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I want to show that if $f$ has jump $0$ then $f(x):A \subseteq \mathbb{R} \rightarrow \mathbb{R} $ is continuous at $c$ from where the jump of $f$ at $x=c\in$ Dom$(f(x))$ defined by

$$o(f, c)=\lim \limits_{\delta \to 0^{+}} (M_{f}(\delta)-m_{f}(\delta))$$

from where

\begin{align*} m_{f}(\delta)&= \inf \{f(x):x\in A, 0<|x-c|<\delta\}\\ M_{f}(\delta)&= \sup \{f(x):x\in A, 0<|x-c|<\delta\} \end{align*}

Assuming that the jump has a value of $0$, then given $\epsilon$ there is a $\delta$ such that if $x\in A$ and $\delta>x>0$ then $|M(f,b,\delta)-m(f,b,\delta)|<\epsilon$; Suppose that for all $x$, $c\in A$, $|x-c|<\delta$.

Now, I don't know how to proceed at all I guess I can come up with something along the lines of

$$|f(x)-f(y)|<M_{f}(\delta)-f(c)<\epsilon$$

The last inequality using the approximation property of the supremum, but how do I arrive at the first inequality? or is there an alternative way to achieve the result? I appreciate any help!

MathFail
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Wrloord
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1 Answers1

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It is false. Let $f(x)=1, x\in (-\infty, 0)\cup (0,\infty)$, and $f(0)=2$, we have

$m_{f}(\delta)= \inf \{f(x):x\in A, 0<|x-0|<\delta\}=1$

$M_{f}(\delta)= \sup \{f(x):x\in A, 0<|x-0|<\delta\}=1$

But $f$ is not continuous at $x=0$.

MathFail
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  • Well, an alternative demonstration seems to be found here https://math.stackexchange.com/questions/504532/prove-that-f-is-continuous-if-and-only-if-textoscf-a-0. – Wrloord May 23 '23 at 01:02
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    But that problem wrote: "(Note: $B_r(a) = {x: |x - a| < r}).$", you wrote $0<|x-c|<\delta$ – MathFail May 23 '23 at 01:08
  • It seems that the implication I mentioned is false after all, but the opposite implication is true, isn't it? – Wrloord May 24 '23 at 14:48