metric space proof open and closed set

show the set {f: ∫f(t)dt>1(integration from 0 to 1) } is an open set in the

metric space ( C[0,1],||.||∞） C[0,1] is f is continuous from 0 to 1.and ||.||∞

is the norm that ||f||∞ =sup | f|

and if A is the subset of C[0,1]

defined by A={f:0<=f<=1} is closed in the norm ||.||∞ norm.

first

one I set U= {f: ∫f(t)dt>1(integration from 0 to 1) },then fixed f in U

s.t.∫f(t)dt>1 and let ∫f(t)dt=r claim B(f,r)is contained in U need to show

for f' in B(f,r) then ∫f'(t)dt>1,but f' in B(f,r) means ||f'-f||∞ =

sup|f'-f|<r then i dont know how to get ∫f'(t)dt>1??

and hows about

the secound part of the question? should i conseder the compementary set???