Good. So you admit that pseudo-metric spaces do not introduce a location concept.
I actually show that even if location is not introduced, still 1=3 according to pseudo-metric space reasoning.
Is it just the distance function concept that escapes you?
It's really simple. A pseudo-metric space is a set A (you may call its members points if that comforts you) and a function D from AxA to R>=0, normally referred to as distance.
Let A be the set of all integer pairs, (m,n). If (a,b) and (c,d) are in A, let D((a,b), (c,d)) be defined as |a-c|.
Don't those meet all the requirements for pseudo-metric space? Are not (1,3) and (1,6) distinct elements of A? Is not the distance between them 0?
According to pseudo-metric space the distinct x=(1,3) and y=(1,6) members are also indistinguishable.
You are still missing that according to pseudo-metric space x and y are indistinguishable
AND distinct, as follows:
Unlike a metric space, points in a pseudometric space need not be distinguishable; that is, one may have d(x,y) = 0 for distinct values x≠y
Points are not any member, they are exactly the smallest existing members, and this property is significant, so your "(you may call its members points if that comforts you)" nonsense simply demonstrates your ignorance about members that have the property of the smallest existing things.
Furthermore, you actually claimed that points are not involved in pseudo-metric space, according to this question:
..and for that matter, where are points mentioned in it all?
It demonstrates that you do not read the considered subject ( in this case
http://en.wikipedia.org/wiki/Pseudometric_space )
before you reply.
Are not (1,3) and (1,6) distinct elements of A? Is not the distance between them 0?
(1,3) and (1,6) are distinct members of A and your abs(1-1)=0 ignores the fact that these members are at least A={(1,3),(1,6)} such that |A|=2, or in other words, you can't break the members into pieces , because according to the notion of set theory, the identity of a given member is defined as a whole.
In this case the wholes are two distinct points that have distance 3 between them in 2-dimensional space.
Moreover, if abs(1-1)=0 is considered as distance, then also abs(3-6)=3 is considered as distance, or in other words, there is no 0 distance between (1,3) and (1,6) members of A set.