Betweenness preserving implies monotonic?
$begingroup$
For this question, we can assume that $f:mathbb{R}rightarrowmathbb{R}$. However, I hope that an answer can generalize to arbitrary linearly ordered sets.
I assume that everyone will know what I mean by $f$ being weakly increasing, weakly decreasing, or monotonic. However, now let me introduce a different notion:
Let me say that for three real numbers $x$, $y$, and $z$ that $y$ lies between $x$ and $z$ if either of the compound inequalities below hold
$$xleq yleq zquadtext{or}quad zleq yleq x,.$$
Let me say also that $f$ preserves betweenness if whenever $y$ lies between $x$ and $z$ then $f(y)$ lies between $f(x)$ and $f(z)$.
Of course, a monotonic function preserves betweenness.
But the question is:
If a function preserves betweenness is it necessary monotonic? I.e., must it either be weakly increasing or weakly decreasing?
It seems like it should be, but then I recall Darboux functions and become hesitant to jump to this conclusion. If a betweenness-preserving non-monotonic function exists, it would be kind of odd.
It's readily apparent that it cannot have any strict local extrema. Also, if it has a local extremum at $x$, then there will be $a$ and $b$ such that $a<x<b$ where $f$ is constant on either $(a,x]$ or $[x,b)$ or both. However, any other properties that we investigate seem to require an intractable number of case analyses.
I'm reminded of some ghastly functions like those that satisfy the Cauchy functional equation but are not continuous anywhere. But those functions are unbounded on every interval.
real-analysis examples-counterexamples order-theory monotone-functions
asked Dec 12 '18 at 5:39
Robert WolfeRobert Wolfe
5,84922763
$endgroup$
add a comment |
$begingroup$
For this question, we can assume that $f:mathbb{R}rightarrowmathbb{R}$. However, I hope that an answer can generalize to arbitrary linearly ordered sets.
I assume that everyone will know what I mean by $f$ being weakly increasing, weakly decreasing, or monotonic. However, now let me introduce a different notion:
Let me say that for three real numbers $x$, $y$, and $z$ that $y$ lies between $x$ and $z$ if either of the compound inequalities below hold
$$xleq yleq zquadtext{or}quad zleq yleq x,.$$
Let me say also that $f$ preserves betweenness if whenever $y$ lies between $x$ and $z$ then $f(y)$ lies between $f(x)$ and $f(z)$.
Of course, a monotonic function preserves betweenness.
But the question is:
If a function preserves betweenness is it necessary monotonic? I.e., must it either be weakly increasing or weakly decreasing?
It seems like it should be, but then I recall Darboux functions and become hesitant to jump to this conclusion. If a betweenness-preserving non-monotonic function exists, it would be kind of odd.
It's readily apparent that it cannot have any strict local extrema. Also, if it has a local extremum at $x$, then there will be $a$ and $b$ such that $a<x<b$ where $f$ is constant on either $(a,x]$ or $[x,b)$ or both. However, any other properties that we investigate seem to require an intractable number of case analyses.
I'm reminded of some ghastly functions like those that satisfy the Cauchy functional equation but are not continuous anywhere. But those functions are unbounded on every interval.
real-analysis examples-counterexamples order-theory monotone-functions
asked Dec 12 '18 at 5:39
Robert WolfeRobert Wolfe
5,84922763
$endgroup$
1
$begingroup$
Note that this is false for partially ordered sets, where you can have incomparable $a,b$ both $<c$ and map this betweenness preservingly but not monotonically to $[3]={1<2<3}$ via $amapsto 1$, $bmapsto 3$, $cmapsto 2$.
$endgroup$
– Christoph
Dec 12 '18 at 8:57
add a comment |
$begingroup$
For this question, we can assume that $f:mathbb{R}rightarrowmathbb{R}$. However, I hope that an answer can generalize to arbitrary linearly ordered sets.
I assume that everyone will know what I mean by $f$ being weakly increasing, weakly decreasing, or monotonic. However, now let me introduce a different notion:
Let me say that for three real numbers $x$, $y$, and $z$ that $y$ lies between $x$ and $z$ if either of the compound inequalities below hold
$$xleq yleq zquadtext{or}quad zleq yleq x,.$$
Let me say also that $f$ preserves betweenness if whenever $y$ lies between $x$ and $z$ then $f(y)$ lies between $f(x)$ and $f(z)$.
Of course, a monotonic function preserves betweenness.
But the question is:
If a function preserves betweenness is it necessary monotonic? I.e., must it either be weakly increasing or weakly decreasing?
It seems like it should be, but then I recall Darboux functions and become hesitant to jump to this conclusion. If a betweenness-preserving non-monotonic function exists, it would be kind of odd.
It's readily apparent that it cannot have any strict local extrema. Also, if it has a local extremum at $x$, then there will be $a$ and $b$ such that $a<x<b$ where $f$ is constant on either $(a,x]$ or $[x,b)$ or both. However, any other properties that we investigate seem to require an intractable number of case analyses.
I'm reminded of some ghastly functions like those that satisfy the Cauchy functional equation but are not continuous anywhere. But those functions are unbounded on every interval.
real-analysis examples-counterexamples order-theory monotone-functions
asked Dec 12 '18 at 5:39
Robert WolfeRobert Wolfe
5,84922763
$endgroup$
For this question, we can assume that $f:mathbb{R}rightarrowmathbb{R}$. However, I hope that an answer can generalize to arbitrary linearly ordered sets.
I assume that everyone will know what I mean by $f$ being weakly increasing, weakly decreasing, or monotonic. However, now let me introduce a different notion:
Let me say that for three real numbers $x$, $y$, and $z$ that $y$ lies between $x$ and $z$ if either of the compound inequalities below hold
$$xleq yleq zquadtext{or}quad zleq yleq x,.$$
Let me say also that $f$ preserves betweenness if whenever $y$ lies between $x$ and $z$ then $f(y)$ lies between $f(x)$ and $f(z)$.
Of course, a monotonic function preserves betweenness.
But the question is:
If a function preserves betweenness is it necessary monotonic? I.e., must it either be weakly increasing or weakly decreasing?
It seems like it should be, but then I recall Darboux functions and become hesitant to jump to this conclusion. If a betweenness-preserving non-monotonic function exists, it would be kind of odd.
It's readily apparent that it cannot have any strict local extrema. Also, if it has a local extremum at $x$, then there will be $a$ and $b$ such that $a<x<b$ where $f$ is constant on either $(a,x]$ or $[x,b)$ or both. However, any other properties that we investigate seem to require an intractable number of case analyses.
I'm reminded of some ghastly functions like those that satisfy the Cauchy functional equation but are not continuous anywhere. But those functions are unbounded on every interval.
real-analysis examples-counterexamples order-theory monotone-functions
real-analysis examples-counterexamples order-theory monotone-functions
asked Dec 12 '18 at 5:39
Robert WolfeRobert Wolfe
5,84922763
asked Dec 12 '18 at 5:39
Robert WolfeRobert Wolfe
5,84922763
asked Dec 12 '18 at 5:39
Robert WolfeRobert Wolfe
5,84922763
asked Dec 12 '18 at 5:39
Robert WolfeRobert Wolfe
5,84922763
asked Dec 12 '18 at 5:39
Robert WolfeRobert Wolfe
5,84922763
5,84922763
1
$begingroup$
Note that this is false for partially ordered sets, where you can have incomparable $a,b$ both $<c$ and map this betweenness preservingly but not monotonically to $[3]={1<2<3}$ via $amapsto 1$, $bmapsto 3$, $cmapsto 2$.
$endgroup$
– Christoph
Dec 12 '18 at 8:57
add a comment |
1
$begingroup$
Note that this is false for partially ordered sets, where you can have incomparable $a,b$ both $<c$ and map this betweenness preservingly but not monotonically to $[3]={1<2<3}$ via $amapsto 1$, $bmapsto 3$, $cmapsto 2$.
$endgroup$
– Christoph
Dec 12 '18 at 8:57
1
1
$begingroup$
Note that this is false for partially ordered sets, where you can have incomparable $a,b$ both $<c$ and map this betweenness preservingly but not monotonically to $[3]={1<2<3}$ via $amapsto 1$, $bmapsto 3$, $cmapsto 2$.
$endgroup$
– Christoph
Dec 12 '18 at 8:57
$begingroup$
Note that this is false for partially ordered sets, where you can have incomparable $a,b$ both $<c$ and map this betweenness preservingly but not monotonically to $[3]={1<2<3}$ via $amapsto 1$, $bmapsto 3$, $cmapsto 2$.
$endgroup$
– Christoph
Dec 12 '18 at 8:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, betweenness-preserving implies monotonic.
Suppose $f$ is not monotonic, then there exist $a<b$, $f(a)<f(b)$, and also $c<d$, $f(c)>f(d)$. We note that $f$ is still not monotonic on $F:={a,b,c,d}subsetmathbb R$. Let's write $F$ as $F={a_1<a_2<dots<a_k}$ where clearly $kin{3, 4}$.
Choose $i$ minimal such that $f(a_i)ne f(a_{i+1})$.
Case 1: $f(a_i)<f(a_{i+1})$. Then there must be $jge i+1$ with $f(a_j)>f(a_{j+1})$, or else $f$ is weakly increasing on $F$. Then for the minimal such $j$ we are done: $f(a_i)<f(a_j)>f(a_{j+1})$ and $f(a_j)$ is not between $f(a_i)$ and $f(a_{j+1})$, so $f$ is not betweenness-preserving.
Case 2 is the same.
answered Dec 12 '18 at 6:15
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,096918
$endgroup$
1
$begingroup$
This is sleeker and clearly covers each case.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 14:08
$begingroup$
Thanks @RobertWolfe yes it avoids the issue of whether $a_2=d$ and $a_3=b$ and... the ghastly number of cases!
$endgroup$
– Bjørn Kjos-Hanssen
Dec 12 '18 at 14:34
add a comment |
$begingroup$
Suppose that f is not monotonic. Then $exists x, k$ s.t. $x < k$ and $f(x) < f(k)$ and $exists x_0, k_0$ s.t. $x_0 < k_0$ and $f(x_0) > f(k_0)$.
WLOG, let $k < x_0$. Then we get a contradiction on $f$ preserving boundedness because either $f(k)$ or $f(x_0)$ will not lie between $f(x)$ and $f(k_0)$.
answered Dec 12 '18 at 6:13
T. FoT. Fo
466311
$endgroup$
$begingroup$
I think there's a typo here. But I can redraw the pictures that I think you had in mind. How do you eliminate the possibility that $f(k)=f(x_0)$? In this set of circumstances. But I also have an issue with the "without a loss of generality" part. How about the possibilities of $x<x_0<k<k_0$? or $x_0<x<k_0<k$?
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:21
$begingroup$
@RobertWolfe You're right, there is a small typo. It should say $f(x) < f(k)$ in the first line. Does that help?
$endgroup$
– T. Fo
Dec 12 '18 at 6:27
$begingroup$
@RobertWolfe The WLOG just allows us to consider the nice ordering $x < k < x_0 < k_0$. Other cases are similar: consider the two 'endpoints' and then one of the two points in between will give a contradiction. The proof above is not entirely rigorous, you're correct.
$endgroup$
– T. Fo
Dec 12 '18 at 6:30
$begingroup$
I think the last $f(k)$ should be $f(k_0)$. I'm running through cases.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:34
1
$begingroup$
I now believe this argument is essentially correct. Following @bof comment's, and employing a little metaphor, the graph of such a function will either have a mountain peak, a valley, or an acute trapezoid in it. All of which don't preserve betweenness. I doubt the argument will get better for arbitrary ordered sets.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 7:02
|
show 5 more comments
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
Yes, betweenness-preserving implies monotonic.
Suppose $f$ is not monotonic, then there exist $a<b$, $f(a)<f(b)$, and also $c<d$, $f(c)>f(d)$. We note that $f$ is still not monotonic on $F:={a,b,c,d}subsetmathbb R$. Let's write $F$ as $F={a_1<a_2<dots<a_k}$ where clearly $kin{3, 4}$.
Choose $i$ minimal such that $f(a_i)ne f(a_{i+1})$.
Case 1: $f(a_i)<f(a_{i+1})$. Then there must be $jge i+1$ with $f(a_j)>f(a_{j+1})$, or else $f$ is weakly increasing on $F$. Then for the minimal such $j$ we are done: $f(a_i)<f(a_j)>f(a_{j+1})$ and $f(a_j)$ is not between $f(a_i)$ and $f(a_{j+1})$, so $f$ is not betweenness-preserving.
Case 2 is the same.
answered Dec 12 '18 at 6:15
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,096918
$endgroup$
1
$begingroup$
This is sleeker and clearly covers each case.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 14:08
$begingroup$
Thanks @RobertWolfe yes it avoids the issue of whether $a_2=d$ and $a_3=b$ and... the ghastly number of cases!
$endgroup$
– Bjørn Kjos-Hanssen
Dec 12 '18 at 14:34
add a comment |
$begingroup$
Yes, betweenness-preserving implies monotonic.
Suppose $f$ is not monotonic, then there exist $a<b$, $f(a)<f(b)$, and also $c<d$, $f(c)>f(d)$. We note that $f$ is still not monotonic on $F:={a,b,c,d}subsetmathbb R$. Let's write $F$ as $F={a_1<a_2<dots<a_k}$ where clearly $kin{3, 4}$.
Choose $i$ minimal such that $f(a_i)ne f(a_{i+1})$.
Case 1: $f(a_i)<f(a_{i+1})$. Then there must be $jge i+1$ with $f(a_j)>f(a_{j+1})$, or else $f$ is weakly increasing on $F$. Then for the minimal such $j$ we are done: $f(a_i)<f(a_j)>f(a_{j+1})$ and $f(a_j)$ is not between $f(a_i)$ and $f(a_{j+1})$, so $f$ is not betweenness-preserving.
Case 2 is the same.
answered Dec 12 '18 at 6:15
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,096918
$endgroup$
1
$begingroup$
This is sleeker and clearly covers each case.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 14:08
$begingroup$
Thanks @RobertWolfe yes it avoids the issue of whether $a_2=d$ and $a_3=b$ and... the ghastly number of cases!
$endgroup$
– Bjørn Kjos-Hanssen
Dec 12 '18 at 14:34
add a comment |
$begingroup$
Yes, betweenness-preserving implies monotonic.
Suppose $f$ is not monotonic, then there exist $a<b$, $f(a)<f(b)$, and also $c<d$, $f(c)>f(d)$. We note that $f$ is still not monotonic on $F:={a,b,c,d}subsetmathbb R$. Let's write $F$ as $F={a_1<a_2<dots<a_k}$ where clearly $kin{3, 4}$.
Choose $i$ minimal such that $f(a_i)ne f(a_{i+1})$.
Case 1: $f(a_i)<f(a_{i+1})$. Then there must be $jge i+1$ with $f(a_j)>f(a_{j+1})$, or else $f$ is weakly increasing on $F$. Then for the minimal such $j$ we are done: $f(a_i)<f(a_j)>f(a_{j+1})$ and $f(a_j)$ is not between $f(a_i)$ and $f(a_{j+1})$, so $f$ is not betweenness-preserving.
Case 2 is the same.
answered Dec 12 '18 at 6:15
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,096918
$endgroup$
Yes, betweenness-preserving implies monotonic.
Suppose $f$ is not monotonic, then there exist $a<b$, $f(a)<f(b)$, and also $c<d$, $f(c)>f(d)$. We note that $f$ is still not monotonic on $F:={a,b,c,d}subsetmathbb R$. Let's write $F$ as $F={a_1<a_2<dots<a_k}$ where clearly $kin{3, 4}$.
Choose $i$ minimal such that $f(a_i)ne f(a_{i+1})$.
Case 1: $f(a_i)<f(a_{i+1})$. Then there must be $jge i+1$ with $f(a_j)>f(a_{j+1})$, or else $f$ is weakly increasing on $F$. Then for the minimal such $j$ we are done: $f(a_i)<f(a_j)>f(a_{j+1})$ and $f(a_j)$ is not between $f(a_i)$ and $f(a_{j+1})$, so $f$ is not betweenness-preserving.
Case 2 is the same.
answered Dec 12 '18 at 6:15
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,096918
edited Dec 12 '18 at 8:43
answered Dec 12 '18 at 6:15
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,096918
answered Dec 12 '18 at 6:15
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,096918
answered Dec 12 '18 at 6:15
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,096918
2,096918
1
$begingroup$
This is sleeker and clearly covers each case.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 14:08
$begingroup$
Thanks @RobertWolfe yes it avoids the issue of whether $a_2=d$ and $a_3=b$ and... the ghastly number of cases!
$endgroup$
– Bjørn Kjos-Hanssen
Dec 12 '18 at 14:34
add a comment |
1
$begingroup$
This is sleeker and clearly covers each case.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 14:08
$begingroup$
Thanks @RobertWolfe yes it avoids the issue of whether $a_2=d$ and $a_3=b$ and... the ghastly number of cases!
$endgroup$
– Bjørn Kjos-Hanssen
Dec 12 '18 at 14:34
1
1
$begingroup$
This is sleeker and clearly covers each case.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 14:08
$begingroup$
This is sleeker and clearly covers each case.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 14:08
$begingroup$
Thanks @RobertWolfe yes it avoids the issue of whether $a_2=d$ and $a_3=b$ and... the ghastly number of cases!
$endgroup$
– Bjørn Kjos-Hanssen
Dec 12 '18 at 14:34
$begingroup$
Thanks @RobertWolfe yes it avoids the issue of whether $a_2=d$ and $a_3=b$ and... the ghastly number of cases!
$endgroup$
– Bjørn Kjos-Hanssen
Dec 12 '18 at 14:34
add a comment |
$begingroup$
Suppose that f is not monotonic. Then $exists x, k$ s.t. $x < k$ and $f(x) < f(k)$ and $exists x_0, k_0$ s.t. $x_0 < k_0$ and $f(x_0) > f(k_0)$.
WLOG, let $k < x_0$. Then we get a contradiction on $f$ preserving boundedness because either $f(k)$ or $f(x_0)$ will not lie between $f(x)$ and $f(k_0)$.
answered Dec 12 '18 at 6:13
T. FoT. Fo
466311
$endgroup$
$begingroup$
I think there's a typo here. But I can redraw the pictures that I think you had in mind. How do you eliminate the possibility that $f(k)=f(x_0)$? In this set of circumstances. But I also have an issue with the "without a loss of generality" part. How about the possibilities of $x<x_0<k<k_0$? or $x_0<x<k_0<k$?
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:21
$begingroup$
@RobertWolfe You're right, there is a small typo. It should say $f(x) < f(k)$ in the first line. Does that help?
$endgroup$
– T. Fo
Dec 12 '18 at 6:27
$begingroup$
@RobertWolfe The WLOG just allows us to consider the nice ordering $x < k < x_0 < k_0$. Other cases are similar: consider the two 'endpoints' and then one of the two points in between will give a contradiction. The proof above is not entirely rigorous, you're correct.
$endgroup$
– T. Fo
Dec 12 '18 at 6:30
$begingroup$
I think the last $f(k)$ should be $f(k_0)$. I'm running through cases.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:34
1
$begingroup$
I now believe this argument is essentially correct. Following @bof comment's, and employing a little metaphor, the graph of such a function will either have a mountain peak, a valley, or an acute trapezoid in it. All of which don't preserve betweenness. I doubt the argument will get better for arbitrary ordered sets.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 7:02
|
show 5 more comments
$begingroup$
Suppose that f is not monotonic. Then $exists x, k$ s.t. $x < k$ and $f(x) < f(k)$ and $exists x_0, k_0$ s.t. $x_0 < k_0$ and $f(x_0) > f(k_0)$.
WLOG, let $k < x_0$. Then we get a contradiction on $f$ preserving boundedness because either $f(k)$ or $f(x_0)$ will not lie between $f(x)$ and $f(k_0)$.
answered Dec 12 '18 at 6:13
T. FoT. Fo
466311
$endgroup$
$begingroup$
I think there's a typo here. But I can redraw the pictures that I think you had in mind. How do you eliminate the possibility that $f(k)=f(x_0)$? In this set of circumstances. But I also have an issue with the "without a loss of generality" part. How about the possibilities of $x<x_0<k<k_0$? or $x_0<x<k_0<k$?
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:21
$begingroup$
@RobertWolfe You're right, there is a small typo. It should say $f(x) < f(k)$ in the first line. Does that help?
$endgroup$
– T. Fo
Dec 12 '18 at 6:27
$begingroup$
@RobertWolfe The WLOG just allows us to consider the nice ordering $x < k < x_0 < k_0$. Other cases are similar: consider the two 'endpoints' and then one of the two points in between will give a contradiction. The proof above is not entirely rigorous, you're correct.
$endgroup$
– T. Fo
Dec 12 '18 at 6:30
$begingroup$
I think the last $f(k)$ should be $f(k_0)$. I'm running through cases.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:34
1
$begingroup$
I now believe this argument is essentially correct. Following @bof comment's, and employing a little metaphor, the graph of such a function will either have a mountain peak, a valley, or an acute trapezoid in it. All of which don't preserve betweenness. I doubt the argument will get better for arbitrary ordered sets.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 7:02
|
show 5 more comments
$begingroup$
Suppose that f is not monotonic. Then $exists x, k$ s.t. $x < k$ and $f(x) < f(k)$ and $exists x_0, k_0$ s.t. $x_0 < k_0$ and $f(x_0) > f(k_0)$.
WLOG, let $k < x_0$. Then we get a contradiction on $f$ preserving boundedness because either $f(k)$ or $f(x_0)$ will not lie between $f(x)$ and $f(k_0)$.
answered Dec 12 '18 at 6:13
T. FoT. Fo
466311
$endgroup$
Suppose that f is not monotonic. Then $exists x, k$ s.t. $x < k$ and $f(x) < f(k)$ and $exists x_0, k_0$ s.t. $x_0 < k_0$ and $f(x_0) > f(k_0)$.
WLOG, let $k < x_0$. Then we get a contradiction on $f$ preserving boundedness because either $f(k)$ or $f(x_0)$ will not lie between $f(x)$ and $f(k_0)$.
answered Dec 12 '18 at 6:13
T. FoT. Fo
466311
edited Dec 12 '18 at 6:35
answered Dec 12 '18 at 6:13
T. FoT. Fo
466311
answered Dec 12 '18 at 6:13
T. FoT. Fo
466311
answered Dec 12 '18 at 6:13
T. FoT. Fo
466311
466311
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I think there's a typo here. But I can redraw the pictures that I think you had in mind. How do you eliminate the possibility that $f(k)=f(x_0)$? In this set of circumstances. But I also have an issue with the "without a loss of generality" part. How about the possibilities of $x<x_0<k<k_0$? or $x_0<x<k_0<k$?
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– Robert Wolfe
Dec 12 '18 at 6:21
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@RobertWolfe You're right, there is a small typo. It should say $f(x) < f(k)$ in the first line. Does that help?
$endgroup$
– T. Fo
Dec 12 '18 at 6:27
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@RobertWolfe The WLOG just allows us to consider the nice ordering $x < k < x_0 < k_0$. Other cases are similar: consider the two 'endpoints' and then one of the two points in between will give a contradiction. The proof above is not entirely rigorous, you're correct.
$endgroup$
– T. Fo
Dec 12 '18 at 6:30
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I think the last $f(k)$ should be $f(k_0)$. I'm running through cases.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:34
1
$begingroup$
I now believe this argument is essentially correct. Following @bof comment's, and employing a little metaphor, the graph of such a function will either have a mountain peak, a valley, or an acute trapezoid in it. All of which don't preserve betweenness. I doubt the argument will get better for arbitrary ordered sets.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 7:02
|
show 5 more comments
$begingroup$
I think there's a typo here. But I can redraw the pictures that I think you had in mind. How do you eliminate the possibility that $f(k)=f(x_0)$? In this set of circumstances. But I also have an issue with the "without a loss of generality" part. How about the possibilities of $x<x_0<k<k_0$? or $x_0<x<k_0<k$?
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:21
$begingroup$
@RobertWolfe You're right, there is a small typo. It should say $f(x) < f(k)$ in the first line. Does that help?
$endgroup$
– T. Fo
Dec 12 '18 at 6:27
$begingroup$
@RobertWolfe The WLOG just allows us to consider the nice ordering $x < k < x_0 < k_0$. Other cases are similar: consider the two 'endpoints' and then one of the two points in between will give a contradiction. The proof above is not entirely rigorous, you're correct.
$endgroup$
– T. Fo
Dec 12 '18 at 6:30
$begingroup$
I think the last $f(k)$ should be $f(k_0)$. I'm running through cases.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:34
1
$begingroup$
I now believe this argument is essentially correct. Following @bof comment's, and employing a little metaphor, the graph of such a function will either have a mountain peak, a valley, or an acute trapezoid in it. All of which don't preserve betweenness. I doubt the argument will get better for arbitrary ordered sets.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 7:02
$begingroup$
I think there's a typo here. But I can redraw the pictures that I think you had in mind. How do you eliminate the possibility that $f(k)=f(x_0)$? In this set of circumstances. But I also have an issue with the "without a loss of generality" part. How about the possibilities of $x<x_0<k<k_0$? or $x_0<x<k_0<k$?
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:21
$begingroup$
I think there's a typo here. But I can redraw the pictures that I think you had in mind. How do you eliminate the possibility that $f(k)=f(x_0)$? In this set of circumstances. But I also have an issue with the "without a loss of generality" part. How about the possibilities of $x<x_0<k<k_0$? or $x_0<x<k_0<k$?
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:21
$begingroup$
@RobertWolfe You're right, there is a small typo. It should say $f(x) < f(k)$ in the first line. Does that help?
$endgroup$
– T. Fo
Dec 12 '18 at 6:27
$begingroup$
@RobertWolfe You're right, there is a small typo. It should say $f(x) < f(k)$ in the first line. Does that help?
$endgroup$
– T. Fo
Dec 12 '18 at 6:27
$begingroup$
@RobertWolfe The WLOG just allows us to consider the nice ordering $x < k < x_0 < k_0$. Other cases are similar: consider the two 'endpoints' and then one of the two points in between will give a contradiction. The proof above is not entirely rigorous, you're correct.
$endgroup$
– T. Fo
Dec 12 '18 at 6:30
$begingroup$
@RobertWolfe The WLOG just allows us to consider the nice ordering $x < k < x_0 < k_0$. Other cases are similar: consider the two 'endpoints' and then one of the two points in between will give a contradiction. The proof above is not entirely rigorous, you're correct.
$endgroup$
– T. Fo
Dec 12 '18 at 6:30
$begingroup$
I think the last $f(k)$ should be $f(k_0)$. I'm running through cases.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:34
$begingroup$
I think the last $f(k)$ should be $f(k_0)$. I'm running through cases.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 6:34
1
1
$begingroup$
I now believe this argument is essentially correct. Following @bof comment's, and employing a little metaphor, the graph of such a function will either have a mountain peak, a valley, or an acute trapezoid in it. All of which don't preserve betweenness. I doubt the argument will get better for arbitrary ordered sets.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 7:02
$begingroup$
I now believe this argument is essentially correct. Following @bof comment's, and employing a little metaphor, the graph of such a function will either have a mountain peak, a valley, or an acute trapezoid in it. All of which don't preserve betweenness. I doubt the argument will get better for arbitrary ordered sets.
$endgroup$
– Robert Wolfe
Dec 12 '18 at 7:02
|
show 5 more comments
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Note that this is false for partially ordered sets, where you can have incomparable $a,b$ both $<c$ and map this betweenness preservingly but not monotonically to $[3]={1<2<3}$ via $amapsto 1$, $bmapsto 3$, $cmapsto 2$.
$endgroup$
– Christoph
Dec 12 '18 at 8:57