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In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.

All distance-like functions seem to agree in being

• nonnegative: $d(x,y) \geq 0$
• symmetric: $d(x,y) = d(y,x)$
• reflexive: $d(x,x)=0$

– let me call such functions proto-metrics – but differ in other respects:

• a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.

• an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$

• a dissimilarity function – conceived as some kind of distance, e.g. in a property space – is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality at all.

One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing – without any restriction – two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other.

Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:

Which conditions does a function $F$ have to fulfill such that

$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$

can be seen as a reasonable triangle inequality, thus making the proto-metric $d$ distance-like?

Some conditions on $F$ seem to spring to mind:

• symmetry: $F(x,y)=F(y,x)$
• monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$
• scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$
• associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?)
• $F(x,0)=x$ (Thanks to David Feldman)

Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?

In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.

All distance-like functions seem to agree in being

• nonnegative: $d(x,y) \geq 0$
• symmetric: $d(x,y) = d(y,x)$
• reflexive: $d(x,x)=0$

– let me call such functions proto-metrics – but differ in other respects:

• a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.

• an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$

• a dissimilarity function – conceived as some kind of distance, e.g. in a property space – is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality at all.

One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing – without any restriction – two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other.

Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:

Which conditions does a function $F$ have to fulfill such that

$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$

can be seen as a reasonable triangle inequality, thus making the proto-metric $d$ distance-like?

Some conditions on $F$ seem to spring to mind:

• symmetry: $F(x,y)=F(y,x)$
• monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$
• scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$
• associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?)
• $F(x,0)=x$ (Thanks to David Feldman)

Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?

In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.

All distance-like functions seem to agree in being

• nonnegative: $d(x,y) \geq 0$
• symmetric: $d(x,y) = d(y,x)$
• fulfilling $d(x,x)=0$

– let me call such functions proto-metrics – but differ in other respects:

• a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.

• an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$

• a dissimilarity function – conceived as some kind of distance, e.g. in a property space – is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality at all.

One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing – without any restriction – two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other.

Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:

Which conditions does a function $F$ have to fulfill such that

$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$

can be seen as a reasonable triangle inequality, thus making the proto-metric $d$ distance-like?

Some conditions on $F$ seem to spring to mind:

• symmetry: $F(x,y)=F(y,x)$
• monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$
• scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$
• associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?)
• $F(x,0)=x$ (Thanks to David Feldman)

Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?

In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.

All distance-like functions seem to agree in being

• nonnegative: $d(x,y) \geq 0$
• symmetric: $d(x,y) = d(y,x)$
• reflexive: $d(x,x)=0$

– let me call such functions proto-metrics – but differ in other respects:

• a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.

• an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$

• a dissimilarity function – conceived as some kind of distance, e.g. in a property space – is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality at all.

One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing – without any restriction – two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other.

Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:

Which conditions does a function $F$ have to fulfill such that

$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$

can be seen as a reasonable triangle inequality, thus making the proto-metric $d$ distance-like?

Some conditions on $F$ seem to spring to mind:

• symmetry: $F(x,y)=F(y,x)$
• monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$
• scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$
• associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?)
• $F(x,0)=x$ (Thanks to David Feldman)

Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?

In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.

All distance-like functions seem to agree in being

• nonnegative: $d(x,y) \geq 0$
• symmetric: $d(x,y) = d(y,x)$
• fulfilling $d(x,x)=0$

— let me call such functions proto-metrics — but differ in other respects:

• a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.

• an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$

• a dissimilarity function — conceived as some kind of distance, e.g. in a property space — is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality at all.

One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other.

Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:

Which conditions does a function $F$ have to fulfill such that

$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$

can be seen as a reasonable triangle inequality, thus making the proto-metric $d$ distance-like?

Some conditions on $F$ seem to spring to mind:

• symmetry: $F(x,y)=F(y,x)$
• monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$
• scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$
• associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?)
• $F(x,0)=x$ (Thanks to David Feldman)

Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?

In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.

All distance-like functions seem to agree in being

• nonnegative: $d(x,y) \geq 0$
• symmetric: $d(x,y) = d(y,x)$
• fulfilling $d(x,x)=0$

– let me call such functions proto-metrics – but differ in other respects:

• a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.

• an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$

• a dissimilarity function – conceived as some kind of distance, e.g. in a property space – is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality at all.

One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing – without any restriction – two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other.

Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:

Which conditions does a function $F$ have to fulfill such that

$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$

can be seen as a reasonable triangle inequality, thus making the proto-metric $d$ distance-like?

Some conditions on $F$ seem to spring to mind:

• symmetry: $F(x,y)=F(y,x)$
• monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$
• scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$
• associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?)
• $F(x,0)=x$ (Thanks to David Feldman)

Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?