# Thread: Topological Metric Space with 3 elements

1. ## Topological Metric Space with 3 elements

Can you create a topological metric space from three points (the set X) in the plane? Obviously, d is defined.

A ball at each point is the point itself (satisfies def of neighborhood of a point).
Union of all the balls is X.
Intersection of any two balls is empty (contains no member of X) so there can't be an x in a ball which is a subset of the intersection.

Is it valid to conclude you can't create a topolgical metric space from a finite set consisting of three or more members? Or does a topological metric space have to have an infinite number of members?

EDIT Origin of the question: A metric is defined on a non-empty set, which could have a finite number of members. In a std text developments go on from there without ever considering the possibility that the set is finite (three or more members to satisfy d), leaving me scratching my head.

2. ## Re: Topological Metric Space with 3 elements

Originally Posted by Hartlw
Can you create a topological metric space from three points
The discrete metric is defined as:
$\delta(x,y) = \left\{ {\begin{array}{*{20}c} {1,} & {x \ne y} \\ {0,} & {x = y} \\ \end{array} } \right.$

For any non-empty set $X$ the pair $(X;\delta)$ is a metric space.

3. ## Re: Topological Metric Space with 3 elements

Perhaps another way of asking the question (generally, I know very little about it) is,

does topology deal only with continua,
or, does topology define a continuum?

4. ## Re: Topological Metric Space with 3 elements

Originally Posted by Plato
The discrete metric is defined as:
$\delta(x,y) = \left\{ {\begin{array}{*{20}c} {1,} & {x \ne y} \\ {0,} & {x = y} \\ \end{array} } \right.$

For any non-empty set $X$ the pair $(X;\delta)$ is a metric space.
Thanks. That answers part of my implied question about utility of defining a metric space on a non-empty set.

But can you create a topological metric space from a finite (3 or more) number of elements? My question arose as I tried to work my way through the definiton of a topological metric space starting with a metric space consisting of three points in the plane. My first post shows where I got stuck.

5. ## Re: Topological Metric Space with 3 elements

I don't understand what you mean by a topological metric space. Is it just a space with a topology?

6. ## Re: Topological Metric Space with 3 elements

if you use the topology induced by the discrete metric, and our set is {a,b,c}, here are the neighborhoods:

neighborhoods of a: {{a}, {a,b}, {a,c},{a,b,c}}

neighborhoods of b: {{b}, {a,b}, {b,c}, {a,b,c}}

neighborhoods of c: {{c}, {a,c}, {b,c}, {{a,b,c}}

this is, in fact, the discrete topology on {a,b,c}, the power set of {a,b,c}. note that there are only 2 possible ε-balls for any point x:

N(x) = {y in {a,b,c} : d(x,y) < r} = {x}, when r ≤ 1, and N(x) = {y in {a,b,c} : d(x,y) < r} = {a,b,c}, when r > 1.

note that any two neighborhoods of x (whether x be a,b, or c) contain a neighborhood of x in their intersection....namely {x}.

(the notion that "two neighbohoods must contain a neighborhood in their intersection" is patently false: consider the usual topology on R

where a < b < c < d, the open interval (a,b) has null intersection with (c,d), and this is the "prototypical" metric space we are modelling

the entire concept on, with distance function d(x,y) = |y - x|. what IS true, is that the intersection of two open intervals containing x,

contains a third open interval containing x).

7. ## Re: Topological Metric Space with 3 elements

I feel as though it's important to point out that the only finite $T_1$ spaces (points are closed) are discrete. Thus, the only finite metric spaces are discrete.