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Let the $\kappa$-oscillation of the set be : $\{x \in [a,b]: \text{osc} f \ge \kappa$}

How do we show that it is a closed set? Do we prove its complement is open? If so, what would one call its compliment?

I am totally lost here. If anyone can propose a proof, I will be grateful.

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Its complement (note spelling) is simply $\big\{x\in[a,b]:\operatorname{osc}f<\kappa\big\}$. And yes, proving that this is open is a very reasonable approach. –  Brian M. Scott Jan 28 '13 at 3:59
    
Exactly what definition of oscillation are you using? –  Brian M. Scott Jan 28 '13 at 4:12
    
Thanks for noting the spelling, Brian. But I am lost as to how to go about it. Can you point me to the right direction? Also I was think whether I can use the fact that a closed set contains all its limit points. So I have to construct a sequence (I do not know what it would look like) such that is $x$ is a limit point of a sequence in the set, then $x\in$ set. –  user43901 Jan 28 '13 at 4:12
    
the definition : osc $f = \lim_{r \to 0} \text{diam} f [x+r,x-r]$ –  user43901 Jan 28 '13 at 4:14
    
Theorem 6.27, p.259 in Bruckner, Bruckner, Thomson: Elementary Real Analysis. The book is freely available here. –  Martin Sleziak Jan 28 '13 at 6:26

2 Answers 2

HINT: Suppose that the oscillation of $f$ at $x$ is less than $\kappa$. This means that

$$\lim_{r\to 0}\operatorname{diam}f\big[[x-r,x+r]\big]<\kappa\;.$$

By the definition of limit there is some $r_0>0$ such that $\operatorname{diam}f\big[[x-r,x+r]\big]<\kappa$ for all positive $r\le r_0$. Suppose that $|x-y|<\frac{r_0}2$, then $$\left[y-\frac{r_0}2,y+\frac{r_0}2\right]\subseteq[x-r_0,x+r_0]\;.$$

Can you now use that to show that the oscillation of $f$ at $y$ is less than $\kappa$ and hence that the complement of your set is open?

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Let $A=\{ x\in [a,b] : \text{osc} f \geq \kappa \}$ and take a sequence $x_n\in A$ such that $x_n\to x$. We want to prove $x\in A$. To do this pick an $r>0$ and $x_n$ such that $|x_n-x|<r/2$. Then $[x_n-r/2,x_n+r/2] \subset [x-r,x+r]$ and so $$ \kappa\leq\text{diam}f([x_n-r/2,x_n+r/2]) \leq \text{diam}f([x-r,x+r]) $$ Take the limit and conclude.

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