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This is a revised version of a question I already posted, but which patently was ill posed. Please give me another try.


For comparison's sake, the axioms of a metric:

Axiom A1: $(\forall x)\ d(x,x) = 0$

Axiom A2: $(\forall x,y)\ d(x,y) = 0 \rightarrow x = y$

Axiom A3: $(\forall x,y)\ d(x,y) = d(y,x)$

Axiom A4: $(\forall x,y,z)\ d(x,y) + d(y,z) \geq d(x,z)$


Let $T$ = {X,T} be a topology, $B$ a base of $T$, $x, y, z$ $\in$ X

Definition D0: $x$ is nearer to $y$ than to $z$ with respect to $B$ ($N_Bxyz$) iff $(\exists b \in B)\ x, y \in b \ \& \ z \notin b \ \&\ (\nexists b \in B)\ x, z \in b \ \& \ y \notin b$

Definition D1: $B$ is pre-metric1 iff $(\forall x,y)\ x \neq y \rightarrow N_Bxxy$

Definition D2: $B$ is pre-metric2 iff $(\forall x,y,z)\ ((z \neq x\ \&\ z \neq y) \rightarrow N_Bxyz) \rightarrow x = y$

Definition D3: $B$ is pre-metric3 iff $(\forall x,y,z)\ N_Bzyx \rightarrow (N_Byxz \rightarrow N_Bxyz)$


Definition: $T$ is pre-metrici iff $(\exists B)\ B$ is pre-metrici (i = 1,2,3).

Definition: $B$ is pre-metric iff $B$ is pre-metric1, pre-metric2 and pre-metric3.

Definition: $T$ is pre-metric iff $(\exists B)\ B$ is pre-metric.

Remark: D1 is an analogue of axiom A1, D2 of axiom A2, D3 of axiom A3.

Remark: $T$ is pre-metric1 iff $T$ is T1 [not quite sure].

Remark: If $T$ is induced by a metric, then $T$ is pre-metric.


Question: Can a property pre-metric4 be defined such that $T$ induces a metric iff $T$ is induced by a metric with

Definition: $B$ is metric iff $B$ is pre-metric and pre-metric4.

Definition: $T$ induces a metric iff $(\exists B)\ B$ is metric.

Remark: Property pre-metric4 should be an analogue of A4 (the triangle inequality).

If provably no such property can be defined does this shed a light on the difference (an asymmetry) between topologies and metric spaces? ("It's the triangle inequality, that cannot be captured topologically.")

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One reason I like this improvement is that as written, the topology T can induce multiple metrics, e.g. by picking different bases B. –  Theo Johnson-Freyd Dec 27 '09 at 16:36
    
Yes, premetric_1 is precisely T_1 and doesn't depend on the basis. A metric space (with ball basis) need not satisfy premetric_2; e.g., consider a three point space with d(x,y)=1, d(x,z)=d(y,z)=2. (If no point has a closest point then I think it would hold.) D2 is too strong of an analogue, because A2 has a stronger hypothesis (every basis element that contains x also contains y). –  Jonas Meyer Dec 27 '09 at 20:15
    
I really doubt metrizability can be characterized by a formula "there exists a basis $B$ such that P(B)" with $P$ a first order formula... –  Mariano Suárez-Alvarez Dec 27 '09 at 23:04
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up vote 8 down vote accepted

I shall prove that there can be no characterization of metrizability along the lines that you seek. (This argument fleshes out and fulfills the expectation of Mariano in the comments.)

Your axioms are all stated in the language with two types of objects: points and basis elements. So let us be generous here, and entertain the possibility of any statement of a similar type, expressed in the same language. You are considering structures consisting of a set X and a collection B of subsets of X, to be used to generate a topology on X. Your language allows you to quantify over points and over basis elements, and to say that a given point is an element of a given basis element, and so on. This is a perfectly fine formal language to work in. One can express that B is indeed a basis for a topology in this language, by the assertion first that every point is in some basis element and second that for every two basis element b and c, and every point x in both b in c, there is a basis element d such that x is in d and for all points z, if z is in d, then it is both b and c. Similarly, one can state the Hausdorff condition and other topological properties in this language.

But I claim that metrizability is simply not expressible in this language. To see why, suppose toward contradiction that there were an axiom Phi expressible in the language of points and basic open sets in the way we have described, which held exactly of the bases that generate metric topologies. Consider now the real line (R,<). This is a linear order, and the natural basis for the order topology, the collection of open intervals, is first-order definable in the language consisting only of the order. Thus, the collection of basic open sets is describable in the language of the linear order. For example, to quantify over basic open sets, one says, "there are two points a<b such that..." and then refers to the interval (a,b). Since the real line is indeed a metric topology, the collection of intervals in the real line (R,<) will satisfy your axiom Phi.

Now comes the problem. By the methods of nonstandard analysis, there is an extension of the real line to the nonstandard real line, denoted (R*,<), which contains infinitesimal elements and infinite integers, but which has the amazing property that any statement expressible in the language of < which is true in the real line (R,<) is also true in the nonstandard line (R*,<). In fact, nonstandard analysis is rather powerful, for one can add the field structure and indeed, any structure whatsover to R and get the nonstandard version in R*. In particular, your axiom Phi will remain true for the nonstandard reals (R*,<). That is, your axiom Phi, whatever it is, will remain true for the order topology on the nonstandard real line. But this is a serious problem, because there are nonstandard extensions R* whose order topology is not metrizable. For example, there are instances of R* where the cofinality of the positive infinitesimals is uncountable, and in these instances, there are no nontrivial convergent sequences in the order topology of R*, but the topology is not trivial. Thus, the axiom Phi has made a mistake with R*, contradiction.

Thus, there can be no axiomatization of metrizability along the lines you suggest.

Edit: In case you suggest that the standard basis for the order should not be the witness for Phi in the real line, but instead some other weird basis B is the one satisfying Phi, then we simply consider the nonstandard (R*,<) together with B*. By the transfer principle, B* is a basis for the order topology on (R*,<) and continues to satisfy Phi, but R* is not metrizable, for the same contradiction.

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I wonder if what you are saying amounts to saying that for every formula in the language of a totally ordered set which has a model has another model such that the corresponding topological space is not metrizable? –  Mariano Suárez-Alvarez Jan 4 '10 at 3:12
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In any case, very good :) –  Mariano Suárez-Alvarez Jan 4 '10 at 3:21
    
That is almost right, but not quite correct, since discrete orders are metrizable, and the discreteness of the order IS first order expressible. For the question, it suffices to find a single metrizable linear order with an elementary extension that is not metrizable, and the easiest example of this was the real line and the nonstandard line. This is enough to show that the metrizability of a linear order is not first order expressible. –  Joel David Hamkins Jan 4 '10 at 3:24
    
Thanks! I'm glad you liked it... –  Joel David Hamkins Jan 4 '10 at 3:24
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Thanks for accepting my answer. Yes, nonstandard analysis is quite interesting; I think you will like it, since it gives rigorous substance to the old ideas of Leibnitz on infinitesimals, which mathematicians found so appealing in calculus for hundreds of years. Finally, we have a way of treating them sensibly. And it is relatively easy to get started with. As to the issue of metrizability, I don't think one can avoid the issue of nonstandardness easily without a fundamental change in the kind of axiom you seek. –  Joel David Hamkins Jan 4 '10 at 13:13
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