Given a metric space (X, d), recall Co(X, R) = {f : X -> IR | f is continuous and bounded}. (a) Show that if IIflloo f(x) = 1 n=0 (Note that these are all just real numbers.) For part (c), let 1 denote the constant function 1 : X - R : 1(x) = 1 for every r E X. We say f E Co(X, R) is invertible with respect to the pointwise product if there exists g E Cb(X, R) such that fg = 1, i.e. f(x)g(x) = 1 for every r E X. For instance, on [0, 1] the inverse of f(x) = 1+x2 is g(x) = T (which are indeed continuous and bounded on [0, 1]). (c) Let Inv(X, R) := {f E C.(X, R) | f is invertible]. Show that the constant function 1 is in the interior of Inv(X, R). (Hint: If g : X -+ R satisfies ||1 - gl|.