Subsets of the unit interval The 2019 Stack Overflow Developer Survey Results Are InTwo exercises on set theory about Cantor's equation and the von Neumann hierarchyRegarding functions on $omega_1$Definability in $L(omega_1)$The Diamond Principle plus implies a Kurepa family.Kunen exercise III.8.21Consequence of $V=L$Jech Third Edition Set Theory p191 Theorem 13.21 Jensen. What is the Essence of the Diamond Principle?Cardinality of the set $(aleph_omega_1)^kappa : 1 leq kappa < aleph_omega_1$Enumerating an IntervalBuilding a special cofinal set in a totally ordered set
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Subsets of the unit interval
The 2019 Stack Overflow Developer Survey Results Are InTwo exercises on set theory about Cantor's equation and the von Neumann hierarchyRegarding functions on $omega_1$Definability in $L(omega_1)$The Diamond Principle plus implies a Kurepa family.Kunen exercise III.8.21Consequence of $V=L$Jech Third Edition Set Theory p191 Theorem 13.21 Jensen. What is the Essence of the Diamond Principle?Cardinality of the set $(aleph_omega_1)^kappa : 1 leq kappa < aleph_omega_1$Enumerating an IntervalBuilding a special cofinal set in a totally ordered set
$begingroup$
I have been working on a problem and I need to answer the following question:
Is there a family $F_alpha: alpha in omega_1$ of subsets of the interval $]0,1[$ such that:
(a) $F_alpha=x_1^alpha, x_2^alpha$ with $x_1^alpha< x_2^alpha$;
(b) If $alpha, beta in omega_1$ with $alpha ne beta$, then
$x_1^alpha<x_1^beta<x_2^alpha<x_2^beta$ or $x_1^beta<x_1^alpha<x_2^beta<x_2^alpha$?
I have tried to build such a family without success. Does anybody see the answer?
set-theory order-theory
$endgroup$
migrated from mathoverflow.net Mar 23 at 1:56
This question came from our site for professional mathematicians.
add a comment |
$begingroup$
I have been working on a problem and I need to answer the following question:
Is there a family $F_alpha: alpha in omega_1$ of subsets of the interval $]0,1[$ such that:
(a) $F_alpha=x_1^alpha, x_2^alpha$ with $x_1^alpha< x_2^alpha$;
(b) If $alpha, beta in omega_1$ with $alpha ne beta$, then
$x_1^alpha<x_1^beta<x_2^alpha<x_2^beta$ or $x_1^beta<x_1^alpha<x_2^beta<x_2^alpha$?
I have tried to build such a family without success. Does anybody see the answer?
set-theory order-theory
$endgroup$
migrated from mathoverflow.net Mar 23 at 1:56
This question came from our site for professional mathematicians.
2
$begingroup$
What about the family of all $F_x=x,x+1/2$ for $0<x<1/2$?
$endgroup$
– YCor
Mar 21 at 14:14
$begingroup$
Which I think would naturally occur to almost everyone who posts in this forum. I suspect a cover of (0,1) is wanted but not stated, and thus also some set theoretical assumptions. Gerhard "And Where To Put 1/2?" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 14:31
$begingroup$
@YCor, thanks a lot!
$endgroup$
– Claudia Correa
Mar 21 at 16:43
add a comment |
$begingroup$
I have been working on a problem and I need to answer the following question:
Is there a family $F_alpha: alpha in omega_1$ of subsets of the interval $]0,1[$ such that:
(a) $F_alpha=x_1^alpha, x_2^alpha$ with $x_1^alpha< x_2^alpha$;
(b) If $alpha, beta in omega_1$ with $alpha ne beta$, then
$x_1^alpha<x_1^beta<x_2^alpha<x_2^beta$ or $x_1^beta<x_1^alpha<x_2^beta<x_2^alpha$?
I have tried to build such a family without success. Does anybody see the answer?
set-theory order-theory
$endgroup$
I have been working on a problem and I need to answer the following question:
Is there a family $F_alpha: alpha in omega_1$ of subsets of the interval $]0,1[$ such that:
(a) $F_alpha=x_1^alpha, x_2^alpha$ with $x_1^alpha< x_2^alpha$;
(b) If $alpha, beta in omega_1$ with $alpha ne beta$, then
$x_1^alpha<x_1^beta<x_2^alpha<x_2^beta$ or $x_1^beta<x_1^alpha<x_2^beta<x_2^alpha$?
I have tried to build such a family without success. Does anybody see the answer?
set-theory order-theory
set-theory order-theory
asked Mar 21 at 14:09
Claudia Correa
migrated from mathoverflow.net Mar 23 at 1:56
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Mar 23 at 1:56
This question came from our site for professional mathematicians.
2
$begingroup$
What about the family of all $F_x=x,x+1/2$ for $0<x<1/2$?
$endgroup$
– YCor
Mar 21 at 14:14
$begingroup$
Which I think would naturally occur to almost everyone who posts in this forum. I suspect a cover of (0,1) is wanted but not stated, and thus also some set theoretical assumptions. Gerhard "And Where To Put 1/2?" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 14:31
$begingroup$
@YCor, thanks a lot!
$endgroup$
– Claudia Correa
Mar 21 at 16:43
add a comment |
2
$begingroup$
What about the family of all $F_x=x,x+1/2$ for $0<x<1/2$?
$endgroup$
– YCor
Mar 21 at 14:14
$begingroup$
Which I think would naturally occur to almost everyone who posts in this forum. I suspect a cover of (0,1) is wanted but not stated, and thus also some set theoretical assumptions. Gerhard "And Where To Put 1/2?" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 14:31
$begingroup$
@YCor, thanks a lot!
$endgroup$
– Claudia Correa
Mar 21 at 16:43
2
2
$begingroup$
What about the family of all $F_x=x,x+1/2$ for $0<x<1/2$?
$endgroup$
– YCor
Mar 21 at 14:14
$begingroup$
What about the family of all $F_x=x,x+1/2$ for $0<x<1/2$?
$endgroup$
– YCor
Mar 21 at 14:14
$begingroup$
Which I think would naturally occur to almost everyone who posts in this forum. I suspect a cover of (0,1) is wanted but not stated, and thus also some set theoretical assumptions. Gerhard "And Where To Put 1/2?" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 14:31
$begingroup$
Which I think would naturally occur to almost everyone who posts in this forum. I suspect a cover of (0,1) is wanted but not stated, and thus also some set theoretical assumptions. Gerhard "And Where To Put 1/2?" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 14:31
$begingroup$
@YCor, thanks a lot!
$endgroup$
– Claudia Correa
Mar 21 at 16:43
$begingroup$
@YCor, thanks a lot!
$endgroup$
– Claudia Correa
Mar 21 at 16:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that the condition is about embedding into the open unit interval the order $2 times P$ for two total orders of cardinalities two and $omega_1$. (Unless I got it backwards and really mean $P times 2$.) This means not just two disjoint copies of $P$, but also (by condition (b)) every element of one copy is below every element of the other copy, and the family $F$ defines an order isomorphism between the copies. Thus both copies possess (or both lack) extremal elements. This means one cannot cover the open interval with such an embedding. If one does not need a cover, the other answer and comment provide a solution.
Gerhard "Is 1/2 Out Of Order?" Paseman, 2019.03.21
$endgroup$
$begingroup$
Thanks a lot! You really helped.
$endgroup$
– Claudia Correa
Mar 21 at 15:08
$begingroup$
@Claudia, you're welcome. Is this for a functional analysis course? Gerhard "Is Wondering A Little Bit" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 15:45
$begingroup$
If this was the intended question, you should edit the question and accept this answer.
$endgroup$
– YCor
Mar 21 at 16:10
$begingroup$
@YCor, I wanted exactly what I asked for. Thanks!
$endgroup$
– Claudia Correa
Mar 21 at 16:14
$begingroup$
@Gerhard, it is not for a course. I am trying to solve a problem and I simply couldnt see the answer ;-)
$endgroup$
– Claudia Correa
Mar 21 at 16:16
|
show 2 more comments
$begingroup$
As suggested YCor take $F_x=x,x+frac12$. Then use the axiom of choice to set a well-order $prec$ of $(0;frac12)$. Then you can get such a family $G_alpha$ to be the $F_x$'s enumereted with respect to $prec$.
$endgroup$
$begingroup$
No, you cannot just steal complete ideas and get credit for them as answers.
$endgroup$
– mbsq
Mar 21 at 22:19
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that the condition is about embedding into the open unit interval the order $2 times P$ for two total orders of cardinalities two and $omega_1$. (Unless I got it backwards and really mean $P times 2$.) This means not just two disjoint copies of $P$, but also (by condition (b)) every element of one copy is below every element of the other copy, and the family $F$ defines an order isomorphism between the copies. Thus both copies possess (or both lack) extremal elements. This means one cannot cover the open interval with such an embedding. If one does not need a cover, the other answer and comment provide a solution.
Gerhard "Is 1/2 Out Of Order?" Paseman, 2019.03.21
$endgroup$
$begingroup$
Thanks a lot! You really helped.
$endgroup$
– Claudia Correa
Mar 21 at 15:08
$begingroup$
@Claudia, you're welcome. Is this for a functional analysis course? Gerhard "Is Wondering A Little Bit" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 15:45
$begingroup$
If this was the intended question, you should edit the question and accept this answer.
$endgroup$
– YCor
Mar 21 at 16:10
$begingroup$
@YCor, I wanted exactly what I asked for. Thanks!
$endgroup$
– Claudia Correa
Mar 21 at 16:14
$begingroup$
@Gerhard, it is not for a course. I am trying to solve a problem and I simply couldnt see the answer ;-)
$endgroup$
– Claudia Correa
Mar 21 at 16:16
|
show 2 more comments
$begingroup$
Note that the condition is about embedding into the open unit interval the order $2 times P$ for two total orders of cardinalities two and $omega_1$. (Unless I got it backwards and really mean $P times 2$.) This means not just two disjoint copies of $P$, but also (by condition (b)) every element of one copy is below every element of the other copy, and the family $F$ defines an order isomorphism between the copies. Thus both copies possess (or both lack) extremal elements. This means one cannot cover the open interval with such an embedding. If one does not need a cover, the other answer and comment provide a solution.
Gerhard "Is 1/2 Out Of Order?" Paseman, 2019.03.21
$endgroup$
$begingroup$
Thanks a lot! You really helped.
$endgroup$
– Claudia Correa
Mar 21 at 15:08
$begingroup$
@Claudia, you're welcome. Is this for a functional analysis course? Gerhard "Is Wondering A Little Bit" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 15:45
$begingroup$
If this was the intended question, you should edit the question and accept this answer.
$endgroup$
– YCor
Mar 21 at 16:10
$begingroup$
@YCor, I wanted exactly what I asked for. Thanks!
$endgroup$
– Claudia Correa
Mar 21 at 16:14
$begingroup$
@Gerhard, it is not for a course. I am trying to solve a problem and I simply couldnt see the answer ;-)
$endgroup$
– Claudia Correa
Mar 21 at 16:16
|
show 2 more comments
$begingroup$
Note that the condition is about embedding into the open unit interval the order $2 times P$ for two total orders of cardinalities two and $omega_1$. (Unless I got it backwards and really mean $P times 2$.) This means not just two disjoint copies of $P$, but also (by condition (b)) every element of one copy is below every element of the other copy, and the family $F$ defines an order isomorphism between the copies. Thus both copies possess (or both lack) extremal elements. This means one cannot cover the open interval with such an embedding. If one does not need a cover, the other answer and comment provide a solution.
Gerhard "Is 1/2 Out Of Order?" Paseman, 2019.03.21
$endgroup$
Note that the condition is about embedding into the open unit interval the order $2 times P$ for two total orders of cardinalities two and $omega_1$. (Unless I got it backwards and really mean $P times 2$.) This means not just two disjoint copies of $P$, but also (by condition (b)) every element of one copy is below every element of the other copy, and the family $F$ defines an order isomorphism between the copies. Thus both copies possess (or both lack) extremal elements. This means one cannot cover the open interval with such an embedding. If one does not need a cover, the other answer and comment provide a solution.
Gerhard "Is 1/2 Out Of Order?" Paseman, 2019.03.21
answered Mar 21 at 14:51
Gerhard Paseman
$begingroup$
Thanks a lot! You really helped.
$endgroup$
– Claudia Correa
Mar 21 at 15:08
$begingroup$
@Claudia, you're welcome. Is this for a functional analysis course? Gerhard "Is Wondering A Little Bit" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 15:45
$begingroup$
If this was the intended question, you should edit the question and accept this answer.
$endgroup$
– YCor
Mar 21 at 16:10
$begingroup$
@YCor, I wanted exactly what I asked for. Thanks!
$endgroup$
– Claudia Correa
Mar 21 at 16:14
$begingroup$
@Gerhard, it is not for a course. I am trying to solve a problem and I simply couldnt see the answer ;-)
$endgroup$
– Claudia Correa
Mar 21 at 16:16
|
show 2 more comments
$begingroup$
Thanks a lot! You really helped.
$endgroup$
– Claudia Correa
Mar 21 at 15:08
$begingroup$
@Claudia, you're welcome. Is this for a functional analysis course? Gerhard "Is Wondering A Little Bit" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 15:45
$begingroup$
If this was the intended question, you should edit the question and accept this answer.
$endgroup$
– YCor
Mar 21 at 16:10
$begingroup$
@YCor, I wanted exactly what I asked for. Thanks!
$endgroup$
– Claudia Correa
Mar 21 at 16:14
$begingroup$
@Gerhard, it is not for a course. I am trying to solve a problem and I simply couldnt see the answer ;-)
$endgroup$
– Claudia Correa
Mar 21 at 16:16
$begingroup$
Thanks a lot! You really helped.
$endgroup$
– Claudia Correa
Mar 21 at 15:08
$begingroup$
Thanks a lot! You really helped.
$endgroup$
– Claudia Correa
Mar 21 at 15:08
$begingroup$
@Claudia, you're welcome. Is this for a functional analysis course? Gerhard "Is Wondering A Little Bit" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 15:45
$begingroup$
@Claudia, you're welcome. Is this for a functional analysis course? Gerhard "Is Wondering A Little Bit" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 15:45
$begingroup$
If this was the intended question, you should edit the question and accept this answer.
$endgroup$
– YCor
Mar 21 at 16:10
$begingroup$
If this was the intended question, you should edit the question and accept this answer.
$endgroup$
– YCor
Mar 21 at 16:10
$begingroup$
@YCor, I wanted exactly what I asked for. Thanks!
$endgroup$
– Claudia Correa
Mar 21 at 16:14
$begingroup$
@YCor, I wanted exactly what I asked for. Thanks!
$endgroup$
– Claudia Correa
Mar 21 at 16:14
$begingroup$
@Gerhard, it is not for a course. I am trying to solve a problem and I simply couldnt see the answer ;-)
$endgroup$
– Claudia Correa
Mar 21 at 16:16
$begingroup$
@Gerhard, it is not for a course. I am trying to solve a problem and I simply couldnt see the answer ;-)
$endgroup$
– Claudia Correa
Mar 21 at 16:16
|
show 2 more comments
$begingroup$
As suggested YCor take $F_x=x,x+frac12$. Then use the axiom of choice to set a well-order $prec$ of $(0;frac12)$. Then you can get such a family $G_alpha$ to be the $F_x$'s enumereted with respect to $prec$.
$endgroup$
$begingroup$
No, you cannot just steal complete ideas and get credit for them as answers.
$endgroup$
– mbsq
Mar 21 at 22:19
add a comment |
$begingroup$
As suggested YCor take $F_x=x,x+frac12$. Then use the axiom of choice to set a well-order $prec$ of $(0;frac12)$. Then you can get such a family $G_alpha$ to be the $F_x$'s enumereted with respect to $prec$.
$endgroup$
$begingroup$
No, you cannot just steal complete ideas and get credit for them as answers.
$endgroup$
– mbsq
Mar 21 at 22:19
add a comment |
$begingroup$
As suggested YCor take $F_x=x,x+frac12$. Then use the axiom of choice to set a well-order $prec$ of $(0;frac12)$. Then you can get such a family $G_alpha$ to be the $F_x$'s enumereted with respect to $prec$.
$endgroup$
As suggested YCor take $F_x=x,x+frac12$. Then use the axiom of choice to set a well-order $prec$ of $(0;frac12)$. Then you can get such a family $G_alpha$ to be the $F_x$'s enumereted with respect to $prec$.
answered Mar 21 at 14:21
user660476
$begingroup$
No, you cannot just steal complete ideas and get credit for them as answers.
$endgroup$
– mbsq
Mar 21 at 22:19
add a comment |
$begingroup$
No, you cannot just steal complete ideas and get credit for them as answers.
$endgroup$
– mbsq
Mar 21 at 22:19
$begingroup$
No, you cannot just steal complete ideas and get credit for them as answers.
$endgroup$
– mbsq
Mar 21 at 22:19
$begingroup$
No, you cannot just steal complete ideas and get credit for them as answers.
$endgroup$
– mbsq
Mar 21 at 22:19
add a comment |
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2
$begingroup$
What about the family of all $F_x=x,x+1/2$ for $0<x<1/2$?
$endgroup$
– YCor
Mar 21 at 14:14
$begingroup$
Which I think would naturally occur to almost everyone who posts in this forum. I suspect a cover of (0,1) is wanted but not stated, and thus also some set theoretical assumptions. Gerhard "And Where To Put 1/2?" Paseman, 2019.03.21.
$endgroup$
– Gerhard Paseman
Mar 21 at 14:31
$begingroup$
@YCor, thanks a lot!
$endgroup$
– Claudia Correa
Mar 21 at 16:43