Suppose k>2, x, y in R^k, |x-y| = d > 0, and r > 0.
Prove if 2r > d, there are infinitely many z in R^k such that
|z-x| = |z-y| = r
(In Principles of Mathematical Analysis, it is problem 16(a) on page 23.)
|ax| = |a||x|
|x-z| < or = |x-y| + |y-z|
|x+y| < or = |x| + |y|
The Attempt at a Solution
I'm not quite sure how to tackle this proof. Here's the general outline I have:
- Show that there exists at least one z.
- Suppose there existed one and only one z. Is there a contradiction? Or, can I find z', a linear combination of z that also works and then from z' use the same rule to get z'', ad infinitum?
As you can see, I don't really know how to tackle "show there are infinitely many solutions" proofs.
One thought I had was, well, suppose z satisfies the necessary conditions.
Can I show that there exists c in R or d in R^k such that z' = cz + d also satisfies the necessary conditions?
But all I can get is |z' - y| = |z - y + d| < or = |z-y| + |d|
And |z' - y| has to = |z-y| = r, but that doesn't tell me anything.
Can someone give a hint or two to point me in the right direction? How should I tackle this problem?