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Let $f$ be a real valued function on $\mathbb R$. Consider the functions $$ w_j(x) = \sup\left\{|f(u) − f(v)| : u, v \in \left[x − \frac{1}{j}, x + \frac{1}{j}\right]\right\} $$ where $j$ is a positive integer and $x\in\mathbb R$. Define next, $$ A_{j,n} = \left\{x\in\mathbb R : w_j(x) < \frac{1}{ n}\right\} \quad\text{and}\quad A_n = \bigcup_{j=1}^n A_{j,n} \quad n=1,2,\ldots $$ Now let $$ C = \{x \in\mathbb R : f \text{ is continuous at }x\} $$ how do we write $C$ in terms of $A_n$?

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What do you mean by wj(x) = sup{|f(u) − f(v)| : u, v ∈ [x − 1 j , x + 1 j ]} , specifically with [ x - 1 j , x + 1 j ]? And what about the last 'n' in Aj,n = {x ∈ R : wj(x) < 1 n}? – AndreasT Jan 18 '13 at 17:13
@AndreasT its x-1\j, x+1\j – anonymous Jan 18 '13 at 17:25
@anonymous like that? – Julian Kuelshammer Jan 18 '13 at 17:26
@JulianKuelshammer yes – anonymous Jan 18 '13 at 17:29
Are you sure that you don’t want $$A_n=\bigcup_{j=n}^\infty A_{j,n}\;?$$ – Brian M. Scott Jan 18 '13 at 18:06
up vote 1 down vote accepted

HINT: With the definitions revised so that $$A_n=\bigcup_{j\ge 1}A_{j,n}\;,$$ we have $x\in A_n$ if and only if there is some $j\in\Bbb Z^+$ such that $x\in A_{j,n}$. Intuitively speaking, this says that $x\in A_n$ if and only if $|f(u)-f(v)|<\frac1n$ whenever $u$ and $v$ are sufficiently close to $x$. (It’s a little more complicated than that, since a set of real numbers less than $\frac1n$ doesn’t necessarily have a supremum strictly less than $\frac1n$, but that’s good enough to get the right intuition.) If we could say that for every positive integer $n$, it would sound a lot like saying that $f$ is continuous at $x$. That suggests that you should try to prove that $$C=\bigcap_{n\ge 1}A_n\;.\tag{1}$$

Once you get the idea that $(1)$ ought to be true, proving it is pretty straightforward. Show that if $f$ is continuous at $x$, then $x\in A_n$ for every $n\in\Bbb Z^+$. Then show that if $f$ is not continuous at $x$, there is an $n\in\Bbb Z^+$ such that $x\notin A_n$.

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