A question in metric space Rn

[miki]

Hello all,

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

[Chip Eastham]

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.

regards, chip

[TCL]

Use the inequality:
|r(z,Y)-r(x,Y)| <= r(x,z)

-TCL

[miki]

> -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?

Thanks a lot,
Miki

[José Carlos Santos]

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,

|r(z,Y) - r(x,Y)| <= d(x,z).

Best regards,

Jose Carlos Santos

[MeAmI.org]

MeAmI.org scribed:

Is it possible to adapt the a and b variables?

R and N?

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