-- rational distances
quasi
is in Q for all points p in Q^2?
quasi
smn
S=R^2 smn
quasi
wrote:
Yep, I forgot to bar that.
Ok, consided it barred ...
Does there exist a closed, nonempty proper subset S of R^2 such that
quasi
No, that can also be easily achieved.
For example, let S = {(x,y) in R^2 | y <=0}.
Ok, here's a fixed version ...
Does there exist a nonempty compact subset S of R^2 such that d(p,S)
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?
>
D = { (x,y) | x^2 + y^2 <= 1 }
Let r = inf{ r | S subset rD }
Some point p in S /\ rD.
Let L be the tangent to rD at p.
L divides the plane into a region that cnntains S and one
that doesn't. In the latter region you can find some points
s in QxQ with d(s,S) = d(p,s) not in Q.
From this you can generalized from compact sets to bounded sets.
In addition notice d(p,nulset) not in Q.
Robert Israel
> On Thu, 2 Oct 2008, quasi wrote:
>
> > 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?
> >
> D = { (x,y) | x^2 + y^2 <= 1 }
> Let r = inf{ r | S subset rD }
Too many r's...
> Some point p in S /\ rD.
I presume you want p with ||p|| = r = sup {||s||: s in S}.
> Let L be the tangent to rD at p.
>
> L divides the plane into a region that cnntains S and one
> that doesn't. In the latter region you can find some points
> s in QxQ with d(s,S) = d(p,s) not in Q.
How so? The only points s where d(s,S) = d(p,s) might be the
ray {tp: 1 <= t < infty}, and this might not intersect Q^2.
> From this you can generalized from compact sets to bounded sets.
> In addition notice d(p,nulset) not in Q.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
quasi
<ma...@hevanet.remove.com> wrote:
>On Thu, 2 Oct 2008, quasi wrote:
>
>> 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?
>
>D = { (x,y) | x^2 + y^2 <= 1 }
>Let r = inf{ r | S subset rD }
>
>Some point p in S /\ rD.
You're mixing up letters -- in the stated problem, p is in Q^2.
Then later you take s in Q^2 (rather than s in S).
OK, so aside from that mixup, when you say
"some point p in S /\ rD",
I assume you meant
some point p in S intersect (the circular boundary of rD).
Right?
>Let L be the tangent to rD at p.
>
>L divides the plane into a region that cnntains S and one
>that doesn't. In the latter region you can find some points
> s in QxQ with d(s,S) = d(p,s) not in Q.
Seem plausible.
But that's the key -- can you always find some such points in Q^2?
What if S is convex?
>From this you can generalized from compact sets to bounded sets.
Sure, but for distances to sets, it suffices to consider closed sets.
>In addition notice d(p,nulset) not in Q.
The problem explicitly requires that S be nonempty.
quasi
LudovicoVan
> On Thu, 02 Oct 2008 01:16:12 -0400, quasi <qu...@null.set> wrote:
> >wrote:
> >>On Oct 1, 9:14 pm, quasi <qu...@null.set> wrote:
> >>> Does there exist a closed, nonempty subset S of R^2 such that d(p,S)
> >>> is in Q for all points p in Q^2?
>
> >>> quasi
>
> >>S=R^2 smn
>
> >Yep, I forgot to bar that.
>
> >Ok, consided it barred ...
>
> >Does there exist a closed, nonempty proper subset S of R^2 such that
> >d(p,S) is in Q for all points p in Q^2?
>
> No, that can also be easily achieved.
>
> For example, let S = {(x,y) in R^2 | y <=0}.
This seems to work because you don't have to take square roots, and
anyway require that the frontier of S is only made of rational points,
which I suppose means the frontier stays always parallel to one or the
other axis.
> Ok, here's a fixed version ...
>
> 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?
As soon as you make S bounded, I wouldn't see how you can fulfil the
required property at all, because you'll have anyway to take square
roots for the distance from almost all points...
-LV
William Elliot
> <ma...@hevanet.remove.com> wrote:
> >On Thu, 2 Oct 2008, quasi wrote:
> >
> >> 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?
> >
> >D = { (x,y) | x^2 + y^2 <= 1 }
> >Let r = inf{ r | S subset rD }
> >
> >Some point p in S /\ rD.
>
> You're mixing up letters -- in the stated problem, p is in Q^2.
>
> Then later you take s in Q^2 (rather than s in S).
>
> OK, so aside from that mixup, when you say
>
> "some point p in S /\ rD",
>
> I assume you meant
>
> some point p in S intersect (the circular boundary of rD).
>
> Right?
>
> >Let L be the tangent to rD at p.
> >
> >L divides the plane into a region that cnntains S and one
> >that doesn't. In the latter region you can find some points
> > s in QxQ with d(s,S) = d(p,s) not in Q.
>
> Seem plausible.
>
> But that's the key -- can you always find some such points in Q^2?
>
> What if S is convex?
>
What's the problem? It's still in rD.
Oh, the point p may not be unique. What the heck?
S could be rD itself and all of bd rD would be 'p' points.
> >From this you can generalized from compact sets to bounded sets.
>
> Sure, but for distances to sets, it suffices to consider closed sets.
>
> >In addition notice d(p,nulset) not in Q.
>
> The problem explicitly requires that S be nonempty.
>
It's slight generalization. ;-)
William Elliot
> William Elliot <ma...@hevanet.remove.com> writes:
>
> > On Thu, 2 Oct 2008, quasi wrote:
> >
> > > 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?
> > >
> > D = { (x,y) | x^2 + y^2 <= 1 }
> > Let r = inf{ r | S subset rD }
>
> Too many r's...
>
> > Some point p in S /\ rD.
>
> I presume you want p with ||p|| = r = sup {||s||: s in S}.
>
It works out to that.
> > Let L be the tangent to rD at p.
> >
> > L divides the plane into a region that cnntains S and one
> > that doesn't. In the latter region you can find some points
> > s in QxQ with d(s,S) = d(p,s) not in Q.
>
> How so? The only points s where d(s,S) = d(p,s) might be the
> ray {tp: 1 <= t < infty}, and this might not intersect Q^2.
>
Take the normal to L at p, and vary the slope by +-epsilon.
Some of the outside points in QxQ and the wedge between the
two lines will have irrational distance to p.
----
David Bernier
I've been thinking at looking at possible
closest points in a fixed S. For a
filled-in square, say [0,1]x[0,1] ,
the four corner (0,0) , (0, 1), (1,0) and (1, 1)
will each be closest points in S for points
in Q^2 lyeing on half-rays in R^2 for many
directions. We could say they're extremal points
of S.
For a general non-empty compact S in R^2, we
could say (a, b) in S is extremal if there exists
a closed half-plane in R^2 containing S as a
subset such that the line which is the
topological boundary of the half-plane
intersects (or meets) S at one and only one
point of S, namely (a, b).
For a closed disk, all points on the boundary
circle of the disk are extremal, so there are
2^aleph_0 such points.
It's not easy to see what's possible for
extremal sets of points. For convex
polygonal shapes, maybe some extremal point
and an x in Q^2 are at an irrational distance,
for some x. It seems related to
rational coordinate points on the unit circle,
number theory, etc.
David Bernier
The World Wide Wade
William Elliot <ma...@hevanet.remove.com> wrote:
But the distance to p for those points may not be the distance to S,
as easy examples show (take S to be a circle for example).
William Elliot
> William Elliot <ma...@hevanet.remove.com> wrote:
> > On Thu, 2 Oct 2008, Robert Israel wrote:
> > > William Elliot <ma...@hevanet.remove.com> writes:
> > > > On Thu, 2 Oct 2008, quasi wrote:
> > > >
> > > > > 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?
> > > > >
> > > > D = { (x,y) | x^2 + y^2 <= 1 }
> > > > Let r = inf{ r | S subset rD }
> > >
> > > I presume you want p with ||p|| = r = sup {||s||: s in S}.
> > It works out to that.
> >
> > > > Let L be the tangent to rD at p.
> > > >
> > > > L divides the plane into a region that cnntains S and one
> > > > that doesn't. In the latter region you can find some points
> > > > s in QxQ with d(s,S) = d(p,s) not in Q.
> > >
> > > How so? The only points s where d(s,S) = d(p,s) might be the
> > > ray {tp: 1 <= t < infty}, and this might not intersect Q^2.
> > >
> > Take the normal to L at p, and vary the slope by +-epsilon.
> > Some of the outside points in QxQ and the wedge between the
> > two lines will have irrational distance to p.
>
> But the distance to p for those points may not be the distance to S,
> as easy examples show (take S to be a circle for example).
>
The epsilon has to be small enought.
One could even require the distance from the center of rD be
2r, 3r or more. Yet still one needs to require a small distance
from the normal.
For example, consider the region from 5r to 6r,
within r^-100 of the normal. Oh oh. If r < 1, then just to be
sure, from 5/r to 6/r within r^100 of the norm and if r = 1, to
within 1/1000 from the norm and 10 from the center.
The World Wide Wade
William Elliot <ma...@hevanet.remove.com> wrote:
No, you always get the closest point on the circle by drawing a ray
through the center to the point in question.
William Elliot
> William Elliot <ma...@hevanet.remove.com> wrote:
> > > > > >
> > > > > > > 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?
> > > > > > >
> > > > > > D = { (x,y) | x^2 + y^2 <= 1 }
> > > > > > Let r = inf{ r | S subset rD }
> > > > >
> > > > > > Some point p in S /\ rD.
> >
> > > > > >
> > > > > > doesn't. In the latter region you can find some points
> > > > > > s in QxQ with d(s,S) = d(p,s) not in Q.
> > > > >
> > > > > How so? The only points s where d(s,S) = d(p,s) might be the
> > > > > ray {tp: 1 <= t < infty}, and this might not intersect Q^2.
> > > > >
> > > > Take the normal to L at p, and vary the slope by +-epsilon.
> > > > Some of the outside points in QxQ and the wedge between the
> > > > two lines will have irrational distance to p.
> > >
> > > But the distance to p for those points may not be the distance to S,
> > > as easy examples show (take S to be a circle for example).
> > >
> > One could even require the distance from the center of rD be
> > 2r, 3r or more. Yet still one needs to require a small distance
> > from the normal.
> >
> > For example, consider the region from 5r to 6r,
> > within r^-100 of the normal. Oh oh. If r < 1, then just to be
> > sure, from 5/r to 6/r within r^100 of the norm and if r = 1, to
> > within 1/1000 from the norm and 10 from the center.
>
> No, you always get the closest point on the circle by drawing a ray
> through the center to the point in question.
>
on B = S /\ bd rD. If p on B and c center of rD, then the ray from c
through p may or may not hit QxQ. If it doesn't, if all of the points in
B have radius rays that miss QxQ and none of them are isolated, they are
locally dense within bd rD. That is, S /\ bd rD contains an arc which
will provide a nice point p who's radius ray from c through p will hit QxQ
outside of rD.
Does that cinch it? No more holes in my proof?
Is there a short cut?
quasi
It's not clear to me what you claim to have proved.
How about a stated proposition together with a complete proof?
quasi
William Elliot
> 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).
Find a point x in V /\ QxQ with irrational d(x,p)
--
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).
----
quasi
<ma...@hevanet.remove.com> 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).
If you are claiming that V must have nonempty interior, you certainly
haven't proved it. Moreover, I don't actually believe it has to be
true. I think it's possible to construct a nonempty compact subset S
such that B has an isolated point, but the set V is just a ray,
emanating outward from p.
>Find a point x in V /\ QxQ with irrational d(x,p)
If V is just a ray, such a point x may not exist.
>No p in B is isolated point of B.
>Then B contains a multi point arc A.
That doesn't follow.
>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).
But without an arc, your proof won't float.
quasi
Gerry Myerson
I believe that should be "ark."
--
GM
David Bernier
If we let w = (3/5, 4/5), then d( (0, 0), w) = 1.
It seems that if a point x is on the half-ray
from w away from (0, 0) and with slope 3/4, and
also x is in Q^2, then d(w, x) is rational.
If w is identified with 3/5 + i*4/5 , then for
any integer n>0, | w^n | = 1, and Re( w^n) \in Q,
Im(w^n) \in Q. I think the points w^n enjoy the
same half-ray property as w. Also, the w^(-n),
for n>0. The closure of {w^n | n in Z} is
the unit circle T, which is compact. But T doesn't
have the "rational distances" property.
I don't know that it helps, but I thought I'd
mention it anyway.
David Bernier
Timothy Murphy
>>> > > 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?
It wasn't clear to me if a proof of the result has been given.
But it seems to me much easier than the arguments given.
The distance d(p,S) from a point p to S is a continuous function of p.
It is not constant, so it cannot always be rational.
Maybe I mis-read the question?
--
Timothy Murphy
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
quasi
wrote:
>quasi wrote:
>
>>>> > > 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?
>
>It wasn't clear to me if a proof of the result has been given.
>But it seems to me much easier than the arguments given.
>
>The distance d(p,S) from a point p to S is a continuous function of p.
>It is not constant, so it cannot always be rational.
>
>Maybe I mis-read the question?
Yes, probably you misread it.
d(p,S) is only required to be rational on Q^2, that is, on rational
points p.
Intuitively, the answer to the question is "no", that is, no such set
S exists, and if so, the proof might be easy, but I haven't seen a
valid proof yet.
quasi
Robert Israel
> quasi wrote:
>
> >>> > > 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?
>
> It wasn't clear to me if a proof of the result has been given.
> But it seems to me much easier than the arguments given.
>
> The distance d(p,S) from a point p to S is a continuous function of p.
> It is not constant, so it cannot always be rational.
>
> Maybe I mis-read the question?
Yes, you did. Note the "p in Q^2". There are plenty of continuous
functions from Q^2 to Q.
The World Wide Wade
William Elliot <ma...@hevanet.remove.com> 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.
William Elliot
> William Elliot <ma...@hevanet.remove.com> 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.
>
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.
> > 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?
The World Wide Wade
William Elliot <ma...@hevanet.remove.com> wrote:
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?
William Elliot
> 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?
Let C = bd B((0,0),2); S = bd B((0,1),1)
C and S are tangent at p = (0,2).
Now if I'm off the y axis and ... ok, I get it.
David Bernier
I've been thinking about points in Q^2 on the
line y = x. Suppose P = { (n, n), n>=1 and n in N}.
Does there exist a nonempty compact subset S
of R^2 contained in the unit disk such that
d(p, S) is in Q for all p in P?
I haven't found any S with that (weaker)
property so far ...
David Bernier
quasi
<davi...@videotron.ca> wrote:
>quasi wrote:
>>
>> 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?
>
>I've been thinking about points in Q^2 on the
>line y = x. Suppose P = { (n, n), n>=1 and n in N}.
>
>Does there exist a nonempty compact subset S
>of R^2 contained in the unit disk such that
>d(p, S) is in Q for all p in P?
>
>I haven't found any S with that (weaker) property so far ...
Good idea.
quasi
quasi
Unfortunately, I don't think it works.
That is, I believe that the answer to your question is "yes", while I
think the answer to the original question is "no".
I have in mind a construction of a set S which would satisfy the
requirements of your question.
I don't have it fully fleshed out, and I don't have time right now to
work out all the details, but here's the plan of attack ...
For n in N, let p_n = (n,n)
Define a sequence of points q_1, q_2, q_3, ... in the interior of the
unit disk such that,
(1) The sequence q_1, q_2, q_3, ... converges to q.
(2) d(p_n,q_n) is rational, for all n.
(3) d(p_n,q_n) < d(p_n,q_k) if k != n.
Then let S = {q_n | n in N} U {q}.
Perhaps someone can pick up this line of attack and finish it?
If not, I'll try to work it out later.
quasi
David Bernier
If S is some compact set as in my question, and S' is the set
obtained by reflecting S about the line y=x, then for any
p in P, d(p, S) = d(p, S'). If T = S \union S', T is compact,
contained inside the unit circle and for any p in P,
d(p, S) = d(p, T).
So if the answer to my question is "yes", then I believe
we can assume without loss of generality that the compact
set S has mirror symmetry about y=x.
David Bernier
** Posted from http://www.teranews.com **
amy666
>
> > quasi wrote:
> >
> > >>> > > 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?
> >
> > It wasn't clear to me if a proof of the result has
> been given.
> > But it seems to me much easier than the arguments
> given.
> >
> > The distance d(p,S) from a point p to S is a
> continuous function of p.
> > It is not constant, so it cannot always be
> rational.
> >
> > Maybe I mis-read the question?
>
> Yes, you did. Note the "p in Q^2". There are plenty
> of continuous
> functions from Q^2 to Q.
but there are even more reals than rationals !!
> --
> Robert Israel
> isr...@math.MyUniversitysInitials.ca
> Department of Mathematics
> http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver,
> BC, Canada
regards
tommy1729
David Bernier
>> Timothy Murphy <t...@maths.tcd.ie> writes:
>>
>>> quasi wrote:
>>>
>>>>>>>> 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?
>>> It wasn't clear to me if a proof of the result has
>> been given.
>>> But it seems to me much easier than the arguments
>> given.
>>> The distance d(p,S) from a point p to S is a
>> continuous function of p.
>>> It is not constant, so it cannot always be
>> rational.
>>> Maybe I mis-read the question?
>> Yes, you did. Note the "p in Q^2". There are plenty
>> of continuous
>> functions from Q^2 to Q.
>
> but there are even more reals than rationals !!
>
[...]
You can have a closed set S of R^2, not compact, non-empty, such
that d(p, S) is rational for any p in Q^2. For example,
if S is the x-axis, S = {(x,y) in R^2 such that y=0},
then d( (p/q, r/s), S ) = | r/s| where
p/q and r/s are arbitrary elements of Q.
So there exists a nonempty closed subset S of
R^2 such that d(p,S) is in Q for all points
p in Q^2. ( e.g. if S is the x-axis).
For subsets of R^2, "compact" and "closed and bounded" are equivalent.
The x-axis isn't bounded.
Returning to the x-axis example, d((0,sqrt(2)), S) = sqrt(2)
which is irrational; but p = (0, sqrt(2)) isn't
an element of Q^2.
Either switching between the words "compact" and "closed"
in quasi's question changes the answer, or it
doesn't.
I hope this helps ...
David Bernier
We could rotate all the points in P by 45 degrees
anti-clockwise and then any S in R^2 could
be rotated in the same manner.
So we could have t_n = (0, n*sqrt(2)) , n>=1.
I'm thinking of an even smooth function
f: [-a, a] ->R with f(0) = sqrt(2)-1.4
and where the radius of curvature R(x) at
(x, f(x)) starts at d(t_1, (0, f(0))) = 1.4,
or maybe something else.
for x = 0, and increases strictly for x>0 ,
so that the curvature decreases strictly.
The closest point to t_n on the arc with x>=0
of the graph of f would have x and y increasing as
n increases. Maybe what we need is for
R(x) for the point (x, f(x)) that should be closest
to t_n to have the property R(x) ~= d(t_n, (x, f(x)),
and in more detail: R(x)< d(t_n, (x, f(x)), and
in fact d(t_n, (x, f(x)) - R(x) increasing
slowly as n increases. The idea is to have
a unique point on the graph of f for 0<=x<=a
closest to each t_n, and to get there to draw
circles centered at t_n which pass close to
(0, sqrt(2)-1.4) but not enclosing that point,
and of course with d(t_n, (x, f(x))) rational.
That just might provide a "no" answer to my question
without resolving your question because your
question seems to involve rays in infinitely many
orientations (or slopes).
David Bernier