Is the Set of Increasing Continuous Functions on [0,1] Closed?

[l888l888l888]

Homework Statement

prove that the set of continuous functions on [0,1] that are increasing is a closed set.

Homework Equations

The Attempt at a Solution

Need to prove the complement is open. So need to prove the set of continuous functions on [0,1] that are non increasing is open. Let f be a continuous 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 continuous functions on [0,1] that are non increasing.

 

[micromass]

Hi l888l888l888!

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 \varepsilon &gt;0 such that every element in [f(b)-\varepsilon,f(b)+\varepsilon] is smaller than f(a)-\varepsilon, do you see why?

 

[l888l888l888]

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. I am basically going about this by looking at this graphically and trying to write it out as a proof

 

[micromass]

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. I am basically going about this by looking at this graphically and trying to write it out as a proof

You have to eliminate the last case by choosing \varepsilon small enough...

 

[l888l888l888]

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.

 

[micromass]

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.

Indeed, this epsilon will make it all work! Btw, it's perfectly ok to say to choose \varepsilon&lt;\frac{f(a)-f(b)}{2}. You don't need to introduce the number c, but it's not wrong if you do.

 

[l888l888l888]

oh ok thanks so much. Your so helpful! :)