Back to the [metric space comment](https://forum.azimuthproject.org/discussion/comment/18435/#Comment_18435) .

A metric space consists of

1) a set \\(X\\) the elements of which are called points and

2) a metric function \\( d : X \times X \rightarrow \mathbb{R}_{\ge 0} \\), where \\( d(x,y) \\) is the metric, a distance between x and y

and the rules

a) \\( \forall x | x \in X \\), we have \\( d(x,x) = 0 \\)

b) \\( \forall x, y | x, y \in X \\), if \\( d(x,y) = 0 \\) then \\( x = y \\)

c) \\( \forall x, y | x, y \in X \\), we have \\( d(x,y) = d(y,x) \\)

d) \\( \forall x, y, z | x, y, z \in X \\), we have \\( d(x,y) + d(y,z) \ge d(x,z) \\)

Is **Cost**-category [I want to call it a **Cost**-set] a metric space?

We have a symmetric monoidal preorder with which to embelish.

$$ \tag{cost} \textbf{Cost} := ( [0,\infty], \ge, 0, +) $$

And a \\( \textbf{Cost} \\)\-category

$$ \tag{cost-set} \mathcal{X} := (\textbf{X}, \textbf{Cost}) $$

With the rules attendant on monoidal categories and \\( \mathcal{V}\text{-category} \\).

i) \\( 0 \ge d(x,x) \\)

j) \\( d(x,y) + d(y,z) \ge d(x,z) \\)

We have

1) a set \\( \textbf{X} := Ob(\mathcal{X}) \\) and

2) a metric function \\( + := d \\) from \\( \textbf{Cost} \\).

$$ \forall x,y \in \textbf{X} , d(x,y) := x + y $$

a) from (i) we have \\( 0 \ge d(x,x) \\) and from **cost-set** we have \\(d \in [0,\infty] \\) giving \\(d(x,x) = 0 \\)

b) It is possible to have two objects in \\( \textbf{X} \\) having the same "cost" so \\(\mathcal{X}\\) is not a metric space.

c) The only rule restricting opposite costs was symmetry, now that we are working not working with symmetric monoidal posets but monoidal preorders that is off the table. In other words they are certainly not required to be equivalent.

d) this is exactly (j)

So \\(\mathcal{X}\\) is almost a metric space if we drop property (b) and (c).

Q: Would (c) be satisfied if \\( \mathcal{V} \\) were symmetric?

Q: Would (b) be satisfied if \\( \mathcal{V} \\) were a poset?

A: No, I can see no reason as the mapping \\( \mathcal{X}(x,y) \rightarrow \mathcal{V} \\) is not required to be 1-to-1.

**Edit** I am going to stop fiddling with this as the [next lecture](https://forum.azimuthproject.org/discussion/2128) addresses it better.

A metric space consists of

1) a set \\(X\\) the elements of which are called points and

2) a metric function \\( d : X \times X \rightarrow \mathbb{R}_{\ge 0} \\), where \\( d(x,y) \\) is the metric, a distance between x and y

and the rules

a) \\( \forall x | x \in X \\), we have \\( d(x,x) = 0 \\)

b) \\( \forall x, y | x, y \in X \\), if \\( d(x,y) = 0 \\) then \\( x = y \\)

c) \\( \forall x, y | x, y \in X \\), we have \\( d(x,y) = d(y,x) \\)

d) \\( \forall x, y, z | x, y, z \in X \\), we have \\( d(x,y) + d(y,z) \ge d(x,z) \\)

Is **Cost**-category [I want to call it a **Cost**-set] a metric space?

We have a symmetric monoidal preorder with which to embelish.

$$ \tag{cost} \textbf{Cost} := ( [0,\infty], \ge, 0, +) $$

And a \\( \textbf{Cost} \\)\-category

$$ \tag{cost-set} \mathcal{X} := (\textbf{X}, \textbf{Cost}) $$

With the rules attendant on monoidal categories and \\( \mathcal{V}\text{-category} \\).

i) \\( 0 \ge d(x,x) \\)

j) \\( d(x,y) + d(y,z) \ge d(x,z) \\)

We have

1) a set \\( \textbf{X} := Ob(\mathcal{X}) \\) and

2) a metric function \\( + := d \\) from \\( \textbf{Cost} \\).

$$ \forall x,y \in \textbf{X} , d(x,y) := x + y $$

a) from (i) we have \\( 0 \ge d(x,x) \\) and from **cost-set** we have \\(d \in [0,\infty] \\) giving \\(d(x,x) = 0 \\)

b) It is possible to have two objects in \\( \textbf{X} \\) having the same "cost" so \\(\mathcal{X}\\) is not a metric space.

c) The only rule restricting opposite costs was symmetry, now that we are working not working with symmetric monoidal posets but monoidal preorders that is off the table. In other words they are certainly not required to be equivalent.

d) this is exactly (j)

So \\(\mathcal{X}\\) is almost a metric space if we drop property (b) and (c).

Q: Would (c) be satisfied if \\( \mathcal{V} \\) were symmetric?

Q: Would (b) be satisfied if \\( \mathcal{V} \\) were a poset?

A: No, I can see no reason as the mapping \\( \mathcal{X}(x,y) \rightarrow \mathcal{V} \\) is not required to be 1-to-1.

**Edit** I am going to stop fiddling with this as the [next lecture](https://forum.azimuthproject.org/discussion/2128) addresses it better.