I'd be happy if someone here will be able to solve my problem.
Assume all metric spaces are Rn.
Let Y be a closed set of a metric space X. Let x and z be some points in X (dont belong to Y) then if
3 * r(x, z) < r(z, Y)
then
2 * r(x, z) < r(x, Y)
where r is the distance function. Moreover, I assume that the distance between a point and a set is the infimum of all distances between the point and set over the points of the set.
> I'd be happy if someone here will be able to solve my problem.
> Assume all metric spaces are Rn.
> Let Y be a closed set of a metric space X. > Let x and z be some points in X (dont belong to Y) > then if
> 3 * r(x, z) < r(z, Y)
> then
> 2 * r(x, z) < r(x, Y)
> where r is the distance function. Moreover, I assume that the distance > between a point and a set is the infimum of all distances between the > point and set over the points of the set.
> Any help would be highly appreciated.
> Miki
Hint: Argue by contradiction using the triangle inequality. For example, if there exists y in Y s.t.
r(x,y) <= 2*r(x,z)
what can you say about r(z,y) ? This just uses properties of metric spaces.
> I'd be happy if someone here will be able to solve my > problem.
> Assume all metric spaces are Rn.
> Let Y be a closed set of a metric space X. > Let x and z be some points in X (dont belong to Y) > then if
> 3 * r(x, z) < r(z, Y)
> then
> 2 * r(x, z) < r(x, Y)
> where r is the distance function. Moreover, I assume > that the distance > between a point and a set is the infimum of all > distances between the > point and set over the points of the set.
> > I'd be happy if someone here will be able to solve my > > problem.
> > Assume all metric spaces are Rn.
> > Let Y be a closed set of a metric space X. > > Let x and z be some points in X (dont belong to Y) > > then if
> > 3 * r(x, z) < r(z, Y)
> > then
> > 2 * r(x, z) < r(x, Y)
> > where r is the distance function. Moreover, I assume > > that the distance > > between a point and a set is the infimum of all > > distances between the > > point and set over the points of the set.
> > Any help would be highly appreciated.
> > Miki
> Use the inequality: > |r(z,Y)-r(x,Y)| <= r(x,z)
> -TCL- Hide quoted text -
> - Show quoted text -
Well, using the inequality you have mentioned indeed solves the problem. But, why is that inequality true? How can I prove it?
>>> I'd be happy if someone here will be able to solve my >>> problem. >>> Assume all metric spaces are Rn. >>> Let Y be a closed set of a metric space X. >>> Let x and z be some points in X (dont belong to Y) >>> then if >>> 3 * r(x, z) < r(z, Y) >>> then >>> 2 * r(x, z) < r(x, Y) >>> where r is the distance function. Moreover, I assume >>> that the distance >>> between a point and a set is the infimum of all >>> distances between the >>> point and set over the points of the set. >>> Any help would be highly appreciated. >>> Miki >> Use the inequality: >> |r(z,Y)-r(x,Y)| <= r(x,z)
>> -TCL- Hide quoted text -
>> - Show quoted text -
> Well, using the inequality you have mentioned indeed solves the > problem. But, why is that inequality true? > How can I prove it?
In order to prove that r(z,Y) - r(x,Y) <= r(x,z), I shall in fact prove that r(z,Y) <= r(x,z) + r(x,Y). Let _y_ be an element of Y. Then
r(z,y) <= r(z,x) + r(x,y).
Since this is true for every _y_, it follows that
r(z,Y) <= r(x,z) + r(x,Y).
By symmetry, r(x,Y) - r(z,Y) <= r(x,z). Therefore,
Yes. b = 95.951(6) , V = 1403.4(2) 3 , space g. P2/n, Z = 2, crystal size 0.22 .... Metric parameters of (PNP)AlCl 2 are reproduced to within 0.05 /3 ...h
José Carlos Santos wrote: > On 04-07-2009 16:02, miki wrote:
> >>> I'd be happy if someone here will be able to solve my > >>> problem. > >>> Assume all metric spaces are Rn. > >>> Let Y be a closed set of a metric space X. > >>> Let x and z be some points in X (dont belong to Y) > >>> then if > >>> 3 * r(x, z) < r(z, Y) > >>> then > >>> 2 * r(x, z) < r(x, Y) > >>> where r is the distance function. Moreover, I assume > >>> that the distance > >>> between a point and a set is the infimum of all > >>> distances between the > >>> point and set over the points of the set. > >>> Any help would be highly appreciated. > >>> Miki > >> Use the inequality: > >> |r(z,Y)-r(x,Y)| <= r(x,z)
> >> -TCL- Hide quoted text -
> >> - Show quoted text -
> > Well, using the inequality you have mentioned indeed solves the > > problem. But, why is that inequality true? > > How can I prove it?
> In order to prove that r(z,Y) - r(x,Y) <= r(x,z), I shall in fact prove > that r(z,Y) <= r(x,z) + r(x,Y). Let _y_ be an element of Y. Then
> r(z,y) <= r(z,x) + r(x,y).
> Since this is true for every _y_, it follows that
> r(z,Y) <= r(x,z) + r(x,Y).
> By symmetry, r(x,Y) - r(z,Y) <= r(x,z). Therefore,
