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I think I have made a rigorous and intuitive definition of discontinuity.

Definition: For a function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$, there is a discontinuity of $a$ at $x_0 \in \mathbf{R}$ if for all $\delta > 0$ around $x_0$; whenever $|x-x_0|<\delta$, there exists an $x$ such that $|f(x)-f(x_0)|$ $\geq a$

How much rigor is my definition?

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  • $\begingroup$ "There is a discontinuity of $\epsilon_1$..." I have never heard anyone use that phrase. Also, "For all $\delta$ around $x_0$" feels wrong too. $\endgroup$
    – JMoravitz
    Jun 11, 2019 at 17:27
  • $\begingroup$ Alternatively, just look at the definition of continuity. If it is not met, then you have a discontinuity. They are mutually exclusive. $\endgroup$
    – The Count
    Jun 11, 2019 at 17:28
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    $\begingroup$ Possible duplicate of The negation of a limit condition, Spivak style.. The answers here are exceptional, please take a look. $\endgroup$
    – graeme
    Jun 11, 2019 at 17:39

3 Answers 3

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Your definition is kind of close, if I'm being generous. To get a full, rigorous notion of discontinuity, you just need to negate the definition of continuity: a function is discontinuous at $x_0$ if for some $\epsilon_0>0$ and any $\delta>0,$ there exists $c$ so that both $|x_0-c|<\delta$ and $|f(x_0)-f(c)|\geq \epsilon_0$.

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  • $\begingroup$ I have edited in accordance with what Jose said in his answer. Please take a look at the edit and say in what way it is not close $\endgroup$
    – Joe
    Jun 11, 2019 at 17:41
  • $\begingroup$ What do you mean by $x_0\in x?$ Or discontinuity of $a$? Pretty much, if you formulate it correctly, it will look exactly like what I wrote. $\endgroup$
    – cmk
    Jun 11, 2019 at 17:43
  • $\begingroup$ Shall I replace it with $x_0 \in \mathbf{R}$? $\endgroup$
    – Joe
    Jun 11, 2019 at 17:44
  • $\begingroup$ That would be a good start! $\endgroup$
    – cmk
    Jun 11, 2019 at 17:45
  • $\begingroup$ Please look at my second edit... Now is everything all right? $\endgroup$
    – Joe
    Jun 11, 2019 at 17:52
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There is zero rigor, because:

  1. “For all $\delta$ around $x_0$” means nothing.
  2. You don't say whar $x_0$ is.
  3. The expression “the greatest value of $\lvert f(x)−f(x_0)\rvert$” is undefined.
  4. The number $\varepsilon_1$ is undefined.
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Your definition doesn’t quite work; “largest” may not exist (what if there is an infinite sequence approaching a bound, for example). Using maximum won’t work, you’d need to use supremum. You could simply replace “greatest value of” with “there exists an $x$ such that”, and then your definition works.

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  • $\begingroup$ Please look at my edit... Now is everything all right? $\endgroup$
    – Joe
    Jun 11, 2019 at 17:52
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    $\begingroup$ The gist is correct. The maths is so garbled it is unusable. $\endgroup$
    – auscrypt
    Jun 11, 2019 at 17:55
  • $\begingroup$ Now what makes it that garbled? I have explained everything simply $\endgroup$
    – Joe
    Jun 11, 2019 at 17:57

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