Stuck on a problem from Rudin Ch. 1 (16)

Guy

Mar 2010
98
21
I'm doing this problem as part of my effort to self-study my way through the first 7 chapters of Rudin.

Suppose k\(\displaystyle \ge 3, x, y \in \mathbb{R}^k, ||x-y|| = d > 0\), and \(\displaystyle r > 0\). Prove (a) that if \(\displaystyle 2r > d\), there are infinitely many \(\displaystyle z \in \mathbb{R}^k\) such that \(\displaystyle ||z - x|| = ||z - y|| = r\), (b) there is exactly one such \(\displaystyle z\) if \(\displaystyle 2r = d\) and (c) there is no such z if \(\displaystyle 2r < d\).

My progress:

I proved (b) and (c) using the condition for attainment in Cauchy-Schwarz. Part (a) I proved after assuming that \(\displaystyle x = 0\) and \(\displaystyle y = (y_1, 0, ..., 0)\). I'm probably just being stupid, but the problem I seem to be running into is that I'm having a hard time undoing my rotation and translation without going beyond what can be done with the material learned so far. Any help would be appreciated. If anyone could give a simple justification (i.e. one that doesn't use anything but definition of the Euclidean norm and the basic algebraic properties of the dot product) for assuming WLOG that \(\displaystyle x = 0\) and \(\displaystyle y = (y_1, 0, ..., 0)\), this would certainly do the trick. Any other permissible solution would be fine too.

Thanks.
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
I'm doing this problem as part of my effort to self-study my way through the first 7 chapters of Rudin.

Suppose k\(\displaystyle \ge 3, x, y \in \mathbb{R}^k, ||x-y|| = d > 0\), and \(\displaystyle r > 0\). Prove (a) that if \(\displaystyle 2r > d\), there are infinitely many \(\displaystyle z \in \mathbb{R}^k\) such that \(\displaystyle ||z - x|| = ||z - y|| = r\), (b) there is exactly one such \(\displaystyle z\) if \(\displaystyle 2r = d\) and (c) there is no such z if \(\displaystyle 2r < d\).

My progress:

I proved (b) and (c) using the condition for attainment in Cauchy-Schwarz. Part (a) I proved after assuming that \(\displaystyle x = 0\) and \(\displaystyle y = (y_1, 0, ..., 0)\). I'm probably just being stupid, but the problem I seem to be running into is that I'm having a hard time undoing my rotation and translation without going beyond what can be done with the material learned so far. Any help would be appreciated. If anyone could give a simple justification (i.e. one that doesn't use anything but definition of the Euclidean norm and the basic algebraic properties of the dot product) for assuming WLOG that \(\displaystyle x = 0\) and \(\displaystyle y = (y_1, 0, ..., 0)\), this would certainly do the trick. Any other permissible solution would be fine too.

Thanks.
Can't you just find one and extend the line from the origin it lies on outwards?
 

Guy

Mar 2010
98
21
I'm not sure I understand. I'm probably misunderstanding your post, but the set of z should be a circle (in R^3), so I don't see what extending a line comes in.
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
I'm not sure I understand. I'm probably misunderstanding your post, but the set of z should be a circle (in R^3), so I don't see what extending a line comes in.
It can't be a sphere, then all the points would be a fixed distance from "one" point.
 

Guy

Mar 2010
98
21
The set of z such that ||z-x|| = r is clearly a sphere with in Euclidean space centered at x. Ditto for the set of z such that ||z - y|| = r. The set of z that satisfy both of these is an intersection of the two spheres, which is (I hope) a circle (for the case k = 3).
 
Sep 2009
64
23
Edinburgh, UK
Hey,

Can't you just take the points along the perpendicular bisector of \(\displaystyle y-x\)? By the triangle inequality, \(\displaystyle d = \|x-y \| \leq \|x-z\| + \|y-z\| = 2r\) for any point \(\displaystyle z\) equidistant from \(\displaystyle x,y\). So you could never have \(\displaystyle 2r < d\), and \(\displaystyle 2r = d\) precisely at the midpoint of \(\displaystyle y-x\), and if we move from this midpoint along the perpendicular to \(\displaystyle y-x\) then we get an infinite number of points equidistant from \(\displaystyle x,y\). That is, there are an infinite number of isosceles triangles whose base is \(\displaystyle y-x\).

That is not very mathematical, but it should be possible to translate it: the essential ingredient appears to be the triangle inequality.
 

Guy

Mar 2010
98
21
Ah yes, I should be able to, for any line that bisects and is orthogonal to y-x, find two points on it such that the equality holds for a given r. Since there are an infinite number of lines that satisfy those conditions, I'll have an infinite number of points for each r.

Thanks for the help, much appreciated.