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Homework Help: Proving a set in closed and nowhere dense

  1. Jul 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Dn={f in C([0,1]) : there exists t in [0,1] for every h in R/{0}, abs((f(t+h)-f(t))/h) <=n}
    prove the Dn's are closed nowhere dense sets. A subset of some set A is closed in A if its complement is open in A.



    2. Relevant equations



    3. The attempt at a solution
    i first defined the complement of Dn as the set in which for all t in [0,1], abs((f(t+h)-f(t))/h)>n. Is trying to prove closed first and then nowhere dense.
     
  2. jcsd
  3. Jul 4, 2011 #2
    Hi l888l888l888! :smile:

    Let's prove that Dn is open. Pick an f in Dn. You know that

    [tex]|f(t+h)-f(t)|>n|h|[/tex]

    You'll need to find an epsilon such that for all g with [itex]\|f-g\|_\infty[/itex] holds that

    [tex]|g(t+h)-g(t)|>n|h|[/tex]

    Start by working out

    [tex]|f(t+h)-f(t)|=|(f(t+h)-g(t+h))+(g(t+h)-g(t))+(g(t)-f(t))|[/tex]
     
  4. Jul 4, 2011 #3
    i think you meant prove Dn is closed. but anyway...
    f(t+h)−f(t)|=|(f(t+h)−g(t+h))+(g(t+h)−g(t))+(g(t)−f(t))|
    <=|f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)|. is that what you meant?
     
  5. Jul 4, 2011 #4
    Yes, I meant to prove that the complement of Dn is open.

    Yes, and you know that

    [tex]n|h|<|f(t+h)-f(t)|[/tex]

    and [itex]\|f-g\|_\infty<\epsilon[/itex]. So what can you deduce from this?
     
  6. Jul 4, 2011 #5
    well if n|h|<|f(t+h)−f(t)| and if |f(t+h)−f(t)|<=|f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)| then n|h|< |f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)|. and if sup |f-g|< epsilon then n|h|< |f(t+h)−g(t+h)| + |g(t+h)−g(t)| + |g(t)−f(t)|< 2epsilon + |g(t+h)−g(t)|?
     
  7. Jul 4, 2011 #6
    Indeed, but let's refine this result. So you know that

    [tex]n|h|<|f(t)+f(t+h)|[/tex]

    So there is actually a number a such that

    [tex]n|h|+a\leq|f(t)+f(t+h)|[/tex]

    So, by the same reasoning, you have

    [tex]n|h|+a<2\epsilon+|g(t+h)-g(t)|[/tex]

    So, can you find an epsilon small enough such that

    [tex]n|h|<|g(t+h)-g(t)|[/tex]
     
  8. Jul 4, 2011 #7
    Right! to be more specific are we letting epsilon to be a/2? also for the nowhere dense part. there are many theorems to use in my book about proving something is nowhere dense. for instance we have " a subset is nowhere dense if int(cl(H)) = empty set or iff T- Cl(H) is dense. which shall i use or do u suggest to use another theorem?
     
  9. Jul 4, 2011 #8
    Indeed.

    Since H is closed, you'll only need to prove that int(H) is closed or that T/H is dense.
    So, what do we need to prove?? We need to show that very every function f, there is a function g such that [itex]\|f-g\|_\infty[/itex] and such that

    [tex]\frac{|(g(t+h)-g(t)|}{|h|}\geq n[/tex]

    Can you see graphically what must happen? We need to find a function g very close to f such that the slope between two close points t and t+h is big.

    Let's do this for a simple case first: let's take f=0. Then we need to find a function g which is always close to zero, but the slope between two very close points must become large. Can you find me such a function?
     
  10. Jul 4, 2011 #9
    so what your saying is basically we need g=mx+b where m the slope is sufficiently large but g~0 (g is close to 0). so mx+b~0.....??? graphically wen i draw it out its hard to explain it on here. g has to intercept the y axis at a negative number????
     
  11. Jul 4, 2011 #10
    Hmm, why do you say that we need g=mx+b?? All we need is that for every point t, there is a point t+h, such that if we draw the line [g(t),g(t+h)] then it's slope is sufficiently large.

    A way to accomplish this is to let g occilate fast...
     
  12. Jul 4, 2011 #11
    AH! Ok so considering a function g that oscillates about the x axis (y=0)? now what?
     
  13. Jul 4, 2011 #12
    Now try to do this for a general f.
     
  14. Jul 4, 2011 #13
    so for every function f there is a functions g that oscillates about f and the slope between g(t+h) and g(t) is sufficiently large and g is still close to f. Also, when you said,
    "We need to show that very every function f, there is a function g such that ∥f−g∥∞ and such that

    |(g(t+h)−g(t)|/|h|≥n" did you mean there is a function g st ||f-g||<some epsilon. that would be the closeness of f and g you are refering to.
     
  15. Jul 4, 2011 #14
    Yes, for every epsilon, there is a function g such that [itex]\|f-g\|_\infty[/itex] and such that

    [tex]\frac{|g(t+h)-g(t)|}{|h|}\geq n[/tex].

    And the function g which we will take is a function that oscillates fast about f.
     
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