# Continuity and Subsequences

Suppose that $f$ is continuous on $[a,b]$. Prove that given $\epsilon>0$, there exist points $x_0=a<x_1<...<x_n=b$ such that if

$E_k=\{y: f(x)=y\ for\ some\ x \in [x_{k-1}, x_k]\}$,

then sup$E_k -$inf$E_k<\epsilon$ for $k=1,2,...,n$.

Just from looking at the question, I suspect that I need to eventually the Bolzano Weierstrass Theorem and Extreme Value Theorem to complete this proof. However, I am having some difficulty setting up the problem so that I can use theorems as needed. Please help! I really appreciate any guidance that I can get.

-
Solving the proof???????????????? How is this possible? –  FromCuba Oct 28 '12 at 8:10
@LJym89,not everyone speaks English as their first language. –  user43081 Oct 28 '12 at 8:13

## 1 Answer

Notice that what the question asks is that continuous functions are Riemann-integrable.

If you have the theorem that states that continuous functions defined on compact sets are uniformly continuous, then it suffices you to choose $$a = x_0 < x_1 < \dots < x_n = b$$ such that $$|x_{i+1} - x_i| < \delta, i = 0, 1, \dots, n-1$$ where $$\forall \varepsilon > 0,\quad \exists \delta > 0,\quad \forall x,y \in [a,b], \quad |x-y| < \delta \quad \Rightarrow \quad |f(x) - f(y)| < \varepsilon,$$ i.e. the definition of uniform continuity. I believe this is a sufficiently good hint, if you want more details you can ask me again.

Hope that helps,

-
Thank you so much! The hint was actually really helpful and led me in exactly the right direction. I combined the Extreme Value Theorem and the theorem that you stated (which I did know), and I was able to find the proof relatively quickly. Thank you again! –  Jess Oct 28 '12 at 8:47
@Jess : How did you use the EVT? I'm curious because I don't need it. –  Patrick Da Silva Oct 29 '12 at 18:14
I used the EVT to show that because $[x_{k-1},x_{k}]$ is a closed, bounded interval and f is continuous on $[x_{k-1},x_{k}]$, f is bound on $[x_{k-1},x_{k}]$ and $\exists x_{M}, x_{m} \in [x_{k-1},x_{k}] \ni f(x_{M})=M=supE_{k} and f(x_{m})=m=infE_{k}$ Then, I stated that as $[x_{k-1},x_{k}]$ is a closed, bounded interval and f is continuous on $[x_{k-1},x_{k}]$, f is uniformly continuous on that interval. I then used the definition of uniform continuity to show that $|f(x_{M})-f(x_{m})|=|sup(E_{k})-inf(E_{k})|<\epsilon$ –  Jess Nov 11 '12 at 5:31
The problem with your argument is that the last line works if the $\delta_k$ for uniform continuity is such that $|x_M - x_m| < \delta_k$, which nothing ensures you that that will happen. You need to do the exact same reasoning from the start, but begin with the interval $[a,b]$ instead ; if you choose your partition with $x_{k-1} - x_k < \delta$ for all $k$, then uniform continuity over $[a,b]$ will ensure you that $|\mathrm{sup} - \mathrm{inf}| < \varepsilon/(b-a)$, so that when you compute the Riemann sum, it is smaller than $(b-a) \varepsilon/(b-a) = \varepsilon$. –  Patrick Da Silva Nov 11 '12 at 20:21