# Metric space proof open and closed set

1. Dec 2, 2012

### cummings12332

1. The problem statement, all variables and given/known data
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.

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

3. 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 dont know how to get ∫f'(t)dt>1??

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

2. Dec 2, 2012

### gopher_p

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 $f\in U, g\not\in U$, $\|f-g\|_{\infty}=1<\frac{3}{2}\Rightarrow g\in B(f, \frac{3}{2})$.

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.

3. Dec 2, 2012

### cummings12332

Many thanks, I have solved the first part out. but for secound part i still dont 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?

4. Dec 3, 2012

5. Dec 3, 2012

### gopher_p

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