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Let $f$ be a real valued function on $\mathbb{R}$. Consider the functions
$$ w_j(x) := \sup\left\{\vert f(u) − f (v)\vert : u,v \in \left[x−\frac{1}{j},x+\frac{1}{j}\right]\right\} $$ where $j$ is a positive integer and $x\in\mathbb{R}$. Then define $$ A_{j,n}: = \left\{x \in \mathbb{R}: w_j(x)<\frac{1}{n}\right\},\qquad n=1,2,\ldots $$ and $$ A_n:=\bigcup_{j=1}^\infty A_{j,n}, \qquad n = 1,2,\ldots $$
Now let $C=\{x \in \mathbb{R}:f \text{ is continuous at }x \}$. Express $C$ in terms of the sets $A_n$.

I am totally confused. Please help me.

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  Julian Kuelshammer Jan 1 '13 at 11:04
What have you tried? Perhaps you can start by noting that if $x$ is a point of continuity, then $\lim_{j\rightarrow\infty}w_j(x) = 0$. Can you prove that this implies: for every $n$ there exists $j$ such that $x\in A_{j,n}$? If the last assertion holds, then $C$ must be a subset of $\cap_n A_n$. Then try to show the other inclusion. –  William Jan 1 '13 at 12:28
will you explain please.i am still not getting it. –  abdakchi Jan 1 '13 at 13:37

1 Answer 1

Let $x \in C$, i.e. $f$ is continuous in $x$. We want to show that $$x \in \bigcap_{n \in \mathbb{N}} A_n \tag{1}$$

Let $\varepsilon>0$. By definition of continuity there exists $\delta(\varepsilon)>0$ such that for all $y \in \mathbb{R}$, $|y-x| \leq \delta(\varepsilon)$ $$|f(y)-f(x)| \leq \varepsilon$$

Now let $u,v \in \mathbb{R}$ such that $|u-x|< \delta(\varepsilon)$, $|v-x| \leq \delta(\varepsilon)$. By triangular inequality:

$$|f(u)-f(v)| \leq |f(u)-f(x)|+|f(x)-f(v)| \leq 2 \varepsilon$$

In particular we can choose $\varepsilon := \frac{1}{2n}$, then $$|f(u)-f(v)| \leq \frac{1}{n}$$ for all $u,v$ such that $|u-x|< \delta(\frac{1}{2n})$, $|v-x| \leq \delta(\frac{1}{2n})$. Now choose $j(n) \in \mathbb{N}$ large enough such that $\frac{1}{j(n)} \leq \delta \left( \frac{1}{2n} \right)$. Then you can easily see that $x \in A_{j(n),n}$. This works for all $n \in \mathbb{N}$, so $(1)$ is proved.

Now let $x \in \bigcap_{n \in \mathbb{N}} A_n$. We want to prove that $f$ is continuous in $x$ (thus $x \in C$). Use the definitions and similar argumentation as above (So let $\varepsilon>0$. Then there exists $n \in \mathbb{N}$ such that $\frac{1}{n} \leq \varepsilon$ and since $x \in A_n$ ... )

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