Quasimetric on the Sorgenfrey Line

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SUMMARY

The discussion focuses on defining a quasimetric on the Sorgenfrey Line, specifically proposing the function d(p,q) = q - p (if q > p) and 0 otherwise. The user aims to demonstrate that the sphere centered at point p with radius e corresponds to the interval [p, p+e). A suggestion is made to redefine the quasimetric as d(x,y) = {y-x, if y>=x; 1 otherwise} to address issues with the original definition, particularly the triangle inequality. The conversation also touches on the relationship between the Sorgenfrey topology and the proposed quasimetric topology.

PREREQUISITES
  • Understanding of quasimetric definitions and properties
  • Familiarity with the Sorgenfrey Line and its topology
  • Knowledge of the triangle inequality in metric spaces
  • Basic concepts of open sets in topology
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  • Study examples of triangle inequality proofs in quasimetric contexts
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Mathematics students, particularly those studying topology and metric spaces, as well as educators looking for examples of quasimetric definitions and their applications in theoretical contexts.

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Homework Statement



Define a quasimetric on the Sorgenfrey Line.

Homework Equations



I know how to show the distance function is always nonnegative, equal to zero if evaluating the distance of a point from itself, and the triangle inequality. I'm having trouble coming up with the function.

The Attempt at a Solution



I have defined d(p,q) = q - p (if q > p) and 0 otherwise. I want to show that the sphere centered at p of radius e is the same as the interval [p, p+e). This will complete the proof.

Suppose x is in the sphere centered at p of radius e. Then either d(p,x) = 0 or d(p,x) < e
(IFF) p <= x or x - p < e (IFF) p <= x or x < p + e.

What bothers me here is the "or". It feels like I need "and" to justify x in [p, p+e). Help?
 
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The function you're proposing doesn't really satisfy the definition of a quasimetric. For example, d(3,1)=0. How about we define d(x,y) = {y-x, if y>=x; 1 otherwise} instead? I think this is a valid quasimetric, although some case-checking is needed to establish the triangle inequality (which I haven't really done in a satisfactory way!).

Now, let's say we're give the basic open set [x,y) in the Sorgenfrey topology. Consider the cases when |y-x|>=1 and |y-x|<1, and show that we can get a ball to sit inside [x,y) in each case.

The other direction, that the Sorgenfrey line is finer than our quasimetric topology, is immediate.
 

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