Bijection is Homeomorphism

Let f : M -> N be a continuous bijection. M is compact. Show that f is a homeomorphism. isn't a homeomorphism by definition a bijection? and since M is compact, will it not be true that N will be compact too?

Differentiable Function

Assume f : R -> R is differentiable and there exists an L < 1 such that for each x in R, f'(x) < L. Prove that there exists a unique z in R such that f (z) = z. [ok, so f'(x)<1, do I do this with contradiction assuming no such z, or can this be done directly with differentiability defs ?]

Difference Quotient

Assume f:(-1,1) --> R and f'(0) exists. If a_n , b_n -> 0 as n->infty, define the difference quotient: D_n = ( f(b_n) - f(a_n) ) / ( b_n - a_n). a) Prove lim [n -> infty] D_n = f'(0) under each condition below: (i) a_n < 0 < b_n . (ii) 0 < a_n < b_n and (b_n) / (b_n - a_n) <= M (iii) f'(x) exists and is continuous forall x in (-1,1). b) Set f(x) = (x^2)sin(1/x) for x neq 0 and f(0)=0. Observe that f is diff everywhere in (-1,1) and f'(0)=0. Find a_n, b_n that tend to 0 in such a way that D_n converges to a limit not equal to f'(0).

The set of functions that are Riemann Integrable.

Let RI be the set of functions that are Riemann Integrable. Disprove with a counterexample or prove the following true. (a) f in RI implies |f| in RI (b) |f| in RI implies f in RI (c) f in RI and 0 < c <= |f(x)| forall x implies 1/f in RI (d) f in RI implies f^2 in RI (e) f^2 in RI implies f in RI (f) f^3 in RI implies f in RI

Series Convergence

Please see the attached file.