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At http://en.wikibooks.org/wiki/Real_Analysis/Metric_Spaces you can find the standard definition of a metric space: a set $X$ given with a function $d:X\times X\to\mathbb{R}$ that satisfies properties 1 through 4. Later on the page it defines open ball and open set, and proves that arbitrary unions and finite intersections of open sets are open. (The page has a few mistakes; in particular, in the second paragraph of the proof, the unions should be intersections.) In other words, the open sets form a topology.

What I find remarkable is that none of properties 1 through 4 are needed for this proof. So consider a set $X$ and an arbitrary function $d:X\times X\to\mathbb{R}$. We can define open balls and open sets using the same definitions, producing a topology on $X$. Such topologies can be very different from those that arise from true metrics; for example, if $d$ is identically 0 we get the indiscrete topology.

Can we prove anything interesting about which topologies can arise this way? In particular, does every finite topology arise this way?

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Careful: if $d$ isn't symmetric then there are two notions of open ball ("left open balls" and "right open balls") so you actually get two topologies. A nice example is the "counterclockwise distance" metric on $S^1$. One of the topologies you get is the "topology of counterclockwise convergence" and the other is the "topology of clockwise convergence." –  Qiaochu Yuan Aug 13 '12 at 22:00
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No, something IS used. In general the intersection of two "open balls" is not "open". –  Anton Petrunin Aug 13 '12 at 23:56
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@Anton: You don´t need anything to show that the open sets form a topology. It is just that the open balls might not be open so they are not a base for this topology. –  Ramiro de la Vega Aug 14 '12 at 11:42
    
@Ramiro, I guess you want to use open balls as prebase, but that is not "the same definition". –  Anton Petrunin Aug 14 '12 at 14:45
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@Anton: I´m using the definition of "open" given by the OP´s link: $U$ is open iff for every $x \in U$ there is $r>0$ such that $B(x,r) \subseteq U$. With this definition it is obvious that the open sets form a topology. A different matter is whether sets of the form $B(x,r)$ are open or not. –  Ramiro de la Vega Aug 14 '12 at 18:37

3 Answers 3

up vote 7 down vote accepted

Some partial answers to OP´s original question:

1) Any first-countable space (and in particular any finite topology) can be obtained in this way.

Let $X$ be first-countable, and for each $x \in X$ fix a local neighborhood base $\{B_n^x:n \in \omega \}$ such that $X=B_0^x$ and $B_{n+1}^x \subseteq B_n^x$. We can now define the (not necessarily symmetric) function $$d(x,y)=\inf \left \{\frac{1}{n+1}:y \in B_n^x \right \}.$$

It is not hard to see that the topology defined by this function (defining balls centered at $x$ as $B(x,\varepsilon):=\{y : d(x,y) < \varepsilon \}$) is the same as the original topology on $X$.

2) One can get non-first-countable topologies.

If we let $X=\{a_{n,m} : n,m \in \omega \} \cup \{b_n :n \in \omega\} \cup \{c\}$ and define $d(b_n,a_{n,m})=d(c,b_n)=\frac{1}{n+1}$, $d(x,x)=0$ and $d(x,y)=1$ otherwise, we obtain Arens´ space which is first-countable at every point except at $c$.

3) Not every topology can be obtained in this way.

If $X=D \cup \{\infty\}$ is the one-point compactification of an uncountable discrete space, then the topology of $X$ cannot come from a "generalized metric". Just note that in this situation the balls $B(\infty,1/n)$ have to be open and hence cofinite; thus their intersection is cocountable and therefore the topology induced by the generilized metric wouldn´t be Hausdorff (but the original topology is).

Edit: here is another example for 3), prompted by Marcos´ comment, which is interesting because it shows that the class of spaces that we are looking at is not closed under subspaces:

The Arens-Fort space, which is just the subspace $\{a_{n,m} : n,m \in \omega \} \cup \{c\}$ of the Arens´ space defined in 2), can´t be obtained by this procedure. The reason is that the balls $B(x,\varepsilon)$ would have to be open (this is not hard to see) but then $\{B(c,1/n):n \in \omega\}$ whould be a countable local base at $c$.

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Notice that this Arens' space is slightly different from the one given in Wikipedia (en.wikipedia.org/wiki/Arens%E2%80%93Fort_space), since in this space the set $\{c\}\cup\{a_{n,m}:n,m\in\omega\}$ is not open. –  Marcos Cossarini Mar 8 '13 at 19:45
    
@Marcos: The $X$ in my answer is the Arens space (usually denoted by $\mathcal{S}_2$), the one given in Wikipedia is the Arens-Fort space which is a (non-open as you point out) subspace of Arens space. Arens-Fort space is not sequential, while Arens space is sequential but not Frechet. –  Ramiro de la Vega Mar 8 '13 at 20:48

This isn't an answer to exactly your question, but it has been proved that all topological spaces come from suitably generalized metric spaces. Specifically in ''All Topologies Come From Generalized Metric Spaces'' by Ralph Kopperman in The American Mathematical Monthly he shows that any topological space can be obtained from a generalized metric space where you weaken the axioms and replace $\mathbb{R}$ by a suitable semi-group with certain properties.

It also was shown in ''Quantales and continuity spaces'' by R. C. Flagg in Algebra Universalis that all topological spaces can be obtained by suitable weakenings of the axioms of a metric space and replacing $\mathbb{R}$ by quantales. This is, to my mind, particularly nice as the resulting generalized objects are essentially just categories enriched in the quantale.

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Along the lines of Nate's "answer", it is also true that every topology can be induced by a quasi-gauge, i.e. by a family of quasi-pseudo-metrics (pseudo- meaning not necessarily Hausdorff; quasi- meaning not necessarily symmetric).

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