1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proving something is closed

  1. Jul 14, 2011 #1
    1. The problem statement, all variables and given/known data
    prove that the set of continous functions on [0,1] that are increasing is a closed set.

    2. Relevant equations

    3. The attempt at a solution
    Need to prove the complement is open. So need to prove the set of continous functions on [0,1] that are non increasing is open. Let f be a continous function that is non increasing on [0,1]. need to choose an epsilon st B(f)={g in C([0,1]): sup|g(x)-f(x)|<epsilon} is contained in the set of continous functions on [0,1] that are non increasing.
  2. jcsd
  3. Jul 14, 2011 #2
    Hi l888l888l888! :smile:

    So f is nonincreasing. So let [a,b] be an interval such that f(b)<f(a). How close must a function g be such that also g(b)<g(a)??

    As a further hint, you must find an [itex]\varepsilon >0[/itex] such that every element in [itex][f(b)-\varepsilon,f(b)+\varepsilon][/itex] is smaller than [itex]f(a)-\varepsilon[/itex], do you see why?
  4. Jul 14, 2011 #3
    well i narrowed it down to a few cases. assuming we have an interval [f(b),f(a)] ...
    if g(b)<f(b) then |g(a)-f(b)| has to be less than |g(b)-f(b)|. if f(b)<g(b)<f(a) then |g(a)-f(a)| has to be less than |g(b)-f(a)| and the last case for g(b)> f(a) i have not figured out. im basically going about this by looking at this graphically and trying to write it out as a proof
  5. Jul 14, 2011 #4
    You have to eliminate the last case by choosing [itex]\varepsilon[/itex] small enough...
  6. Jul 15, 2011 #5
    well in answer to the hint. in order for everything in [f(b)-epsilon, f(b)+epsilon] to be less than f(a)-epsilon, f(b)+epsilon has to be less than f(a)-epsilon. rearranging this inequality we get f(a)-f(b)>2epsilon. ==> (f(a)-f(b))/2 > epsilon ==> epsilon + some number c = (f(a)-f(b))/2 ==> epsilon= (f(a)-f(b))/2 - c.
  7. Jul 15, 2011 #6
    Indeed, this epsilon will make it all work! Btw, it's perfectly ok to say to choose [itex]\varepsilon<\frac{f(a)-f(b)}{2}[/itex]. You don't need to introduce the number c, but it's not wrong if you do.
  8. Jul 15, 2011 #7
    oh ok thanks so much. Your so helpful!!!!!!!!!!!! :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook