Metric space proof open and closed set

[cummings12332]

Homework Statement

show the set {f: ∫f(t)dt>1(integration from 0 to 1) } is an open set in the metric space ( C[0,1],||.||∞）

and if A is the subset of C[0,1] defined by A={f:0<=f<=1} is closed in the norm ||.||∞ norm.

Homework Equations

C[0,1] is f is continuous from 0 to 1.and ||.||∞ is the norm that ||f||∞ =sup | f|

The Attempt at a Solution

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 don't know how to get ∫f'(t)dt>1??

and hows about the secound part of the question? should i conseder the compementary set?

 

[gopher_p]

claim B(f,r)is contained in U

This is not true: For f(x)=\frac{3}{2}, g(x)=\frac{1}{2}, \int_0^1f\ dx=\frac{3}{2}, \int_0^1g\ dx=\frac{1}{2}. So f\in U, g\not\in U, \|f-g\|_{\infty}=1&lt;\frac{3}{2}\Rightarrow g\in B(f, \frac{3}{2}).

You need to adjust the radius of you ball. Think about how you would show (in a way similar to the way that you're attempting) that (1,\infty) is open in \mathbb{R}.

and how about the second part of the question?

There is a theorem about integrals of uniformly convergent sequences of functions that might prove useful.

 

[cummings12332]

This is not true: For f(x)=\frac{3}{2}, g(x)=\frac{1}{2}, \int_0^1f\ dx=\frac{3}{2}, \int_0^1g\ dx=\frac{1}{2}. So f\in U, g\not\in U, \|f-g\|_{\infty}=1&lt;\frac{3}{2}\Rightarrow g\in B(f, \frac{3}{2}).

You need to adjust the radius of you ball. Think about how you would show (in a way similar to the way that you're attempting) that (1,\infty) is open in \mathbb{R}.



There is a theorem about integrals of uniformly convergent sequences of functions that might prove useful.

Many thanks, I have solved the first part out. but for secound part i still don't know how to begin. should i choose a sequence fn in A then check that ||fn-f||<esillope,then what should i do for it?could u give me more details?

 

[cummings12332]

Many thanks, I have solved the first part out. but for secound part ,my idea is choose f in the norm space C[0,1],||.||. then choose a sequence fn in C[0,1] then fn->f for 0<fn<1 then we have 0<f<1 so C[0,1] is closed , does it make sense?QUOTE]

 

[gopher_p]

Which definition of "closed" are you attempting to use? State it precisely.