A question about the proof of...
up vote
2
down vote
favorite
I found the proof by @user642796 here. From his/her personal page, it seems to me that he/she has not visited MSE for a long time. So I post this question here.
Below is the verbatim proof by @user642796.
By definition we know that given an ordinal $alpha$ and a limit ordinal $beta$ that $$alpha^beta = {textstyle sup_{gamma < beta}}: alpha^gamma.$$ But we can actually say a bit more: If $A subseteq beta$ is cofinal in $beta$, then $$sup { alpha^gamma : gamma in A } = alpha^beta.$$ And similarly with the other basic arithmetic operations on ordinals.
The line that you have labelled $***$ (assuming $alpha > 1$) then becomes something to the effect of $$begin{align}
alpha^{beta + gamma}
&= sup { alpha^{beta + delta} : delta < gamma } &&text{(}{ beta+delta : delta < gamma}text{ is cofinal in the limit ord }beta + gammatext{)} \
&= sup { alpha^beta cdot alpha^delta : delta < gamma } &&text{(by induction hypothesis)} \
&= alpha^beta cdot sup { alpha^delta : delta < gamma }
&&text{(}{ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }text{)}
\
&=alpha ^beta cdot alpha^gamma &&text{(by definition)}
end{align}$$
(The assumption that $alpha > 1$ is only needed to ensure that $sup { alpha^delta : delta < gamma }$ is a limit ordinal given that $gamma > 0$ is a limit. The outlying cases where $alpha = 0$ and $alpha = 1$ can easily be taken care of separately.)
My questions:
- I'm able to understand almost everything in the proof except for the most important point: how @user642796 use the property of cofinal to achieve
$$begin{align}&sup { alpha^beta cdot alpha^delta : delta < gamma }\
= &alpha^beta cdot sup { alpha^delta : delta < gamma }
left({ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }right)end{align}$$
- The sentence ${ alpha^delta : delta < gamma }$ is cofinal in the limit ord $sup { alpha^delta : delta < gamma }$ doesn't make sense to me. This sentence somewhat means If $alpha$ is a limit ordinal, then $alpha$ is cofinal in the limit ord $alpha$ ans thus is quite odd.
Thank you so much!
proof-explanation ordinals
This question has an open bounty worth +50
reputation from Le Anh Dung ending in 6 days.
This question has not received enough attention.
It will be great if someone helps me fill in the blank of the proof.
add a comment |
up vote
2
down vote
favorite
I found the proof by @user642796 here. From his/her personal page, it seems to me that he/she has not visited MSE for a long time. So I post this question here.
Below is the verbatim proof by @user642796.
By definition we know that given an ordinal $alpha$ and a limit ordinal $beta$ that $$alpha^beta = {textstyle sup_{gamma < beta}}: alpha^gamma.$$ But we can actually say a bit more: If $A subseteq beta$ is cofinal in $beta$, then $$sup { alpha^gamma : gamma in A } = alpha^beta.$$ And similarly with the other basic arithmetic operations on ordinals.
The line that you have labelled $***$ (assuming $alpha > 1$) then becomes something to the effect of $$begin{align}
alpha^{beta + gamma}
&= sup { alpha^{beta + delta} : delta < gamma } &&text{(}{ beta+delta : delta < gamma}text{ is cofinal in the limit ord }beta + gammatext{)} \
&= sup { alpha^beta cdot alpha^delta : delta < gamma } &&text{(by induction hypothesis)} \
&= alpha^beta cdot sup { alpha^delta : delta < gamma }
&&text{(}{ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }text{)}
\
&=alpha ^beta cdot alpha^gamma &&text{(by definition)}
end{align}$$
(The assumption that $alpha > 1$ is only needed to ensure that $sup { alpha^delta : delta < gamma }$ is a limit ordinal given that $gamma > 0$ is a limit. The outlying cases where $alpha = 0$ and $alpha = 1$ can easily be taken care of separately.)
My questions:
- I'm able to understand almost everything in the proof except for the most important point: how @user642796 use the property of cofinal to achieve
$$begin{align}&sup { alpha^beta cdot alpha^delta : delta < gamma }\
= &alpha^beta cdot sup { alpha^delta : delta < gamma }
left({ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }right)end{align}$$
- The sentence ${ alpha^delta : delta < gamma }$ is cofinal in the limit ord $sup { alpha^delta : delta < gamma }$ doesn't make sense to me. This sentence somewhat means If $alpha$ is a limit ordinal, then $alpha$ is cofinal in the limit ord $alpha$ ans thus is quite odd.
Thank you so much!
proof-explanation ordinals
This question has an open bounty worth +50
reputation from Le Anh Dung ending in 6 days.
This question has not received enough attention.
It will be great if someone helps me fill in the blank of the proof.
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I found the proof by @user642796 here. From his/her personal page, it seems to me that he/she has not visited MSE for a long time. So I post this question here.
Below is the verbatim proof by @user642796.
By definition we know that given an ordinal $alpha$ and a limit ordinal $beta$ that $$alpha^beta = {textstyle sup_{gamma < beta}}: alpha^gamma.$$ But we can actually say a bit more: If $A subseteq beta$ is cofinal in $beta$, then $$sup { alpha^gamma : gamma in A } = alpha^beta.$$ And similarly with the other basic arithmetic operations on ordinals.
The line that you have labelled $***$ (assuming $alpha > 1$) then becomes something to the effect of $$begin{align}
alpha^{beta + gamma}
&= sup { alpha^{beta + delta} : delta < gamma } &&text{(}{ beta+delta : delta < gamma}text{ is cofinal in the limit ord }beta + gammatext{)} \
&= sup { alpha^beta cdot alpha^delta : delta < gamma } &&text{(by induction hypothesis)} \
&= alpha^beta cdot sup { alpha^delta : delta < gamma }
&&text{(}{ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }text{)}
\
&=alpha ^beta cdot alpha^gamma &&text{(by definition)}
end{align}$$
(The assumption that $alpha > 1$ is only needed to ensure that $sup { alpha^delta : delta < gamma }$ is a limit ordinal given that $gamma > 0$ is a limit. The outlying cases where $alpha = 0$ and $alpha = 1$ can easily be taken care of separately.)
My questions:
- I'm able to understand almost everything in the proof except for the most important point: how @user642796 use the property of cofinal to achieve
$$begin{align}&sup { alpha^beta cdot alpha^delta : delta < gamma }\
= &alpha^beta cdot sup { alpha^delta : delta < gamma }
left({ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }right)end{align}$$
- The sentence ${ alpha^delta : delta < gamma }$ is cofinal in the limit ord $sup { alpha^delta : delta < gamma }$ doesn't make sense to me. This sentence somewhat means If $alpha$ is a limit ordinal, then $alpha$ is cofinal in the limit ord $alpha$ ans thus is quite odd.
Thank you so much!
proof-explanation ordinals
I found the proof by @user642796 here. From his/her personal page, it seems to me that he/she has not visited MSE for a long time. So I post this question here.
Below is the verbatim proof by @user642796.
By definition we know that given an ordinal $alpha$ and a limit ordinal $beta$ that $$alpha^beta = {textstyle sup_{gamma < beta}}: alpha^gamma.$$ But we can actually say a bit more: If $A subseteq beta$ is cofinal in $beta$, then $$sup { alpha^gamma : gamma in A } = alpha^beta.$$ And similarly with the other basic arithmetic operations on ordinals.
The line that you have labelled $***$ (assuming $alpha > 1$) then becomes something to the effect of $$begin{align}
alpha^{beta + gamma}
&= sup { alpha^{beta + delta} : delta < gamma } &&text{(}{ beta+delta : delta < gamma}text{ is cofinal in the limit ord }beta + gammatext{)} \
&= sup { alpha^beta cdot alpha^delta : delta < gamma } &&text{(by induction hypothesis)} \
&= alpha^beta cdot sup { alpha^delta : delta < gamma }
&&text{(}{ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }text{)}
\
&=alpha ^beta cdot alpha^gamma &&text{(by definition)}
end{align}$$
(The assumption that $alpha > 1$ is only needed to ensure that $sup { alpha^delta : delta < gamma }$ is a limit ordinal given that $gamma > 0$ is a limit. The outlying cases where $alpha = 0$ and $alpha = 1$ can easily be taken care of separately.)
My questions:
- I'm able to understand almost everything in the proof except for the most important point: how @user642796 use the property of cofinal to achieve
$$begin{align}&sup { alpha^beta cdot alpha^delta : delta < gamma }\
= &alpha^beta cdot sup { alpha^delta : delta < gamma }
left({ alpha^delta : delta < gamma }text{ is cofinal in the limit ord }sup { alpha^delta : delta < gamma }right)end{align}$$
- The sentence ${ alpha^delta : delta < gamma }$ is cofinal in the limit ord $sup { alpha^delta : delta < gamma }$ doesn't make sense to me. This sentence somewhat means If $alpha$ is a limit ordinal, then $alpha$ is cofinal in the limit ord $alpha$ ans thus is quite odd.
Thank you so much!
proof-explanation ordinals
proof-explanation ordinals
edited Nov 8 at 1:29
asked Nov 8 at 1:20
Le Anh Dung
1,2021421
1,2021421
This question has an open bounty worth +50
reputation from Le Anh Dung ending in 6 days.
This question has not received enough attention.
It will be great if someone helps me fill in the blank of the proof.
This question has an open bounty worth +50
reputation from Le Anh Dung ending in 6 days.
This question has not received enough attention.
It will be great if someone helps me fill in the blank of the proof.
add a comment |
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2989363%2fa-question-about-the-proof-of-sup-alpha-beta-alpha-delta-mid-delta-gamma%23new-answer', 'question_page');
}
);
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password