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Let $f$ be a real valued function on $\mathbb{R}$ define $$w_j(x)=\sup\{|f(u)-f(v)|: u,v\in [x-1/j,x+1/j]\}$$ $j\in \mathbb{N}$ and $x\in\mathbb{R}$, Define next $$A_{j,n}=\{x\in\mathbb{R}:w_j(x)<1/n\}$$ $n=1,2,\dots$ and $$A_n=\bigcup_{j=1}^{\infty}A_{j,n}$$ Now let $$C=\{x\in\mathbb{R}: f \text{ is continuos at }x\}$$ How to express $C$ interms of $A_n$? Well, according to the definition, $f$ iss continuos at $x$ iff $w_j(x)=0$ (I guess, I can do that by definition of continuity).

Then I guess, $C=\bigcap A_n$. I am not sure. Thank you.

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up vote 2 down vote accepted

$\def\abs#1{\left|#1\right|}$ Hi, you are right about $C$. We have for $x \in \mathbb R$: \begin{align*} f \text{ is continuous at } x &\iff \forall n \exists j \forall y: \abs{x-y} \le \frac 1j \Rightarrow \abs{f(x) - f(y)} < \frac 1n\\ &\iff \forall n \exists j :\sup_{\abs{x-y} \le \frac 1j} \abs{f(x)-f(y)} < \frac 1n\\ &\iff \forall n \exists j: w_j(x) < \frac 1n\\ &\iff \forall n \exists j :x \in A_{j,n}\\ &\iff x \in \bigcap_n \bigcup_j A_{j,n} \end{align*}

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thank you martini for making thessse precise :) – Un Chien Andalou Jul 10 '12 at 10:39

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