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Math Help - Topological Metric Space with 3 elements

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    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.
    Last edited by Hartlw; October 17th 2011 at 10:58 AM.
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    Re: Topological Metric Space with 3 elements

    Quote Originally Posted by Hartlw View Post
    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.
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    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?

    EDIT: Probably should read "metric topology" instead of "topology."
    Last edited by Hartlw; October 17th 2011 at 11:55 AM.
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    Re: Topological Metric Space with 3 elements

    Quote Originally Posted by Plato View Post
    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.
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    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?
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    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).
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    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.
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