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Metric space proof open and closed set

  1. Dec 2, 2012 #1
    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. jcsd
  3. Dec 2, 2012 #2
    This is not true: For [itex]f(x)=\frac{3}{2}, g(x)=\frac{1}{2}[/itex], [itex]\int_0^1f\ dx=\frac{3}{2}, \int_0^1g\ dx=\frac{1}{2}[/itex]. So [itex]f\in U, g\not\in U[/itex], [itex]\|f-g\|_{\infty}=1<\frac{3}{2}\Rightarrow g\in B(f, \frac{3}{2})[/itex].

    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 [itex](1,\infty)[/itex] is open in [itex]\mathbb{R}[/itex].

    There is a theorem about integrals of uniformly convergent sequences of functions that might prove useful.
     
  4. Dec 2, 2012 #3
    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?
     
  5. Dec 3, 2012 #4
     
  6. Dec 3, 2012 #5
    Which definition of "closed" are you attempting to use? State it precisely.
     
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