Let $(alpha_ximidxi<kappa)$ be a sequence such that ${alpha_ximidxi<kappa}=alpha$. Find an increasing...
Let $alpha$ be a limit ordinal which is not a cardinal, and $kappa=|alpha|$. Then there exists a bijection from $kappa$ to $alpha$, or equivalently, a one-to-one sequence $langle alpha_xi mid xi< kappa rangle$ of ordinals such that ${alpha_xi mid xi< kappa}=alpha$.
My textbook states:
Now we can find (by transfinite recursion) a subsequence which is strictly increasing and has limit $alpha$.
I have tried but to no avail in constructing such subsequence, please shed me some lights.
Thank you so much!
elementary-set-theory cardinals ordinals
asked Dec 27 '18 at 3:15
Le Anh Dung
9621521
add a comment |
Let $alpha$ be a limit ordinal which is not a cardinal, and $kappa=|alpha|$. Then there exists a bijection from $kappa$ to $alpha$, or equivalently, a one-to-one sequence $langle alpha_xi mid xi< kappa rangle$ of ordinals such that ${alpha_xi mid xi< kappa}=alpha$.
My textbook states:
Now we can find (by transfinite recursion) a subsequence which is strictly increasing and has limit $alpha$.
I have tried but to no avail in constructing such subsequence, please shed me some lights.
Thank you so much!
elementary-set-theory cardinals ordinals
asked Dec 27 '18 at 3:15
Le Anh Dung
9621521
2
Pick one, then continue picking the least indexed ordinal which is strictly larger.
– Asaf Karagila♦
Dec 27 '18 at 7:37
@AsafKaragila. He means ordinal sequence.
– William Elliot
Dec 27 '18 at 8:39
Hi @AsafKaragila, is it correct that this subsequence is cofinal in $alpha$?
– Le Anh Dung
Dec 27 '18 at 8:57
1
@William: Yes, and? Your answer is different how?
– Asaf Karagila♦
Dec 27 '18 at 9:19
1
@LeAnhDung: Why wouldn't it be?
– Asaf Karagila♦
Dec 27 '18 at 9:19
add a comment |
Let $alpha$ be a limit ordinal which is not a cardinal, and $kappa=|alpha|$. Then there exists a bijection from $kappa$ to $alpha$, or equivalently, a one-to-one sequence $langle alpha_xi mid xi< kappa rangle$ of ordinals such that ${alpha_xi mid xi< kappa}=alpha$.
My textbook states:
Now we can find (by transfinite recursion) a subsequence which is strictly increasing and has limit $alpha$.
I have tried but to no avail in constructing such subsequence, please shed me some lights.
Thank you so much!
elementary-set-theory cardinals ordinals
asked Dec 27 '18 at 3:15
Le Anh Dung
9621521
Let $alpha$ be a limit ordinal which is not a cardinal, and $kappa=|alpha|$. Then there exists a bijection from $kappa$ to $alpha$, or equivalently, a one-to-one sequence $langle alpha_xi mid xi< kappa rangle$ of ordinals such that ${alpha_xi mid xi< kappa}=alpha$.
My textbook states:
Now we can find (by transfinite recursion) a subsequence which is strictly increasing and has limit $alpha$.
I have tried but to no avail in constructing such subsequence, please shed me some lights.
Thank you so much!
elementary-set-theory cardinals ordinals
elementary-set-theory cardinals ordinals
asked Dec 27 '18 at 3:15
Le Anh Dung
9621521
asked Dec 27 '18 at 3:15
Le Anh Dung
9621521
asked Dec 27 '18 at 3:15
Le Anh Dung
9621521
asked Dec 27 '18 at 3:15
Le Anh Dung
9621521
asked Dec 27 '18 at 3:15
Le Anh Dung
9621521
9621521
2
Pick one, then continue picking the least indexed ordinal which is strictly larger.
– Asaf Karagila♦
Dec 27 '18 at 7:37
@AsafKaragila. He means ordinal sequence.
– William Elliot
Dec 27 '18 at 8:39
Hi @AsafKaragila, is it correct that this subsequence is cofinal in $alpha$?
– Le Anh Dung
Dec 27 '18 at 8:57
1
@William: Yes, and? Your answer is different how?
– Asaf Karagila♦
Dec 27 '18 at 9:19
1
@LeAnhDung: Why wouldn't it be?
– Asaf Karagila♦
Dec 27 '18 at 9:19
add a comment |
2
Pick one, then continue picking the least indexed ordinal which is strictly larger.
– Asaf Karagila♦
Dec 27 '18 at 7:37
@AsafKaragila. He means ordinal sequence.
– William Elliot
Dec 27 '18 at 8:39
Hi @AsafKaragila, is it correct that this subsequence is cofinal in $alpha$?
– Le Anh Dung
Dec 27 '18 at 8:57
1
@William: Yes, and? Your answer is different how?
– Asaf Karagila♦
Dec 27 '18 at 9:19
1
@LeAnhDung: Why wouldn't it be?
– Asaf Karagila♦
Dec 27 '18 at 9:19
2
2
Pick one, then continue picking the least indexed ordinal which is strictly larger.
– Asaf Karagila♦
Dec 27 '18 at 7:37
Pick one, then continue picking the least indexed ordinal which is strictly larger.
– Asaf Karagila♦
Dec 27 '18 at 7:37
@AsafKaragila. He means ordinal sequence.
– William Elliot
Dec 27 '18 at 8:39
@AsafKaragila. He means ordinal sequence.
– William Elliot
Dec 27 '18 at 8:39
Hi @AsafKaragila, is it correct that this subsequence is cofinal in $alpha$?
– Le Anh Dung
Dec 27 '18 at 8:57
Hi @AsafKaragila, is it correct that this subsequence is cofinal in $alpha$?
– Le Anh Dung
Dec 27 '18 at 8:57
1
1
@William: Yes, and? Your answer is different how?
– Asaf Karagila♦
Dec 27 '18 at 9:19
@William: Yes, and? Your answer is different how?
– Asaf Karagila♦
Dec 27 '18 at 9:19
1
1
@LeAnhDung: Why wouldn't it be?
– Asaf Karagila♦
Dec 27 '18 at 9:19
@LeAnhDung: Why wouldn't it be?
– Asaf Karagila♦
Dec 27 '18 at 9:19
add a comment |
1 Answer
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Let K = { a$_t$ : t < $kappa$ }.
Define, using strong transfinite induction
s$_u$ = min{ a$_t$ in K : for all j < u, s$_j$ < a$_t$ }.
answered Dec 27 '18 at 9:11
William Elliot
7,1282519
add a comment |
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1 Answer
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1 Answer
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Let K = { a$_t$ : t < $kappa$ }.
Define, using strong transfinite induction
s$_u$ = min{ a$_t$ in K : for all j < u, s$_j$ < a$_t$ }.
answered Dec 27 '18 at 9:11
William Elliot
7,1282519
add a comment |
Let K = { a$_t$ : t < $kappa$ }.
Define, using strong transfinite induction
s$_u$ = min{ a$_t$ in K : for all j < u, s$_j$ < a$_t$ }.
answered Dec 27 '18 at 9:11
William Elliot
7,1282519
add a comment |
Let K = { a$_t$ : t < $kappa$ }.
Define, using strong transfinite induction
s$_u$ = min{ a$_t$ in K : for all j < u, s$_j$ < a$_t$ }.
answered Dec 27 '18 at 9:11
William Elliot
7,1282519
Let K = { a$_t$ : t < $kappa$ }.
Define, using strong transfinite induction
s$_u$ = min{ a$_t$ in K : for all j < u, s$_j$ < a$_t$ }.
answered Dec 27 '18 at 9:11
William Elliot
7,1282519
answered Dec 27 '18 at 9:11
William Elliot
7,1282519
answered Dec 27 '18 at 9:11
William Elliot
7,1282519
answered Dec 27 '18 at 9:11
William Elliot
7,1282519
7,1282519
add a comment |
add a comment |
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2
Pick one, then continue picking the least indexed ordinal which is strictly larger.
– Asaf Karagila♦
Dec 27 '18 at 7:37
@AsafKaragila. He means ordinal sequence.
– William Elliot
Dec 27 '18 at 8:39
Hi @AsafKaragila, is it correct that this subsequence is cofinal in $alpha$?
– Le Anh Dung
Dec 27 '18 at 8:57
1
@William: Yes, and? Your answer is different how?
– Asaf Karagila♦
Dec 27 '18 at 9:19
1
@LeAnhDung: Why wouldn't it be?
– Asaf Karagila♦
Dec 27 '18 at 9:19