Re: Question re: r=cos[sqrt(2) * theta] on Unit Disk

On Wed, 25 Apr 2007 12:41:06 +0000 (UTC), Dave Seaman
<dseaman@xxxxxxxxxxxx> wrote:

On Wed, 25 Apr 2007 04:55:42 -0700, Leonard M. Wapner wrote:
Hello:

Despite this curve's appearance, it obviously does not "fill" the unit disk.
Far from it!

Would it be correct to say that "almost no points" on the unit disk are on
this graph? I think so. Equivalently, would it be correct to say that if
one randomly selects a point on the unit disk, then the probability of it
being on this curve is zero?

I believe the answer to both questions is "yes". Just checking terminology.

Yes, to both questions. There are only countably many values of theta
corresponding to points on the unit disk.

Did you mean circle, not disk?

Clearly, the curve has only countably many points on any given circle
of radius r centered at the origin, where 0<r<=1.

Even so, how does that show that the curve has measure 0 in R^2? It's
certainly intuitive and I'm sure it's easy, but I don't see it
immediately.

Or perhaps you meant that for any given point on the curve, there are
only countably many values of theta which map to that point? That's
true as well, but I don't see how that fact, by itself, could be used
to show that the curve has measure 0 in R^2.

quasi
.

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