Sufficient and necessary binary relation for given metric
$begingroup$
Problem:
Let $p$ be binary relation on the set $X$, for any $x,y$ $in$ $X$ let:
$$ d(x,y) = begin{cases}
3, & text{for } hspace{0.5cm}xpy \
0, & text{for} hspace{0.5cm} neg xpy
end{cases}
$$
Determine necessary and sufficient condition(characteristic of relation $p$) so that $d$ is metric on the set $X$.
Necessary conditions for metric:
$$d(x,y)=0 Leftrightarrow x=y$$
$$d(x,y)= d(x,y)=0 $$
$$d(x,z)leq d(x,y) + d(y,z)$$
Possible solution:
The binary relation must be symmetric, because of the second condition for metric, but I do not know whether that is necessary and sufficient. It seems to me that the relation $p$ is non-equality, is it possible? Any help is appreciated.
metric-spaces relations
edited Jan 11 at 14:14
Samo Poláček
31
asked Jan 10 at 16:25
nocturnenocturne
689
$endgroup$
add a comment |
$begingroup$
Problem:
Let $p$ be binary relation on the set $X$, for any $x,y$ $in$ $X$ let:
$$ d(x,y) = begin{cases}
3, & text{for } hspace{0.5cm}xpy \
0, & text{for} hspace{0.5cm} neg xpy
end{cases}
$$
Determine necessary and sufficient condition(characteristic of relation $p$) so that $d$ is metric on the set $X$.
Necessary conditions for metric:
$$d(x,y)=0 Leftrightarrow x=y$$
$$d(x,y)= d(x,y)=0 $$
$$d(x,z)leq d(x,y) + d(y,z)$$
Possible solution:
The binary relation must be symmetric, because of the second condition for metric, but I do not know whether that is necessary and sufficient. It seems to me that the relation $p$ is non-equality, is it possible? Any help is appreciated.
metric-spaces relations
edited Jan 11 at 14:14
Samo Poláček
31
asked Jan 10 at 16:25
nocturnenocturne
689
$endgroup$
$begingroup$
Why do you want me to do that?
$endgroup$
– José Carlos Santos
Jan 10 at 16:26
1
$begingroup$
Hint: show $x$ is related to $y$ if and only if $xne y$.
$endgroup$
– Gerry Myerson
Jan 11 at 15:19
$begingroup$
That might be a right solution. Thank you.
$endgroup$
– nocturne
Jan 11 at 16:25
add a comment |
$begingroup$
Problem:
Let $p$ be binary relation on the set $X$, for any $x,y$ $in$ $X$ let:
$$ d(x,y) = begin{cases}
3, & text{for } hspace{0.5cm}xpy \
0, & text{for} hspace{0.5cm} neg xpy
end{cases}
$$
Determine necessary and sufficient condition(characteristic of relation $p$) so that $d$ is metric on the set $X$.
Necessary conditions for metric:
$$d(x,y)=0 Leftrightarrow x=y$$
$$d(x,y)= d(x,y)=0 $$
$$d(x,z)leq d(x,y) + d(y,z)$$
Possible solution:
The binary relation must be symmetric, because of the second condition for metric, but I do not know whether that is necessary and sufficient. It seems to me that the relation $p$ is non-equality, is it possible? Any help is appreciated.
metric-spaces relations
edited Jan 11 at 14:14
Samo Poláček
31
asked Jan 10 at 16:25
nocturnenocturne
689
$endgroup$
Problem:
Let $p$ be binary relation on the set $X$, for any $x,y$ $in$ $X$ let:
$$ d(x,y) = begin{cases}
3, & text{for } hspace{0.5cm}xpy \
0, & text{for} hspace{0.5cm} neg xpy
end{cases}
$$
Determine necessary and sufficient condition(characteristic of relation $p$) so that $d$ is metric on the set $X$.
Necessary conditions for metric:
$$d(x,y)=0 Leftrightarrow x=y$$
$$d(x,y)= d(x,y)=0 $$
$$d(x,z)leq d(x,y) + d(y,z)$$
Possible solution:
The binary relation must be symmetric, because of the second condition for metric, but I do not know whether that is necessary and sufficient. It seems to me that the relation $p$ is non-equality, is it possible? Any help is appreciated.
metric-spaces relations
metric-spaces relations
edited Jan 11 at 14:14
Samo Poláček
31
asked Jan 10 at 16:25
nocturnenocturne
689
edited Jan 11 at 14:14
Samo Poláček
31
asked Jan 10 at 16:25
nocturnenocturne
689
edited Jan 11 at 14:14
Samo Poláček
31
edited Jan 11 at 14:14
Samo Poláček
31
edited Jan 11 at 14:14
Samo Poláček
31
31
asked Jan 10 at 16:25
nocturnenocturne
689
asked Jan 10 at 16:25
nocturnenocturne
689
asked Jan 10 at 16:25
nocturnenocturne
689
689
$begingroup$
Why do you want me to do that?
$endgroup$
– José Carlos Santos
Jan 10 at 16:26
1
$begingroup$
Hint: show $x$ is related to $y$ if and only if $xne y$.
$endgroup$
– Gerry Myerson
Jan 11 at 15:19
$begingroup$
That might be a right solution. Thank you.
$endgroup$
– nocturne
Jan 11 at 16:25
add a comment |
$begingroup$
Why do you want me to do that?
$endgroup$
– José Carlos Santos
Jan 10 at 16:26
1
$begingroup$
Hint: show $x$ is related to $y$ if and only if $xne y$.
$endgroup$
– Gerry Myerson
Jan 11 at 15:19
$begingroup$
That might be a right solution. Thank you.
$endgroup$
– nocturne
Jan 11 at 16:25
$begingroup$
Why do you want me to do that?
$endgroup$
– José Carlos Santos
Jan 10 at 16:26
$begingroup$
Why do you want me to do that?
$endgroup$
– José Carlos Santos
Jan 10 at 16:26
1
1
$begingroup$
Hint: show $x$ is related to $y$ if and only if $xne y$.
$endgroup$
– Gerry Myerson
Jan 11 at 15:19
$begingroup$
Hint: show $x$ is related to $y$ if and only if $xne y$.
$endgroup$
– Gerry Myerson
Jan 11 at 15:19
$begingroup$
That might be a right solution. Thank you.
$endgroup$
– nocturne
Jan 11 at 16:25
$begingroup$
That might be a right solution. Thank you.
$endgroup$
– nocturne
Jan 11 at 16:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$d$ is a metric if and only if $p$ is equal to the relation $ne$, or $$(x,y) in p iff x ne y$$
Assume that $p$ is equal to $ne$ and let's check that $d$ is a metric.
- Clearly $d ge 0$. For $x in X$ we have $x = x$ so $(x,x) notin p$ which implies $d(x,x) = 0$.
Assume $d(x,y) = 0$ for some $x,y in X$. Then $(x,y) notin p$ so $x = y$. - We have $$d(x,y) = 3 iff (x,y) in p iff x ne y iff y ne x iff (y,x) in piff d(y,x) = 3$$
- Since $d$ attains only the values $0,3$ we only have to check that it isn't possible that $d(x,z) = 3$ but $d(x,y) = d(y,z) = 0$. This would be equivalent to $x ne z$ and $x = y = z$, which is a contradiction.
Conversely, if $p$ isn't equal to $ne$, then either
$p$ doesn't contain $ne$, so there exist $x,y in X, x ne y$ such that $(x,y) notin p$. It follows $d(x,y) = 0$ but $x ne y$ so $d$ isn't a metric.
$ne$ doesn't contain $p$ so there exists $x in X$ such that $(x,x) in p$. This implies $d(x,x) = 3$ so $d$ isn't a metric.
answered Jan 11 at 16:58
mechanodroidmechanodroid
28k62447
$endgroup$
$begingroup$
Are you sure that this is the right solution? I mean it is looking good, but I got confused by that notation $neg xpy$, is it equal to the notation $x neg p y$?
$endgroup$
– nocturne
Jan 11 at 17:56
1
$begingroup$
@nocturne Yes, more precisely it is $xp^cy$ where $p^c = Xtimes X setminus p$. $$neg xpy iff neg big[(x,y) in pbig] iff (x,y) notin p iff (x,y) in p^c iff xp^cy$$
$endgroup$
– mechanodroid
Jan 11 at 19:58
add a comment |
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$begingroup$
$d$ is a metric if and only if $p$ is equal to the relation $ne$, or $$(x,y) in p iff x ne y$$
Assume that $p$ is equal to $ne$ and let's check that $d$ is a metric.
- Clearly $d ge 0$. For $x in X$ we have $x = x$ so $(x,x) notin p$ which implies $d(x,x) = 0$.
Assume $d(x,y) = 0$ for some $x,y in X$. Then $(x,y) notin p$ so $x = y$. - We have $$d(x,y) = 3 iff (x,y) in p iff x ne y iff y ne x iff (y,x) in piff d(y,x) = 3$$
- Since $d$ attains only the values $0,3$ we only have to check that it isn't possible that $d(x,z) = 3$ but $d(x,y) = d(y,z) = 0$. This would be equivalent to $x ne z$ and $x = y = z$, which is a contradiction.
Conversely, if $p$ isn't equal to $ne$, then either
$p$ doesn't contain $ne$, so there exist $x,y in X, x ne y$ such that $(x,y) notin p$. It follows $d(x,y) = 0$ but $x ne y$ so $d$ isn't a metric.
$ne$ doesn't contain $p$ so there exists $x in X$ such that $(x,x) in p$. This implies $d(x,x) = 3$ so $d$ isn't a metric.
answered Jan 11 at 16:58
mechanodroidmechanodroid
28k62447
$endgroup$
$begingroup$
Are you sure that this is the right solution? I mean it is looking good, but I got confused by that notation $neg xpy$, is it equal to the notation $x neg p y$?
$endgroup$
– nocturne
Jan 11 at 17:56
1
$begingroup$
@nocturne Yes, more precisely it is $xp^cy$ where $p^c = Xtimes X setminus p$. $$neg xpy iff neg big[(x,y) in pbig] iff (x,y) notin p iff (x,y) in p^c iff xp^cy$$
$endgroup$
– mechanodroid
Jan 11 at 19:58
add a comment |
$begingroup$
$d$ is a metric if and only if $p$ is equal to the relation $ne$, or $$(x,y) in p iff x ne y$$
Assume that $p$ is equal to $ne$ and let's check that $d$ is a metric.
- Clearly $d ge 0$. For $x in X$ we have $x = x$ so $(x,x) notin p$ which implies $d(x,x) = 0$.
Assume $d(x,y) = 0$ for some $x,y in X$. Then $(x,y) notin p$ so $x = y$. - We have $$d(x,y) = 3 iff (x,y) in p iff x ne y iff y ne x iff (y,x) in piff d(y,x) = 3$$
- Since $d$ attains only the values $0,3$ we only have to check that it isn't possible that $d(x,z) = 3$ but $d(x,y) = d(y,z) = 0$. This would be equivalent to $x ne z$ and $x = y = z$, which is a contradiction.
Conversely, if $p$ isn't equal to $ne$, then either
$p$ doesn't contain $ne$, so there exist $x,y in X, x ne y$ such that $(x,y) notin p$. It follows $d(x,y) = 0$ but $x ne y$ so $d$ isn't a metric.
$ne$ doesn't contain $p$ so there exists $x in X$ such that $(x,x) in p$. This implies $d(x,x) = 3$ so $d$ isn't a metric.
answered Jan 11 at 16:58
mechanodroidmechanodroid
28k62447
$endgroup$
$begingroup$
Are you sure that this is the right solution? I mean it is looking good, but I got confused by that notation $neg xpy$, is it equal to the notation $x neg p y$?
$endgroup$
– nocturne
Jan 11 at 17:56
1
$begingroup$
@nocturne Yes, more precisely it is $xp^cy$ where $p^c = Xtimes X setminus p$. $$neg xpy iff neg big[(x,y) in pbig] iff (x,y) notin p iff (x,y) in p^c iff xp^cy$$
$endgroup$
– mechanodroid
Jan 11 at 19:58
add a comment |
$begingroup$
$d$ is a metric if and only if $p$ is equal to the relation $ne$, or $$(x,y) in p iff x ne y$$
Assume that $p$ is equal to $ne$ and let's check that $d$ is a metric.
- Clearly $d ge 0$. For $x in X$ we have $x = x$ so $(x,x) notin p$ which implies $d(x,x) = 0$.
Assume $d(x,y) = 0$ for some $x,y in X$. Then $(x,y) notin p$ so $x = y$. - We have $$d(x,y) = 3 iff (x,y) in p iff x ne y iff y ne x iff (y,x) in piff d(y,x) = 3$$
- Since $d$ attains only the values $0,3$ we only have to check that it isn't possible that $d(x,z) = 3$ but $d(x,y) = d(y,z) = 0$. This would be equivalent to $x ne z$ and $x = y = z$, which is a contradiction.
Conversely, if $p$ isn't equal to $ne$, then either
$p$ doesn't contain $ne$, so there exist $x,y in X, x ne y$ such that $(x,y) notin p$. It follows $d(x,y) = 0$ but $x ne y$ so $d$ isn't a metric.
$ne$ doesn't contain $p$ so there exists $x in X$ such that $(x,x) in p$. This implies $d(x,x) = 3$ so $d$ isn't a metric.
answered Jan 11 at 16:58
mechanodroidmechanodroid
28k62447
$endgroup$
$d$ is a metric if and only if $p$ is equal to the relation $ne$, or $$(x,y) in p iff x ne y$$
Assume that $p$ is equal to $ne$ and let's check that $d$ is a metric.
- Clearly $d ge 0$. For $x in X$ we have $x = x$ so $(x,x) notin p$ which implies $d(x,x) = 0$.
Assume $d(x,y) = 0$ for some $x,y in X$. Then $(x,y) notin p$ so $x = y$. - We have $$d(x,y) = 3 iff (x,y) in p iff x ne y iff y ne x iff (y,x) in piff d(y,x) = 3$$
- Since $d$ attains only the values $0,3$ we only have to check that it isn't possible that $d(x,z) = 3$ but $d(x,y) = d(y,z) = 0$. This would be equivalent to $x ne z$ and $x = y = z$, which is a contradiction.
Conversely, if $p$ isn't equal to $ne$, then either
$p$ doesn't contain $ne$, so there exist $x,y in X, x ne y$ such that $(x,y) notin p$. It follows $d(x,y) = 0$ but $x ne y$ so $d$ isn't a metric.
$ne$ doesn't contain $p$ so there exists $x in X$ such that $(x,x) in p$. This implies $d(x,x) = 3$ so $d$ isn't a metric.
answered Jan 11 at 16:58
mechanodroidmechanodroid
28k62447
answered Jan 11 at 16:58
mechanodroidmechanodroid
28k62447
answered Jan 11 at 16:58
mechanodroidmechanodroid
28k62447
answered Jan 11 at 16:58
mechanodroidmechanodroid
28k62447
28k62447
$begingroup$
Are you sure that this is the right solution? I mean it is looking good, but I got confused by that notation $neg xpy$, is it equal to the notation $x neg p y$?
$endgroup$
– nocturne
Jan 11 at 17:56
1
$begingroup$
@nocturne Yes, more precisely it is $xp^cy$ where $p^c = Xtimes X setminus p$. $$neg xpy iff neg big[(x,y) in pbig] iff (x,y) notin p iff (x,y) in p^c iff xp^cy$$
$endgroup$
– mechanodroid
Jan 11 at 19:58
add a comment |
$begingroup$
Are you sure that this is the right solution? I mean it is looking good, but I got confused by that notation $neg xpy$, is it equal to the notation $x neg p y$?
$endgroup$
– nocturne
Jan 11 at 17:56
1
$begingroup$
@nocturne Yes, more precisely it is $xp^cy$ where $p^c = Xtimes X setminus p$. $$neg xpy iff neg big[(x,y) in pbig] iff (x,y) notin p iff (x,y) in p^c iff xp^cy$$
$endgroup$
– mechanodroid
Jan 11 at 19:58
$begingroup$
Are you sure that this is the right solution? I mean it is looking good, but I got confused by that notation $neg xpy$, is it equal to the notation $x neg p y$?
$endgroup$
– nocturne
Jan 11 at 17:56
$begingroup$
Are you sure that this is the right solution? I mean it is looking good, but I got confused by that notation $neg xpy$, is it equal to the notation $x neg p y$?
$endgroup$
– nocturne
Jan 11 at 17:56
1
1
$begingroup$
@nocturne Yes, more precisely it is $xp^cy$ where $p^c = Xtimes X setminus p$. $$neg xpy iff neg big[(x,y) in pbig] iff (x,y) notin p iff (x,y) in p^c iff xp^cy$$
$endgroup$
– mechanodroid
Jan 11 at 19:58
$begingroup$
@nocturne Yes, more precisely it is $xp^cy$ where $p^c = Xtimes X setminus p$. $$neg xpy iff neg big[(x,y) in pbig] iff (x,y) notin p iff (x,y) in p^c iff xp^cy$$
$endgroup$
– mechanodroid
Jan 11 at 19:58
add a comment |
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$begingroup$
Why do you want me to do that?
$endgroup$
– José Carlos Santos
Jan 10 at 16:26
1
$begingroup$
Hint: show $x$ is related to $y$ if and only if $xne y$.
$endgroup$
– Gerry Myerson
Jan 11 at 15:19
$begingroup$
That might be a right solution. Thank you.
$endgroup$
– nocturne
Jan 11 at 16:25