Re: -- rational distances

In article <Pine.BSI.4.58.0810060055320.8114@xxxxxxxxxxxxxxxxx>,
William Elliot <marsh@xxxxxxxxxxxxxxxxxx> wrote:

On Sun, 5 Oct 2008, The World Wide Wade wrote:
William Elliot <marsh@xxxxxxxxxxxxxxxxxx> wrote:
On Sat, 4 Oct 2008, quasi wrote:
On Sat, 4 Oct 2008 00:05:15 -0700, William Elliot

Does there exist a nonempty compact subset S of R^2 such
that d(p,S) is in Q for all points p in Q^2?

No.

D = { (x,y) | x^2 + y^2 <= 1 }
Let r = inf{ r | S subset rD }

B = S /\ bd rD isn't empty.

There are two cases.
Some p in B is isolated point of B.
Then there is a region V outside of rD with
for all x in V, d(x,p) = d(x,S).

False. Let S be the union of two closed discs of radius 1 with centers
(0,0) and (1,0). Then r = 2 and p = (2,0) is the only point in S /\ bd
rD. The points x where d(x,p) = d(x,S) are precisely the points on the
ray [2, oo) x {0}, and there the rational distance criterion is
satisfied.

If S = B((0,n),1/k), then r = n + 1/k and p = (0,n + 1/k).
Thus I can make B with as large of a radius as I want and
bd S with as much curvature as I want. By your reasoning
even in this case, for example when n = 10^10 and k = 10^-100,
V is still just the horizontal ray from p. Why is this so?
That if bd S and B are tangent, then V is the exterior normal
ray to the point of tangency.

Let x be exterior to a closed disc S, draw the the ray from the center
of S to x, which intersects the boundary at some p ... isn't it clear
that p is the point in S closest to x?

Find a point x in V /\ QxQ with irrational d(x,p)

Oh no, no, no. I'll let you do that. ;-)

Is the second part correct?

--
No p in B is isolated point of B.
Then B contains a multi point arc A.

For all p in B, let B_p be the portion of the ray
from the center of rD through p, that's outside of rD.

For all p in A, x in B_p, d(x,p) = d(x,S).
Find some p in A, x in B_p /\ QxQ with irrational d(x,p).

----
.

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  • Re: -- rational distances
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