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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?

Joe
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3 Answers3

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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 $|f(x_0)-f(c)|\geq \epsilon_0$.

cmk
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  • 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 – Joe Jun 11 '19 at 17:41
  • 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. – cmk Jun 11 '19 at 17:43
  • Shall I replace it with $x_0 \in \mathbf{R}$? – Joe Jun 11 '19 at 17:44
  • That would be a good start! – cmk Jun 11 '19 at 17:45
  • Please look at my second edit... Now is everything all right? – Joe Jun 11 '19 at 17:52
  • It's very close! You say $|x-x_0|<\delta$ before saying that $x$ exists, so those statements should flip. Also, I'm not sure what you mean by $\delta>0$ near $x_0$. Presumably, you mean a ball around $x_0,$ but that's what the second statement is about; $\delta$ should prescribe the radius. I think this is what you're trying to get at, though. I also think you're trying to say for all $\delta$ small enough (all radii small enough), but in fact we can consider any $\delta>0$ (although the case of $\delta$ small is what we're really interested in). – cmk Jun 11 '19 at 18:00
  • @Joe also, I don't know what you mean by"discontinuity of $a$. – cmk Jun 11 '19 at 18:02
  • There is a sudden jump of $a$ (at $x_0$) in the function. That is what discontinuity means, right? – Joe Jun 11 '19 at 18:06
  • Okay, I see what you're trying to say. We don't usually quantify the jump, and in fact you can have discontinuities without jumps (see https://en.wikipedia.org/wiki/Classification_of_discontinuities). We say that there is a discontinuity at a. – cmk Jun 11 '19 at 18:08
  • $|x-x_0|<\delta$ says about the set of different values of $x$ we are interested in. And from those values of $x$ we say there exists an $x$ such that $|f(x)-f(x_0)| \geq a$. So why is there a need to flip the statements. – Joe Jun 11 '19 at 18:19
  • You said "whenever $|x-x_0|<\delta,$ there exists an $x$ such that..." I'm saying this is strange because you say $|x-x_0|<\delta$ before introducing $x$. – cmk Jun 11 '19 at 18:21
  • Also, it comes down to the logical structure. For continuity, it is $(\forall \epsilon>0)(\exists\delta>0)(\forall x)(|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon).$ Negating this gives $(\exists \epsilon_0>0)(\forall\delta>0)(\exists x)(|x-x_0|<\delta \wedge |f(x)-f(x_0)|\geq \epsilon_0)$. – cmk Jun 11 '19 at 18:24
  • I have introduced $x$ in the statement: "For a function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$" Or shall I need to say "For a function $f(x)$ of $x$ from $\mathbf{R}$ to $\mathbf{R}$" in the beginning. – Joe Jun 11 '19 at 18:24
  • You just say a function $f$. We don't specify $x$, yet. You specified $x_0$ at the beginning, so the function will be discontinuous at $x_0$. We don't specify anything else. – cmk Jun 11 '19 at 18:26
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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.
José Carlos Santos
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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.

auscrypt
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