Excess axioms in the definition of a metric
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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:
$forall x, yin X, rho(x, y) = 0 iff x = y$,
$forall x, yin X, rho(x, y) = rho(y, x)$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
$forall x, yin X, 0 leq rho(x, y)$.
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
$forall x, yin X, rho(x, y) = 0 iff x = y$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
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 Dec 23 '18 at 18:28
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334
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show 9 more comments
$begingroup$
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:
$forall x, yin X, rho(x, y) = 0 iff x = y$,
$forall x, yin X, rho(x, y) = rho(y, x)$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
$forall x, yin X, 0 leq rho(x, y)$.
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
$forall x, yin X, rho(x, y) = 0 iff x = y$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
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 Dec 23 '18 at 18:28
hoverlesshoverless
334
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1
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You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
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– Paul Frost
Dec 23 '18 at 18:33
1
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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.
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– Xander Henderson
Dec 23 '18 at 18:39
1
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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.
$endgroup$
– Rob Arthan
Dec 23 '18 at 18:41
1
$begingroup$
Your way of stating the triangle inequality is not standard. The standard definition maintains the "alphabetical order" of $x,y,z$: $forall x, y, zin X, rho(x, z)leq rho(x,y) + rho(y,z)$. This standard version is used for interesting examples of "asymmetric metrics".
$endgroup$
– Lee Mosher
Dec 23 '18 at 18:45
2
$begingroup$
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.
$endgroup$
– Rob Arthan
Dec 23 '18 at 18:56
|
show 9 more comments
$begingroup$
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:
$forall x, yin X, rho(x, y) = 0 iff x = y$,
$forall x, yin X, rho(x, y) = rho(y, x)$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
$forall x, yin X, 0 leq rho(x, y)$.
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
$forall x, yin X, rho(x, y) = 0 iff x = y$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
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 Dec 23 '18 at 18:28
hoverlesshoverless
334
$endgroup$
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:
$forall x, yin X, rho(x, y) = 0 iff x = y$,
$forall x, yin X, rho(x, y) = rho(y, x)$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
$forall x, yin X, 0 leq rho(x, y)$.
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
$forall x, yin X, rho(x, y) = 0 iff x = y$,
$forall x, y, zin X, rho(x, y)leq rho(z, x) + rho(z, y)$,
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
asked Dec 23 '18 at 18:28
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asked Dec 23 '18 at 18:28
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edited Dec 23 '18 at 21:49
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asked Dec 23 '18 at 18:28
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asked Dec 23 '18 at 18:28
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asked Dec 23 '18 at 18:28
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You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
$endgroup$
– Paul Frost
Dec 23 '18 at 18:33
1
$begingroup$
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.
$endgroup$
– Xander Henderson
Dec 23 '18 at 18:39
1
$begingroup$
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.
$endgroup$
– Rob Arthan
Dec 23 '18 at 18:41
1
$begingroup$
Your way of stating the triangle inequality is not standard. The standard definition maintains the "alphabetical order" of $x,y,z$: $forall x, y, zin X, rho(x, z)leq rho(x,y) + rho(y,z)$. This standard version is used for interesting examples of "asymmetric metrics".
$endgroup$
– Lee Mosher
Dec 23 '18 at 18:45
2
$begingroup$
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.
$endgroup$
– Rob Arthan
Dec 23 '18 at 18:56
|
show 9 more comments
1
$begingroup$
You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
$endgroup$
– Paul Frost
Dec 23 '18 at 18:33
1
$begingroup$
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.
$endgroup$
– Xander Henderson
Dec 23 '18 at 18:39
1
$begingroup$
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.
$endgroup$
– Rob Arthan
Dec 23 '18 at 18:41
1
$begingroup$
Your way of stating the triangle inequality is not standard. The standard definition maintains the "alphabetical order" of $x,y,z$: $forall x, y, zin X, rho(x, z)leq rho(x,y) + rho(y,z)$. This standard version is used for interesting examples of "asymmetric metrics".
$endgroup$
– Lee Mosher
Dec 23 '18 at 18:45
2
$begingroup$
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.
$endgroup$
– Rob Arthan
Dec 23 '18 at 18:56
1
1
$begingroup$
You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
$endgroup$
– Paul Frost
Dec 23 '18 at 18:33
$begingroup$
You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
$endgroup$
– Paul Frost
Dec 23 '18 at 18:33
1
1
$begingroup$
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.
$endgroup$
– Xander Henderson
Dec 23 '18 at 18:39
$begingroup$
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.
$endgroup$
– Xander Henderson
Dec 23 '18 at 18:39
1
1
$begingroup$
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.
$endgroup$
– Rob Arthan
Dec 23 '18 at 18:41
$begingroup$
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.
$endgroup$
– Rob Arthan
Dec 23 '18 at 18:41
1
1
$begingroup$
Your way of stating the triangle inequality is not standard. The standard definition maintains the "alphabetical order" of $x,y,z$: $forall x, y, zin X, rho(x, z)leq rho(x,y) + rho(y,z)$. This standard version is used for interesting examples of "asymmetric metrics".
$endgroup$
– Lee Mosher
Dec 23 '18 at 18:45
$begingroup$
Your way of stating the triangle inequality is not standard. The standard definition maintains the "alphabetical order" of $x,y,z$: $forall x, y, zin X, rho(x, z)leq rho(x,y) + rho(y,z)$. This standard version is used for interesting examples of "asymmetric metrics".
$endgroup$
– Lee Mosher
Dec 23 '18 at 18:45
2
2
$begingroup$
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.
$endgroup$
– Rob Arthan
Dec 23 '18 at 18:56
$begingroup$
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.
$endgroup$
– Rob Arthan
Dec 23 '18 at 18:56
|
show 9 more comments
1 Answer
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Your statement of the triangle inequality is not quite the same as the usual one, which says $d(x, y) le d(x, z) + d(z, y)$. The usual definition supports the intuition that $d(x, y)$ measures the time (or cost or work) of getting from $x$ to $y$ with no assumption that getting from $x$ to $y$ involves the same amount of time (or ...) as the return journey.
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 many 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 Dec 23 '18 at 18:58
Rob ArthanRob Arthan
29.6k42967
$endgroup$
$begingroup$
You might want to also include your earlier comment above (about the OP's axiom 3 not being the usual form of the triangle inequality) explicitly in this answer. Without that context, your answer seems a bit disconnected from the OP's question.
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– Ilmari Karonen
Dec 23 '18 at 22:04
add a comment |
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$begingroup$
Your statement of the triangle inequality is not quite the same as the usual one, which says $d(x, y) le d(x, z) + d(z, y)$. The usual definition supports the intuition that $d(x, y)$ measures the time (or cost or work) of getting from $x$ to $y$ with no assumption that getting from $x$ to $y$ involves the same amount of time (or ...) as the return journey.
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 many 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 Dec 23 '18 at 18:58
Rob ArthanRob Arthan
29.6k42967
$endgroup$
$begingroup$
You might want to also include your earlier comment above (about the OP's axiom 3 not being the usual form of the triangle inequality) explicitly in this answer. Without that context, your answer seems a bit disconnected from the OP's question.
$endgroup$
– Ilmari Karonen
Dec 23 '18 at 22:04
add a comment |
$begingroup$
Your statement of the triangle inequality is not quite the same as the usual one, which says $d(x, y) le d(x, z) + d(z, y)$. The usual definition supports the intuition that $d(x, y)$ measures the time (or cost or work) of getting from $x$ to $y$ with no assumption that getting from $x$ to $y$ involves the same amount of time (or ...) as the return journey.
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 many 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 Dec 23 '18 at 18:58
Rob ArthanRob Arthan
29.6k42967
$endgroup$
$begingroup$
You might want to also include your earlier comment above (about the OP's axiom 3 not being the usual form of the triangle inequality) explicitly in this answer. Without that context, your answer seems a bit disconnected from the OP's question.
$endgroup$
– Ilmari Karonen
Dec 23 '18 at 22:04
add a comment |
$begingroup$
Your statement of the triangle inequality is not quite the same as the usual one, which says $d(x, y) le d(x, z) + d(z, y)$. The usual definition supports the intuition that $d(x, y)$ measures the time (or cost or work) of getting from $x$ to $y$ with no assumption that getting from $x$ to $y$ involves the same amount of time (or ...) as the return journey.
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 many 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 Dec 23 '18 at 18:58
Rob ArthanRob Arthan
29.6k42967
$endgroup$
Your statement of the triangle inequality is not quite the same as the usual one, which says $d(x, y) le d(x, z) + d(z, y)$. The usual definition supports the intuition that $d(x, y)$ measures the time (or cost or work) of getting from $x$ to $y$ with no assumption that getting from $x$ to $y$ involves the same amount of time (or ...) as the return journey.
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 many 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 Dec 23 '18 at 18:58
Rob ArthanRob Arthan
29.6k42967
edited Dec 23 '18 at 23:14
answered Dec 23 '18 at 18:58
Rob ArthanRob Arthan
29.6k42967
answered Dec 23 '18 at 18:58
Rob ArthanRob Arthan
29.6k42967
answered Dec 23 '18 at 18:58
Rob ArthanRob Arthan
29.6k42967
29.6k42967
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You might want to also include your earlier comment above (about the OP's axiom 3 not being the usual form of the triangle inequality) explicitly in this answer. Without that context, your answer seems a bit disconnected from the OP's question.
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– Ilmari Karonen
Dec 23 '18 at 22:04
add a comment |
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You might want to also include your earlier comment above (about the OP's axiom 3 not being the usual form of the triangle inequality) explicitly in this answer. Without that context, your answer seems a bit disconnected from the OP's question.
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– Ilmari Karonen
Dec 23 '18 at 22:04
$begingroup$
You might want to also include your earlier comment above (about the OP's axiom 3 not being the usual form of the triangle inequality) explicitly in this answer. Without that context, your answer seems a bit disconnected from the OP's question.
$endgroup$
– Ilmari Karonen
Dec 23 '18 at 22:04
$begingroup$
You might want to also include your earlier comment above (about the OP's axiom 3 not being the usual form of the triangle inequality) explicitly in this answer. Without that context, your answer seems a bit disconnected from the OP's question.
$endgroup$
– Ilmari Karonen
Dec 23 '18 at 22:04
add a comment |
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1
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You consider pseudometrics because you do not require $rho(x,y) = 0 Rightarrow x = y$.
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– Paul Frost
Dec 23 '18 at 18:33
1
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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.
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– Xander Henderson
Dec 23 '18 at 18:39
1
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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.
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– Rob Arthan
Dec 23 '18 at 18:41
1
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Your way of stating the triangle inequality is not standard. The standard definition maintains the "alphabetical order" of $x,y,z$: $forall x, y, zin X, rho(x, z)leq rho(x,y) + rho(y,z)$. This standard version is used for interesting examples of "asymmetric metrics".
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– Lee Mosher
Dec 23 '18 at 18:45
2
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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.
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– Rob Arthan
Dec 23 '18 at 18:56