# Homework Help: Proving something is closed

1. Jul 14, 2011

### l888l888l888

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. Jul 14, 2011

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

3. Jul 14, 2011

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

4. Jul 14, 2011

### micromass

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

5. Jul 15, 2011

### 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.

6. Jul 15, 2011

### micromass

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

7. Jul 15, 2011