4. The Extreme Value Theorem states that if f : D C R -> R is continuous on a set K, where @ # K C D and K is compact, then the restriction fly (i.e., the function with domain K agreeing with f on K ) attains both a global minimum at some point p e K and also a global maximum at some point q e K. Nothing prevents the global minimum value from being attained at many different points pe K, etc.; the theorem says only that it must be attained at some pek.] Using the Extreme Value Theorem, together with the results above, prove that if K and K' are two disjoint (i.e., Kn K' = 0), nonempty, compact subsets of R", and if we define their distance by dist (K. K') := inf { ||x - yl| | x e K, ye K'}, then we must have dist (K, K') >0. Hint: Consider the function f (x y) := |x -yl = v(ri -y)? + . .. + (Ca -ya)? defined on D = Rad, and consider the set K x K' C R . Explain why K x K is nonempty and compact, and why f is continuous on K x K'. Show that there is a point p q e K x K, i.e., there is a point pe K and a point q e K, where flxxx is globally minimized. Thus, dist (K, K') = inf { |x - y| | xe K, ye K'} = min{|x - yl| | xe K. ye K'} = lp -all. What can we infer from the fact that Kin K' = 0