Cameron Buie
Jan 2 at 23:32
$begingroup$
I have already told you the reference book I cannot copy the whole book on SE just to make it clear for you!
$endgroup$
– FreeMind
Jan 3 at 15:05
1
$begingroup$
The whole book isn't necessary; only (a description of) the definitions in question will be needed. Without them, the only people who can help you are those who have access to the text and choose to avail themselves of it.
$endgroup$
– Cameron Buie
Jan 3 at 18:10
add a comment |
2
$begingroup$
What definitions are you given for the ordinal arithmetic operations? It will make a big difference.
$endgroup$
– Cameron Buie
Jan 2 at 23:32
$begingroup$
I have already told you the reference book I cannot copy the whole book on SE just to make it clear for you!
$endgroup$
– FreeMind
Jan 3 at 15:05
1
$begingroup$
The whole book isn't necessary; only (a description of) the definitions in question will be needed. Without them, the only people who can help you are those who have access to the text and choose to avail themselves of it.
$endgroup$
– Cameron Buie
Jan 3 at 18:10
2
2
$begingroup$
What definitions are you given for the ordinal arithmetic operations? It will make a big difference.
$endgroup$
– Cameron Buie
Jan 2 at 23:32
$begingroup$
What definitions are you given for the ordinal arithmetic operations? It will make a big difference.
$endgroup$
– Cameron Buie
Jan 2 at 23:32
$begingroup$
I have already told you the reference book I cannot copy the whole book on SE just to make it clear for you!
$endgroup$
– FreeMind
Jan 3 at 15:05
$begingroup$
I have already told you the reference book I cannot copy the whole book on SE just to make it clear for you!
$endgroup$
– FreeMind
Jan 3 at 15:05
1
1
$begingroup$
The whole book isn't necessary; only (a description of) the definitions in question will be needed. Without them, the only people who can help you are those who have access to the text and choose to avail themselves of it.
$endgroup$
– Cameron Buie
Jan 3 at 18:10
$begingroup$
The whole book isn't necessary; only (a description of) the definitions in question will be needed. Without them, the only people who can help you are those who have access to the text and choose to avail themselves of it.
$endgroup$
– Cameron Buie
Jan 3 at 18:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I am going to explain how to prove the continuity of addition, the rest facts can be proved in similar fashion. First of all, we should keep in mind the monotonicity of addition in the second variable. We now that $$alpha + beta = sup{alpha + rhocolon rho<beta}.$$
WLOG we can consider $beta$ to be a limit ordinal. Hence for every $nu<gamma$ we have $beta_nu < beta$, hence $alpha + beta_nu < alpha + beta$, hence by the definition of supremum $$sup{alpha + beta_nucolon nu<gamma} leq alpha + beta.$$
It remains to prove the converse inequality. Let $rho<beta$. There exists $nu_0<gamma$ such that $rho<beta_{nu_0}$ (otherwise all $beta_nu$ would be less or equal than $rho<beta$). By monotonicity $alpha + rho < alpha + beta_{nu_0}$, so $alpha+rho<sup{alpha+beta_nucolon nu<gamma}$, again by the definition of supremum we get
$$
alpha+beta leq sup{alpha+beta_nucolon nu<gamma}.
$$
These two inequalities give us the desired equality.
answered Jan 4 at 7:29
Ilya VlasovIlya Vlasov
170213
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I am going to explain how to prove the continuity of addition, the rest facts can be proved in similar fashion. First of all, we should keep in mind the monotonicity of addition in the second variable. We now that $$alpha + beta = sup{alpha + rhocolon rho<beta}.$$
WLOG we can consider $beta$ to be a limit ordinal. Hence for every $nu<gamma$ we have $beta_nu < beta$, hence $alpha + beta_nu < alpha + beta$, hence by the definition of supremum $$sup{alpha + beta_nucolon nu<gamma} leq alpha + beta.$$
It remains to prove the converse inequality. Let $rho<beta$. There exists $nu_0<gamma$ such that $rho<beta_{nu_0}$ (otherwise all $beta_nu$ would be less or equal than $rho<beta$). By monotonicity $alpha + rho < alpha + beta_{nu_0}$, so $alpha+rho<sup{alpha+beta_nucolon nu<gamma}$, again by the definition of supremum we get
$$
alpha+beta leq sup{alpha+beta_nucolon nu<gamma}.
$$
These two inequalities give us the desired equality.
answered Jan 4 at 7:29
Ilya VlasovIlya Vlasov
170213
$endgroup$
add a comment |
$begingroup$
I am going to explain how to prove the continuity of addition, the rest facts can be proved in similar fashion. First of all, we should keep in mind the monotonicity of addition in the second variable. We now that $$alpha + beta = sup{alpha + rhocolon rho<beta}.$$
WLOG we can consider $beta$ to be a limit ordinal. Hence for every $nu<gamma$ we have $beta_nu < beta$, hence $alpha + beta_nu < alpha + beta$, hence by the definition of supremum $$sup{alpha + beta_nucolon nu<gamma} leq alpha + beta.$$
It remains to prove the converse inequality. Let $rho<beta$. There exists $nu_0<gamma$ such that $rho<beta_{nu_0}$ (otherwise all $beta_nu$ would be less or equal than $rho<beta$). By monotonicity $alpha + rho < alpha + beta_{nu_0}$, so $alpha+rho<sup{alpha+beta_nucolon nu<gamma}$, again by the definition of supremum we get
$$
alpha+beta leq sup{alpha+beta_nucolon nu<gamma}.
$$
These two inequalities give us the desired equality.
answered Jan 4 at 7:29
Ilya VlasovIlya Vlasov
170213
$endgroup$
add a comment |
$begingroup$
I am going to explain how to prove the continuity of addition, the rest facts can be proved in similar fashion. First of all, we should keep in mind the monotonicity of addition in the second variable. We now that $$alpha + beta = sup{alpha + rhocolon rho<beta}.$$
WLOG we can consider $beta$ to be a limit ordinal. Hence for every $nu<gamma$ we have $beta_nu < beta$, hence $alpha + beta_nu < alpha + beta$, hence by the definition of supremum $$sup{alpha + beta_nucolon nu<gamma} leq alpha + beta.$$
It remains to prove the converse inequality. Let $rho<beta$. There exists $nu_0<gamma$ such that $rho<beta_{nu_0}$ (otherwise all $beta_nu$ would be less or equal than $rho<beta$). By monotonicity $alpha + rho < alpha + beta_{nu_0}$, so $alpha+rho<sup{alpha+beta_nucolon nu<gamma}$, again by the definition of supremum we get
$$
alpha+beta leq sup{alpha+beta_nucolon nu<gamma}.
$$
These two inequalities give us the desired equality.
answered Jan 4 at 7:29
Ilya VlasovIlya Vlasov
170213
$endgroup$
I am going to explain how to prove the continuity of addition, the rest facts can be proved in similar fashion. First of all, we should keep in mind the monotonicity of addition in the second variable. We now that $$alpha + beta = sup{alpha + rhocolon rho<beta}.$$
WLOG we can consider $beta$ to be a limit ordinal. Hence for every $nu<gamma$ we have $beta_nu < beta$, hence $alpha + beta_nu < alpha + beta$, hence by the definition of supremum $$sup{alpha + beta_nucolon nu<gamma} leq alpha + beta.$$
It remains to prove the converse inequality. Let $rho<beta$. There exists $nu_0<gamma$ such that $rho<beta_{nu_0}$ (otherwise all $beta_nu$ would be less or equal than $rho<beta$). By monotonicity $alpha + rho < alpha + beta_{nu_0}$, so $alpha+rho<sup{alpha+beta_nucolon nu<gamma}$, again by the definition of supremum we get
$$
alpha+beta leq sup{alpha+beta_nucolon nu<gamma}.
$$
These two inequalities give us the desired equality.
answered Jan 4 at 7:29
Ilya VlasovIlya Vlasov
170213
answered Jan 4 at 7:29
Ilya VlasovIlya Vlasov
170213
answered Jan 4 at 7:29
Ilya VlasovIlya Vlasov
170213
answered Jan 4 at 7:29
Ilya VlasovIlya Vlasov
170213
170213
add a comment |
add a comment |
2
$begingroup$
What definitions are you given for the ordinal arithmetic operations? It will make a big difference.
$endgroup$
– Cameron Buie
Jan 2 at 23:32
$begingroup$
I have already told you the reference book I cannot copy the whole book on SE just to make it clear for you!
$endgroup$
– FreeMind
Jan 3 at 15:05
1
$begingroup$
The whole book isn't necessary; only (a description of) the definitions in question will be needed. Without them, the only people who can help you are those who have access to the text and choose to avail themselves of it.
$endgroup$
– Cameron Buie
Jan 3 at 18:10