Mixing
Lexicographic Preference Relation on the QxR
Clash Royale CLAN TAG#URR8PPP
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I would like to ask for your help. I recently learned that the Lexicographic Preference relation can be represented by a utility function $u:XtomathbbR$ on $mathbbQtimesmathbbR$ (but not $mathbbRtimesmathbbQ$).
To recall, a lexicographic preference relation says that on $mathbbR^2$, $xsucceq y$x if and only if $x_1>y_1$ or $x_1=y_1$ and $x_2geq y_2$ where $x=(x_1,x_2)$ and $y=(y_1,y_2)$.
Thus, I would have liked to see a proof of this statement and how is it possible to build such a utility representation. I guess that we must assume first that there exist a utility representation for the Lexicographic preference on $mathbbQtimesmathbbQ$ since the latter is countable, but I am lost after.
Infinitely many thanks for your help!
preferences decision-theory
edited Sep 9 at 0:08
Michael Greinecker
3,004718
asked Sep 8 at 22:42
Rororo
241
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Rororo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
4
down vote
favorite
I would like to ask for your help. I recently learned that the Lexicographic Preference relation can be represented by a utility function $u:XtomathbbR$ on $mathbbQtimesmathbbR$ (but not $mathbbRtimesmathbbQ$).
To recall, a lexicographic preference relation says that on $mathbbR^2$, $xsucceq y$x if and only if $x_1>y_1$ or $x_1=y_1$ and $x_2geq y_2$ where $x=(x_1,x_2)$ and $y=(y_1,y_2)$.
Thus, I would have liked to see a proof of this statement and how is it possible to build such a utility representation. I guess that we must assume first that there exist a utility representation for the Lexicographic preference on $mathbbQtimesmathbbQ$ since the latter is countable, but I am lost after.
Infinitely many thanks for your help!
preferences decision-theory
edited Sep 9 at 0:08
Michael Greinecker
3,004718
asked Sep 8 at 22:42
Rororo
241
New contributor
Rororo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I would like to ask for your help. I recently learned that the Lexicographic Preference relation can be represented by a utility function $u:XtomathbbR$ on $mathbbQtimesmathbbR$ (but not $mathbbRtimesmathbbQ$).
To recall, a lexicographic preference relation says that on $mathbbR^2$, $xsucceq y$x if and only if $x_1>y_1$ or $x_1=y_1$ and $x_2geq y_2$ where $x=(x_1,x_2)$ and $y=(y_1,y_2)$.
Thus, I would have liked to see a proof of this statement and how is it possible to build such a utility representation. I guess that we must assume first that there exist a utility representation for the Lexicographic preference on $mathbbQtimesmathbbQ$ since the latter is countable, but I am lost after.
Infinitely many thanks for your help!
preferences decision-theory
edited Sep 9 at 0:08
Michael Greinecker
3,004718
asked Sep 8 at 22:42
Rororo
241
New contributor
Rororo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I would like to ask for your help. I recently learned that the Lexicographic Preference relation can be represented by a utility function $u:XtomathbbR$ on $mathbbQtimesmathbbR$ (but not $mathbbRtimesmathbbQ$).
To recall, a lexicographic preference relation says that on $mathbbR^2$, $xsucceq y$x if and only if $x_1>y_1$ or $x_1=y_1$ and $x_2geq y_2$ where $x=(x_1,x_2)$ and $y=(y_1,y_2)$.
Thus, I would have liked to see a proof of this statement and how is it possible to build such a utility representation. I guess that we must assume first that there exist a utility representation for the Lexicographic preference on $mathbbQtimesmathbbQ$ since the latter is countable, but I am lost after.
Infinitely many thanks for your help!
preferences decision-theory
preferences decision-theory
edited Sep 9 at 0:08
Michael Greinecker
3,004718
asked Sep 8 at 22:42
Rororo
241
New contributor
Rororo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 9 at 0:08
Michael Greinecker
3,004718
asked Sep 8 at 22:42
Rororo
241
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Rororo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Sep 9 at 0:08
Michael Greinecker
3,004718
edited Sep 9 at 0:08
Michael Greinecker
3,004718
edited Sep 9 at 0:08
Michael Greinecker
3,004718
3,004718
asked Sep 8 at 22:42
Rororo
241
New contributor
Rororo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 8 at 22:42
Rororo
241
asked Sep 8 at 22:42
Rororo
241
241
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Rororo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Rororo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
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Note first that for each nontrivial (more than one point) compact interval $I$, there exists a strictly increasing function from $mathbbR$ to $I$.
Let $langle q_1,q_2, q_3,ldotsrangle$ be an enumeration of $mathbbQ$. We will define $u:mathbbQtimesmathbbRtomathbbR$ inductively, one $q_n$ after another.
Let $I_1$ be a nontrivial compact interval and $u_1:mathbbRto I_1$ be strictly increasing.
Now assume there are disjoint nontrivial compact intervals $I_1,ldots,I_n$ that ordered on the line like $q_1,ldots,q_n$, and strictly increasing functions $u_m:mathbbRto I_m$ for $mleq n$. Since these intervals are closed and disjoint, there is space between any two consecutive intervals. So one can put a nontrivial compact interval $I_n+1$ in a place corresponding to $q_n+1$ among $q_1,ldots, q_n$. Choose also some strictly increasing function $u_n+1:mathbbRto I_n+1$.
This will give you a disjoint sequence $langle I_1, I_2, ldotsrangle$ of nontrivial intervals and a sequence of functions $langle u_1, u_2, ldotsrangle$ with $u_n:mathbbRto I$ be strictly increasing. Now define $u:mathbbQtimesmathbbRto mathbbR$ by $u(q_n,r)=u_n(r)$.
You can verify that $u$ represents the lexicographic ordering on $mathbbQtimesmathbbR$.
Warning: The existence of a countably infinite number of disjoint nontrivial intervals might appear very strange, but is perfectly fine.
answered Sep 9 at 0:37
Michael Greinecker
3,004718
Thanks very much for this answer. But I am not sure I completely got the idea of the proof.
– Rororo
Sep 9 at 17:17
The, maybe somewhat surprising, idea is that one can order nontrivial compact intervals in exactly the same way the rationals are ordered. This gives you the ordering of the first coordinate, the numbers within the intervals give you the ordering of the second coordinate.
– Michael Greinecker
Sep 9 at 17:23
When you assume that "there are disjoint nontrivial compact intervals I1,…,In ordered on the line like q1,…,qn" does this mean that we consider nontrivial compact intervals with q1 and all the reals that may be "coupled" with q1, then I2 being the nontrivial compact interval with all the reals attached to q2 ? And finally while defining u(q_n,r)=u_n(r) are we done ? Shouldn't we define a function like u=2^(-i) where i would be the number of elements in the compact interval for ex. ? infinitely many thanks again!
– Rororo
Sep 9 at 17:32
The ordering assumption means that if $q_l < q_m$ (as rational numbers, not in the order of the inidces), then for all $r_lin I_l$ and $r_min I_m$, we have $r_l<r_m$. The reason that the function $u$ is constructed inductively, and not in the familiar way of "counting smaller elements (in a potentially weighted way)," is that I don't know how one can specify the lenghts and positions of the intervals $I_n$ a priori.
– Michael Greinecker
Sep 9 at 17:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Note first that for each nontrivial (more than one point) compact interval $I$, there exists a strictly increasing function from $mathbbR$ to $I$.
Let $langle q_1,q_2, q_3,ldotsrangle$ be an enumeration of $mathbbQ$. We will define $u:mathbbQtimesmathbbRtomathbbR$ inductively, one $q_n$ after another.
Let $I_1$ be a nontrivial compact interval and $u_1:mathbbRto I_1$ be strictly increasing.
Now assume there are disjoint nontrivial compact intervals $I_1,ldots,I_n$ that ordered on the line like $q_1,ldots,q_n$, and strictly increasing functions $u_m:mathbbRto I_m$ for $mleq n$. Since these intervals are closed and disjoint, there is space between any two consecutive intervals. So one can put a nontrivial compact interval $I_n+1$ in a place corresponding to $q_n+1$ among $q_1,ldots, q_n$. Choose also some strictly increasing function $u_n+1:mathbbRto I_n+1$.
This will give you a disjoint sequence $langle I_1, I_2, ldotsrangle$ of nontrivial intervals and a sequence of functions $langle u_1, u_2, ldotsrangle$ with $u_n:mathbbRto I$ be strictly increasing. Now define $u:mathbbQtimesmathbbRto mathbbR$ by $u(q_n,r)=u_n(r)$.
You can verify that $u$ represents the lexicographic ordering on $mathbbQtimesmathbbR$.
Warning: The existence of a countably infinite number of disjoint nontrivial intervals might appear very strange, but is perfectly fine.
answered Sep 9 at 0:37
Michael Greinecker
3,004718
Thanks very much for this answer. But I am not sure I completely got the idea of the proof.
– Rororo
Sep 9 at 17:17
The, maybe somewhat surprising, idea is that one can order nontrivial compact intervals in exactly the same way the rationals are ordered. This gives you the ordering of the first coordinate, the numbers within the intervals give you the ordering of the second coordinate.
– Michael Greinecker
Sep 9 at 17:23
When you assume that "there are disjoint nontrivial compact intervals I1,…,In ordered on the line like q1,…,qn" does this mean that we consider nontrivial compact intervals with q1 and all the reals that may be "coupled" with q1, then I2 being the nontrivial compact interval with all the reals attached to q2 ? And finally while defining u(q_n,r)=u_n(r) are we done ? Shouldn't we define a function like u=2^(-i) where i would be the number of elements in the compact interval for ex. ? infinitely many thanks again!
– Rororo
Sep 9 at 17:32
The ordering assumption means that if $q_l < q_m$ (as rational numbers, not in the order of the inidces), then for all $r_lin I_l$ and $r_min I_m$, we have $r_l<r_m$. The reason that the function $u$ is constructed inductively, and not in the familiar way of "counting smaller elements (in a potentially weighted way)," is that I don't know how one can specify the lenghts and positions of the intervals $I_n$ a priori.
– Michael Greinecker
Sep 9 at 17:39
add a comment |Â
up vote
3
down vote
Note first that for each nontrivial (more than one point) compact interval $I$, there exists a strictly increasing function from $mathbbR$ to $I$.
Let $langle q_1,q_2, q_3,ldotsrangle$ be an enumeration of $mathbbQ$. We will define $u:mathbbQtimesmathbbRtomathbbR$ inductively, one $q_n$ after another.
Let $I_1$ be a nontrivial compact interval and $u_1:mathbbRto I_1$ be strictly increasing.
Now assume there are disjoint nontrivial compact intervals $I_1,ldots,I_n$ that ordered on the line like $q_1,ldots,q_n$, and strictly increasing functions $u_m:mathbbRto I_m$ for $mleq n$. Since these intervals are closed and disjoint, there is space between any two consecutive intervals. So one can put a nontrivial compact interval $I_n+1$ in a place corresponding to $q_n+1$ among $q_1,ldots, q_n$. Choose also some strictly increasing function $u_n+1:mathbbRto I_n+1$.
This will give you a disjoint sequence $langle I_1, I_2, ldotsrangle$ of nontrivial intervals and a sequence of functions $langle u_1, u_2, ldotsrangle$ with $u_n:mathbbRto I$ be strictly increasing. Now define $u:mathbbQtimesmathbbRto mathbbR$ by $u(q_n,r)=u_n(r)$.
You can verify that $u$ represents the lexicographic ordering on $mathbbQtimesmathbbR$.
Warning: The existence of a countably infinite number of disjoint nontrivial intervals might appear very strange, but is perfectly fine.
answered Sep 9 at 0:37
Michael Greinecker
3,004718
Thanks very much for this answer. But I am not sure I completely got the idea of the proof.
– Rororo
Sep 9 at 17:17
The, maybe somewhat surprising, idea is that one can order nontrivial compact intervals in exactly the same way the rationals are ordered. This gives you the ordering of the first coordinate, the numbers within the intervals give you the ordering of the second coordinate.
– Michael Greinecker
Sep 9 at 17:23
When you assume that "there are disjoint nontrivial compact intervals I1,…,In ordered on the line like q1,…,qn" does this mean that we consider nontrivial compact intervals with q1 and all the reals that may be "coupled" with q1, then I2 being the nontrivial compact interval with all the reals attached to q2 ? And finally while defining u(q_n,r)=u_n(r) are we done ? Shouldn't we define a function like u=2^(-i) where i would be the number of elements in the compact interval for ex. ? infinitely many thanks again!
– Rororo
Sep 9 at 17:32
The ordering assumption means that if $q_l < q_m$ (as rational numbers, not in the order of the inidces), then for all $r_lin I_l$ and $r_min I_m$, we have $r_l<r_m$. The reason that the function $u$ is constructed inductively, and not in the familiar way of "counting smaller elements (in a potentially weighted way)," is that I don't know how one can specify the lenghts and positions of the intervals $I_n$ a priori.
– Michael Greinecker
Sep 9 at 17:39
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Note first that for each nontrivial (more than one point) compact interval $I$, there exists a strictly increasing function from $mathbbR$ to $I$.
Let $langle q_1,q_2, q_3,ldotsrangle$ be an enumeration of $mathbbQ$. We will define $u:mathbbQtimesmathbbRtomathbbR$ inductively, one $q_n$ after another.
Let $I_1$ be a nontrivial compact interval and $u_1:mathbbRto I_1$ be strictly increasing.
Now assume there are disjoint nontrivial compact intervals $I_1,ldots,I_n$ that ordered on the line like $q_1,ldots,q_n$, and strictly increasing functions $u_m:mathbbRto I_m$ for $mleq n$. Since these intervals are closed and disjoint, there is space between any two consecutive intervals. So one can put a nontrivial compact interval $I_n+1$ in a place corresponding to $q_n+1$ among $q_1,ldots, q_n$. Choose also some strictly increasing function $u_n+1:mathbbRto I_n+1$.
This will give you a disjoint sequence $langle I_1, I_2, ldotsrangle$ of nontrivial intervals and a sequence of functions $langle u_1, u_2, ldotsrangle$ with $u_n:mathbbRto I$ be strictly increasing. Now define $u:mathbbQtimesmathbbRto mathbbR$ by $u(q_n,r)=u_n(r)$.
You can verify that $u$ represents the lexicographic ordering on $mathbbQtimesmathbbR$.
Warning: The existence of a countably infinite number of disjoint nontrivial intervals might appear very strange, but is perfectly fine.
answered Sep 9 at 0:37
Michael Greinecker
3,004718
Note first that for each nontrivial (more than one point) compact interval $I$, there exists a strictly increasing function from $mathbbR$ to $I$.
Let $langle q_1,q_2, q_3,ldotsrangle$ be an enumeration of $mathbbQ$. We will define $u:mathbbQtimesmathbbRtomathbbR$ inductively, one $q_n$ after another.
Let $I_1$ be a nontrivial compact interval and $u_1:mathbbRto I_1$ be strictly increasing.
Now assume there are disjoint nontrivial compact intervals $I_1,ldots,I_n$ that ordered on the line like $q_1,ldots,q_n$, and strictly increasing functions $u_m:mathbbRto I_m$ for $mleq n$. Since these intervals are closed and disjoint, there is space between any two consecutive intervals. So one can put a nontrivial compact interval $I_n+1$ in a place corresponding to $q_n+1$ among $q_1,ldots, q_n$. Choose also some strictly increasing function $u_n+1:mathbbRto I_n+1$.
This will give you a disjoint sequence $langle I_1, I_2, ldotsrangle$ of nontrivial intervals and a sequence of functions $langle u_1, u_2, ldotsrangle$ with $u_n:mathbbRto I$ be strictly increasing. Now define $u:mathbbQtimesmathbbRto mathbbR$ by $u(q_n,r)=u_n(r)$.
You can verify that $u$ represents the lexicographic ordering on $mathbbQtimesmathbbR$.
Warning: The existence of a countably infinite number of disjoint nontrivial intervals might appear very strange, but is perfectly fine.
answered Sep 9 at 0:37
Michael Greinecker
3,004718
answered Sep 9 at 0:37
Michael Greinecker
3,004718
answered Sep 9 at 0:37
Michael Greinecker
3,004718
answered Sep 9 at 0:37
Michael Greinecker
3,004718
3,004718
Thanks very much for this answer. But I am not sure I completely got the idea of the proof.
– Rororo
Sep 9 at 17:17
The, maybe somewhat surprising, idea is that one can order nontrivial compact intervals in exactly the same way the rationals are ordered. This gives you the ordering of the first coordinate, the numbers within the intervals give you the ordering of the second coordinate.
– Michael Greinecker
Sep 9 at 17:23
When you assume that "there are disjoint nontrivial compact intervals I1,…,In ordered on the line like q1,…,qn" does this mean that we consider nontrivial compact intervals with q1 and all the reals that may be "coupled" with q1, then I2 being the nontrivial compact interval with all the reals attached to q2 ? And finally while defining u(q_n,r)=u_n(r) are we done ? Shouldn't we define a function like u=2^(-i) where i would be the number of elements in the compact interval for ex. ? infinitely many thanks again!
– Rororo
Sep 9 at 17:32
The ordering assumption means that if $q_l < q_m$ (as rational numbers, not in the order of the inidces), then for all $r_lin I_l$ and $r_min I_m$, we have $r_l<r_m$. The reason that the function $u$ is constructed inductively, and not in the familiar way of "counting smaller elements (in a potentially weighted way)," is that I don't know how one can specify the lenghts and positions of the intervals $I_n$ a priori.
– Michael Greinecker
Sep 9 at 17:39
add a comment |Â
Thanks very much for this answer. But I am not sure I completely got the idea of the proof.
– Rororo
Sep 9 at 17:17
The, maybe somewhat surprising, idea is that one can order nontrivial compact intervals in exactly the same way the rationals are ordered. This gives you the ordering of the first coordinate, the numbers within the intervals give you the ordering of the second coordinate.
– Michael Greinecker
Sep 9 at 17:23
When you assume that "there are disjoint nontrivial compact intervals I1,…,In ordered on the line like q1,…,qn" does this mean that we consider nontrivial compact intervals with q1 and all the reals that may be "coupled" with q1, then I2 being the nontrivial compact interval with all the reals attached to q2 ? And finally while defining u(q_n,r)=u_n(r) are we done ? Shouldn't we define a function like u=2^(-i) where i would be the number of elements in the compact interval for ex. ? infinitely many thanks again!
– Rororo
Sep 9 at 17:32
The ordering assumption means that if $q_l < q_m$ (as rational numbers, not in the order of the inidces), then for all $r_lin I_l$ and $r_min I_m$, we have $r_l<r_m$. The reason that the function $u$ is constructed inductively, and not in the familiar way of "counting smaller elements (in a potentially weighted way)," is that I don't know how one can specify the lenghts and positions of the intervals $I_n$ a priori.
– Michael Greinecker
Sep 9 at 17:39
Thanks very much for this answer. But I am not sure I completely got the idea of the proof.
– Rororo
Sep 9 at 17:17
Thanks very much for this answer. But I am not sure I completely got the idea of the proof.
– Rororo
Sep 9 at 17:17
The, maybe somewhat surprising, idea is that one can order nontrivial compact intervals in exactly the same way the rationals are ordered. This gives you the ordering of the first coordinate, the numbers within the intervals give you the ordering of the second coordinate.
– Michael Greinecker
Sep 9 at 17:23
The, maybe somewhat surprising, idea is that one can order nontrivial compact intervals in exactly the same way the rationals are ordered. This gives you the ordering of the first coordinate, the numbers within the intervals give you the ordering of the second coordinate.
– Michael Greinecker
Sep 9 at 17:23
When you assume that "there are disjoint nontrivial compact intervals I1,…,In ordered on the line like q1,…,qn" does this mean that we consider nontrivial compact intervals with q1 and all the reals that may be "coupled" with q1, then I2 being the nontrivial compact interval with all the reals attached to q2 ? And finally while defining u(q_n,r)=u_n(r) are we done ? Shouldn't we define a function like u=2^(-i) where i would be the number of elements in the compact interval for ex. ? infinitely many thanks again!
– Rororo
Sep 9 at 17:32
When you assume that "there are disjoint nontrivial compact intervals I1,…,In ordered on the line like q1,…,qn" does this mean that we consider nontrivial compact intervals with q1 and all the reals that may be "coupled" with q1, then I2 being the nontrivial compact interval with all the reals attached to q2 ? And finally while defining u(q_n,r)=u_n(r) are we done ? Shouldn't we define a function like u=2^(-i) where i would be the number of elements in the compact interval for ex. ? infinitely many thanks again!
– Rororo
Sep 9 at 17:32
The ordering assumption means that if $q_l < q_m$ (as rational numbers, not in the order of the inidces), then for all $r_lin I_l$ and $r_min I_m$, we have $r_l<r_m$. The reason that the function $u$ is constructed inductively, and not in the familiar way of "counting smaller elements (in a potentially weighted way)," is that I don't know how one can specify the lenghts and positions of the intervals $I_n$ a priori.
– Michael Greinecker
Sep 9 at 17:39
The ordering assumption means that if $q_l < q_m$ (as rational numbers, not in the order of the inidces), then for all $r_lin I_l$ and $r_min I_m$, we have $r_l<r_m$. The reason that the function $u$ is constructed inductively, and not in the familiar way of "counting smaller elements (in a potentially weighted way)," is that I don't know how one can specify the lenghts and positions of the intervals $I_n$ a priori.
– Michael Greinecker
Sep 9 at 17:39
add a comment |Â
Rororo is a new contributor. Be nice, and check out our Code of Conduct.
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