Excess axioms in the definition of a metric
First, I am aware of how to show the following, my question concerns the reasoning of the standard metric definition.
Traditionally, the axioms defining a metric $rho:Xtimes X rightarrow mathbb{R}$ where $X$ is a nonempty set, are given as:
$forall x, yin X, rho(x, y) = 0 Rightarrow x = y$,
$forall x, yin X, rho(x, y) = rho(y, x)$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
$forall x, yin X, 0 leq rho(x, y)$.
It is easy to see that axiom 4 follows from axioms 1 and 3. It is also fairly trivial to show that axiom 2 follows from axioms 1 and 3. It then seems logical to only require the two axioms
$forall x, yin X, rho(x, y) = 0 Rightarrow x = y$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
as a definition for a metric. Is there a good reason that this is not the case, besides historical continuity?
Edit: Thanks to Paul Frost for mentioning my original definition was for pseudometrics... The question has been updated.
Edit 2: As commenters have asked for a proof:
Axioms 1, 3 imply axiom 2: First, consider the triangle inequality for general $x, y, zin X$. Then, $$rho(x, y) leq rho(z, x) + rho(z, y),$$ next, set $z = y$ to obtain $$rho(x, y) leq rho(y, x) + rho(y, y)Rightarrow rho(x, y) leq rho(y, x)$$ by axiom 1. Similarly, $$rho(y, x) leq rho(z, y) + rho(z, x),$$ and setting $z=x$ yields $$rho(y, x) leq rho(x, y) + rho(x, x)Rightarrow rho(y, x) leq rho(x, y)$$ so $rho(x, y) = rho(y, x)$.
We may then show Axioms 1, 2, and 3 imply 4. See this answer. It may be somewhat incorrect to say axioms 1 and 3 imply 4, but I don't see this as a problem, as 1 and 3 imply 2, and 1, 2, 3 imply 4. If I'm incorrect in that assumption, please let me know.
metric-spaces definition axioms
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show 8 more comments
First, I am aware of how to show the following, my question concerns the reasoning of the standard metric definition.
Traditionally, the axioms defining a metric $rho:Xtimes X rightarrow mathbb{R}$ where $X$ is a nonempty set, are given as:
$forall x, yin X, rho(x, y) = 0 Rightarrow x = y$,
$forall x, yin X, rho(x, y) = rho(y, x)$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
$forall x, yin X, 0 leq rho(x, y)$.
It is easy to see that axiom 4 follows from axioms 1 and 3. It is also fairly trivial to show that axiom 2 follows from axioms 1 and 3. It then seems logical to only require the two axioms
$forall x, yin X, rho(x, y) = 0 Rightarrow x = y$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
as a definition for a metric. Is there a good reason that this is not the case, besides historical continuity?
Edit: Thanks to Paul Frost for mentioning my original definition was for pseudometrics... The question has been updated.
Edit 2: As commenters have asked for a proof:
Axioms 1, 3 imply axiom 2: First, consider the triangle inequality for general $x, y, zin X$. Then, $$rho(x, y) leq rho(z, x) + rho(z, y),$$ next, set $z = y$ to obtain $$rho(x, y) leq rho(y, x) + rho(y, y)Rightarrow rho(x, y) leq rho(y, x)$$ by axiom 1. Similarly, $$rho(y, x) leq rho(z, y) + rho(z, x),$$ and setting $z=x$ yields $$rho(y, x) leq rho(x, y) + rho(x, x)Rightarrow rho(y, x) leq rho(x, y)$$ so $rho(x, y) = rho(y, x)$.
We may then show Axioms 1, 2, and 3 imply 4. See this answer. It may be somewhat incorrect to say axioms 1 and 3 imply 4, but I don't see this as a problem, as 1 and 3 imply 2, and 1, 2, 3 imply 4. If I'm incorrect in that assumption, please let me know.
metric-spaces definition axioms
asked 1 hour ago
hoverless
284
New contributor
hoverless is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
– Paul Frost
58 mins ago
You're right, let me edit that requirement. My mistake!
– hoverless
55 mins ago
1
You say that it is "easy" to see that axiom 4 follows from axioms 1 and 3. Perhaps you would like to share your proof.
– Xander Henderson
51 mins ago
Your statement Is not quite the usual triangle inequality, which is $forall x, y, zin X.rho(x, y) le rho(x, z) + rho(z, y)$. Whether that affects your proof that symmetry follows from other axioms is unclear, because you haven't given that proof. Please give the proof.
– Rob Arthan
50 mins ago
2
There are even very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generalyl make the axiomatisation harder to understand and harder to generalise.
– Rob Arthan
35 mins ago
|
show 8 more comments
First, I am aware of how to show the following, my question concerns the reasoning of the standard metric definition.
Traditionally, the axioms defining a metric $rho:Xtimes X rightarrow mathbb{R}$ where $X$ is a nonempty set, are given as:
$forall x, yin X, rho(x, y) = 0 Rightarrow x = y$,
$forall x, yin X, rho(x, y) = rho(y, x)$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
$forall x, yin X, 0 leq rho(x, y)$.
It is easy to see that axiom 4 follows from axioms 1 and 3. It is also fairly trivial to show that axiom 2 follows from axioms 1 and 3. It then seems logical to only require the two axioms
$forall x, yin X, rho(x, y) = 0 Rightarrow x = y$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
as a definition for a metric. Is there a good reason that this is not the case, besides historical continuity?
Edit: Thanks to Paul Frost for mentioning my original definition was for pseudometrics... The question has been updated.
Edit 2: As commenters have asked for a proof:
Axioms 1, 3 imply axiom 2: First, consider the triangle inequality for general $x, y, zin X$. Then, $$rho(x, y) leq rho(z, x) + rho(z, y),$$ next, set $z = y$ to obtain $$rho(x, y) leq rho(y, x) + rho(y, y)Rightarrow rho(x, y) leq rho(y, x)$$ by axiom 1. Similarly, $$rho(y, x) leq rho(z, y) + rho(z, x),$$ and setting $z=x$ yields $$rho(y, x) leq rho(x, y) + rho(x, x)Rightarrow rho(y, x) leq rho(x, y)$$ so $rho(x, y) = rho(y, x)$.
We may then show Axioms 1, 2, and 3 imply 4. See this answer. It may be somewhat incorrect to say axioms 1 and 3 imply 4, but I don't see this as a problem, as 1 and 3 imply 2, and 1, 2, 3 imply 4. If I'm incorrect in that assumption, please let me know.
metric-spaces definition axioms
asked 1 hour ago
hoverless
284
New contributor
hoverless is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
First, I am aware of how to show the following, my question concerns the reasoning of the standard metric definition.
Traditionally, the axioms defining a metric $rho:Xtimes X rightarrow mathbb{R}$ where $X$ is a nonempty set, are given as:
$forall x, yin X, rho(x, y) = 0 Rightarrow x = y$,
$forall x, yin X, rho(x, y) = rho(y, x)$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
$forall x, yin X, 0 leq rho(x, y)$.
It is easy to see that axiom 4 follows from axioms 1 and 3. It is also fairly trivial to show that axiom 2 follows from axioms 1 and 3. It then seems logical to only require the two axioms
$forall x, yin X, rho(x, y) = 0 Rightarrow x = y$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
as a definition for a metric. Is there a good reason that this is not the case, besides historical continuity?
Edit: Thanks to Paul Frost for mentioning my original definition was for pseudometrics... The question has been updated.
Edit 2: As commenters have asked for a proof:
Axioms 1, 3 imply axiom 2: First, consider the triangle inequality for general $x, y, zin X$. Then, $$rho(x, y) leq rho(z, x) + rho(z, y),$$ next, set $z = y$ to obtain $$rho(x, y) leq rho(y, x) + rho(y, y)Rightarrow rho(x, y) leq rho(y, x)$$ by axiom 1. Similarly, $$rho(y, x) leq rho(z, y) + rho(z, x),$$ and setting $z=x$ yields $$rho(y, x) leq rho(x, y) + rho(x, x)Rightarrow rho(y, x) leq rho(x, y)$$ so $rho(x, y) = rho(y, x)$.
We may then show Axioms 1, 2, and 3 imply 4. See this answer. It may be somewhat incorrect to say axioms 1 and 3 imply 4, but I don't see this as a problem, as 1 and 3 imply 2, and 1, 2, 3 imply 4. If I'm incorrect in that assumption, please let me know.
metric-spaces definition axioms
metric-spaces definition axioms
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asked 1 hour ago
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edited 38 mins ago
asked 1 hour ago
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asked 1 hour ago
hoverless
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asked 1 hour ago
hoverless
284
284
New contributor
hoverless is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
hoverless is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
hoverless is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
– Paul Frost
58 mins ago
You're right, let me edit that requirement. My mistake!
– hoverless
55 mins ago
1
You say that it is "easy" to see that axiom 4 follows from axioms 1 and 3. Perhaps you would like to share your proof.
– Xander Henderson
51 mins ago
Your statement Is not quite the usual triangle inequality, which is $forall x, y, zin X.rho(x, y) le rho(x, z) + rho(z, y)$. Whether that affects your proof that symmetry follows from other axioms is unclear, because you haven't given that proof. Please give the proof.
– Rob Arthan
50 mins ago
2
There are even very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generalyl make the axiomatisation harder to understand and harder to generalise.
– Rob Arthan
35 mins ago
|
show 8 more comments
1
You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
– Paul Frost
58 mins ago
You're right, let me edit that requirement. My mistake!
– hoverless
55 mins ago
1
You say that it is "easy" to see that axiom 4 follows from axioms 1 and 3. Perhaps you would like to share your proof.
– Xander Henderson
51 mins ago
Your statement Is not quite the usual triangle inequality, which is $forall x, y, zin X.rho(x, y) le rho(x, z) + rho(z, y)$. Whether that affects your proof that symmetry follows from other axioms is unclear, because you haven't given that proof. Please give the proof.
– Rob Arthan
50 mins ago
2
There are even very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generalyl make the axiomatisation harder to understand and harder to generalise.
– Rob Arthan
35 mins ago
1
1
You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
– Paul Frost
58 mins ago
You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
– Paul Frost
58 mins ago
You're right, let me edit that requirement. My mistake!
– hoverless
55 mins ago
You're right, let me edit that requirement. My mistake!
– hoverless
55 mins ago
1
1
You say that it is "easy" to see that axiom 4 follows from axioms 1 and 3. Perhaps you would like to share your proof.
– Xander Henderson
51 mins ago
You say that it is "easy" to see that axiom 4 follows from axioms 1 and 3. Perhaps you would like to share your proof.
– Xander Henderson
51 mins ago
Your statement Is not quite the usual triangle inequality, which is $forall x, y, zin X.rho(x, y) le rho(x, z) + rho(z, y)$. Whether that affects your proof that symmetry follows from other axioms is unclear, because you haven't given that proof. Please give the proof.
– Rob Arthan
50 mins ago
Your statement Is not quite the usual triangle inequality, which is $forall x, y, zin X.rho(x, y) le rho(x, z) + rho(z, y)$. Whether that affects your proof that symmetry follows from other axioms is unclear, because you haven't given that proof. Please give the proof.
– Rob Arthan
50 mins ago
2
2
There are even very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generalyl make the axiomatisation harder to understand and harder to generalise.
– Rob Arthan
35 mins ago
There are even very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generalyl make the axiomatisation harder to understand and harder to generalise.
– Rob Arthan
35 mins ago
|
show 8 more comments
1 Answer
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There are very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. See Lee Mosher's comment for much more interesting examples.
The answer to your question necessarily involves a matter of mathematical opinion. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generally make the axiomatisation harder to understand and harder to generalise.
answered 33 mins ago
Rob Arthan
29k42866
add a comment |
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There are very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. See Lee Mosher's comment for much more interesting examples.
The answer to your question necessarily involves a matter of mathematical opinion. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generally make the axiomatisation harder to understand and harder to generalise.
answered 33 mins ago
Rob Arthan
29k42866
add a comment |
There are very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. See Lee Mosher's comment for much more interesting examples.
The answer to your question necessarily involves a matter of mathematical opinion. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generally make the axiomatisation harder to understand and harder to generalise.
answered 33 mins ago
Rob Arthan
29k42866
add a comment |
There are very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. See Lee Mosher's comment for much more interesting examples.
The answer to your question necessarily involves a matter of mathematical opinion. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generally make the axiomatisation harder to understand and harder to generalise.
answered 33 mins ago
Rob Arthan
29k42866
There are very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. See Lee Mosher's comment for much more interesting examples.
The answer to your question necessarily involves a matter of mathematical opinion. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generally make the axiomatisation harder to understand and harder to generalise.
answered 33 mins ago
Rob Arthan
29k42866
edited 16 mins ago
answered 33 mins ago
Rob Arthan
29k42866
answered 33 mins ago
Rob Arthan
29k42866
answered 33 mins ago
Rob Arthan
29k42866
29k42866
add a comment |
add a comment |
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1
You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
– Paul Frost
58 mins ago
You're right, let me edit that requirement. My mistake!
– hoverless
55 mins ago
1
You say that it is "easy" to see that axiom 4 follows from axioms 1 and 3. Perhaps you would like to share your proof.
– Xander Henderson
51 mins ago
Your statement Is not quite the usual triangle inequality, which is $forall x, y, zin X.rho(x, y) le rho(x, z) + rho(z, y)$. Whether that affects your proof that symmetry follows from other axioms is unclear, because you haven't given that proof. Please give the proof.
– Rob Arthan
50 mins ago
2
There are even very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) mapsto xy^{-1}$) but this will generalyl make the axiomatisation harder to understand and harder to generalise.
– Rob Arthan
35 mins ago