On Sat, 19 Apr 2008, ~greg wrote:
> In such settings the ability to fill in the details is taken for
granted.
> Time isn't wasted on it. That's only done later, in private, and only
> when necessary, eg for publication.
Such as posting to a newsgroup. take the time to proof before posting.
Even twice perhaps. It the long run it'll save you time and us time and
also perhaps you some embarrassment.
> I had hoped that this newsgroup would be more like an informal
> brain-storming senimar.
>
Your brain is storming. You want me to off the cuff, oh yea yea, I see.
Not only that but also, off the top of my head hear say and fairy tales or
do you want some actual facts and math?
If you're in too much of a hurry to do the math, then storm your brains
in a geometric art group.
> I started from the observation that a probe into the integer-lattice
> will hit a lattice point iff its slope is irrational.
> Unless the "lattice" consists of disks of fixed finite radius,
> in which case any probe, of rational or irrational slope,
> will hit some of them.
>
> It then it occured to me to go outward from there,
> into a consideration of a sort of intermediate case,
> in which the radii diminish outward from the origin,
> ---but never actually become 0 (--which I should have said clearly.)
> Then a probe of irrational slope would again miss all the disks,
> - provided their radii diminished in a specific way.
>
> Likewise, given ANY FINITE SET L of probes of irrational slopes,
> then radii can be defined in such a way that all the probes in L
> miss all the disks.
>
> Then it occured to me to ask if ANY EXAMPLE of a COUNTABLE OR
> UNCOUNTABLE SET L, of probes of irrational slopes can be constructed
> such that each probe in L misses all the disks, provided their radii
> diminish in an appropriate way.
>
> And you gave me A COUNTER-EXAMPLE
>
> But I wasn't looking for that.
> I was looking for ANY EXAMPLE WHERE IT IS TRUE.
>
That wasn't what you stated. Now that you have changed the inquiry more
thought, none of that quick splash dash thought, but real thoughtful
thought that may take more time and bother than instant gratification
media tutored consumers are want to take.
> In your counter-example the slopes of the lines in L approach 1/2.
> Which is rational. So the probe-line of slope 1/2 is not in L.
> It's just a limit of the set L.
>
> And you say
> > > No matter how small of a disk you put around (1,1)
> > > infinitely many of L will pass through the disk.
>
> And that's true.
>
> And more generally, if any limit-line of the set L has
> a rational slope a/b, then no matter how small a disk
> is put around the specific lattice point (b,a),
> then infinitely many lines in L will pass through it.
>
> The upshot is that the countable or uncountable example
> of a set L of probe lines of irrational slope
> such as I am looking for must neither contain (obviously)
> - NOR APPROACH -
> any probe line of rational slope.
>
Good insight.
> The set of slopes W, of the lines in a set L such as I am looking for
> WOULD HAVE TO BE A SET OF IRRATIONAL NUMBERS
> SUCH THAT ALL ITS LIMIT POINTS ARE ALSO IRRATIONAL,
> (whether or not any or all of them actually belong to W).
>
Hey hey, I can hear you from here. YOU DON"T HAVE TO SHOUT, that's
consider rude and lacking in netiquette.
> Because I don't know the answer to the question:
> -- are there sets of irrational numbers
> -- all of whose cluster points are also irrational?
>
{ pi }, { pi, e }, { pi, e, sqr 2 },
{ n.pi | n in N }, { pi + 1/n | n in N }
{ pi + 1/n + 1/m | n,m in N }
{ m.pi + 1/n + 1/k | n,m,k in N }
Thus you're left pondering the question if there's an uncountable
set of reals whose closure is disjoint from the rationals, Q.
That is an interesting question. A weaker interesting question is:
does an uncountable set of reals have uncountably many limit points?
According to my notes, such a set has uncountably many limit
points in the set itself. The other question I have no answer.
> Even if there does exist such a set, I am not 100% certain
> that it could completely get around the gist of your
> counter-example. But now I think it could. If it exists.
>
> If such a set exists, then it gets around your specific
> counter-example of a rational limit slope a/b implying
> the existence of a specific lattice point (b,a) whose disk
> would have to have a radius equal to zero.
>
By convention, there isn't any.
> But, while probes of irrational slope w, which are limits
> of the set L, do have probes of rational slope a/b approaching
> w arbitrarily closely, nevertheless the lattice point (b,a) in this case
> is not fixed. As a/b->w, the disk on (b,a) does have to vanish.
> But that's already true of all the probes of irrational slope in L
anyway!
>
> So, that line of thinking does not automatically
> supply any dis-proof of the possible existence
> of the kind of set I'm looking for.
> ~
>
> I am sorry if I sounded testy before.
>
Were you to do a rewright, posting in a newsgroup with a wider audience,
some having much greater learning than I, you should first clarify
your terminology with some like:
A probe is a line of the real plane through the origin.
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