Consider the subset A C0 1 of all strictly increasing contin

Consider the subset A C[0, 1] of all strictly increasing continuous functions f: [0, 1] rightarrow [0,1]. Is it closed? Is it open?

Solution

It is closed.

To show that it is closed in sup-norm, you only have to show that if fnfn converges to ff uniformly and fnFfnF, then the limit is in FF. We know that integral behaves well w.r.t. uniform convergence.(f=limfn)

Proof of the fact that the condition (x,y[0,1])|f(x)f(y)||xy|(x,y[0,1])|f(x)f(y)||xy| is preserved by uniform convergence is standard.

Example of closed subset is

The setA={x:t[0,1]|x(t)|t22+1}is closed inC([0,1]).The setA={x:t[0,1]|x(t)|t22+1}is closed inC([0,1]).

proof :

Let >0>0 and let (x)nA(x)nA, xnx0xnx0. Then for nNnN sup[0,1]|xn(t)x0(t)|sup[0,1]|xn(t)x0(t)|for some NN and we have

|x0(t)||x0(t)xN(t)|+|xN(t)|+t22+1|x0(t)||x0(t)xN(t)|+|xN(t)|+t22+1

Then |x0(t)|t22+1|x0(t)|t22+1 for all t[0,1]