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Homework Help: Real Analysis (Rudin) exercise with inequalities

  1. Jul 21, 2010 #1
    1. The problem statement, all variables and given/known data
    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.)
    2. Relevant equations
    |ax| = |a||x|
    |x-z| < or = |x-y| + |y-z|
    |x+y| < or = |x| + |y|

    3. 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?
  2. jcsd
  3. Jul 21, 2010 #2
    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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