I need to prove the following claim using the triangle inequality:

d(Xm+Ym,p+q)< or equal to d(Xm,p)+(Ym,q)

Hi jen,

Please explain us what the terms mean so we can help you. And also could you explain in details what you need to prove and what you need to use in order to prove it.

Thank you.

Cheers!

Let K be a compact subset of Rn, and let p be an element of Rn and p is not in the set K.
(a) Prove that the function f : K-->R defined by
f (q) = d (p, q) , q E K,
is continuous on K. Hence, deduce that among the points in K, there is one, q0,
closest to p.
(b) Prove, by giving an example, that q0 is not necessarily unique. Could there be
infinitely many such q0?

Hi,

To prove that a function is continuous you can use the Weierstrass definition:

For any number ε > 0, however small, there exists some number δ > 0 such that for all x in the domain of ƒ with c − δ < x < c + δ, the value of ƒ(x) satisfies

f(c) - ε < f(x) < f(c) + ε

To put that for your function, we need to know who's d(p, q).

Please tell us who is d(p,q) so we can go further.

Cheers!

q=x and p=c?

Yes jen.

Do you need any more help?

Regards!