Real Analysis (Rudin) exercise with inequalities

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Homework Statement


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.)

Homework Equations


|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?
 

Answers and Replies

  • #2
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Once you show that some z satisfies the equation, you can use the definition of the distance function in R^k to yield two quadratic equations in at least three unknowns. Expand with the binomial theorem and you cancel out the components of z and thus there are two linear equations in at k>2 unknowns, which has infinite solutions.
 

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