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This very interesting question asks about 2 definitions of continuity

Definition.
Let $\Bbb{R}$ be the real numbers.
Let $\Bbb{R}^+$ be the positive real numbers.
Let $f: [a , b] \to \Bbb{R}$ be an $\Bbb{R}$-valued function on the closed-closed interval $[a,b] \subseteq \Bbb{R}$.
Let $\text{Dom}(f)$ be the domain of $f: [a , b] \to \Bbb{R}$.

Now the function $f: [a , b] \to \Bbb{R}$ is continuous at $a \in \operatorname{Dom}(f)$ iff

  1. $\forall \epsilon \in \Bbb{R}^+\ \exists \delta \in \Bbb{R}^+ \ \forall x\in \text{Dom}(f) \left< 0<|x-a|<\delta\implies |f(x)-f(a)|<\epsilon \right>\ $ or
  2. $\forall \epsilon \in \Bbb{R}^+\ \exists \delta \in \Bbb{R}^+ \ \forall x\in \text{Dom}(f) \left< |x-a|<\delta\implies |f(x)-f(a)|<\epsilon \right>\ $

(I think definition 0 is using punctured neighborhoods, meaning neighborhoods of $a$ that exclude $a$, and I'm curious as to what the consequences of this are.)

Questions.

  • What are some examples of functions in the "symmetric difference" of the 2 definitions? (The functions that are continuous under definition 0 and not continuous under definition 1, and the functions that are not continuous under definition 0 and continuous under definition 1.)
  • Is there a "characterization" of the functions in the "symmetric difference"?
  • What are some theorems/results in the "symmetric difference" of the 2 definitions? (Theorems/results that are true for continuous functions when using definition 0 and false when using definition 1, and theorems/results that are false for continuous functions when using definition 0 and true when using definition 1.)
  • In what ways does the development of calculus/analysis on $\Bbb{R}$ differ when using each definition?
K.defaoite
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étale-cohomology
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    If $|x-a|=0$, then $x=a$ and therefore also $|f(x)-f(a)|= 0$. Am I missing something? – NiklasvMoers Nov 04 '21 at 20:27
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    We would find the differences if there were a $ y \neq x $ such that $ d (x, y) = 0 $. That is, in $\mathbb{R}$ there is no difference if we equip it with a norm. More generally, there is no difference in any normed space. In non-normed spaces it can be interesting. – Martín Vacas Vignolo Nov 04 '21 at 20:29
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    What exactly is $| \cdot |$ supposed to mean if not a norm? In metric spaces, we have $d(x,y)=0$ iff $x=y$. – NiklasvMoers Nov 04 '21 at 20:34
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    The question you link to asks about two definitions of limit, not of continuity. While the two definitions would amount to different things for limits (which do not require the function to be defined at the point), they are equivalent for continuity, which does require the function to be defined at the point. Limits as $x\to a$ do not actually care what the value of $f$ is at $a$ (if any), while continuity at $a$ does. – Arturo Magidin Nov 04 '21 at 20:35
  • @NiklasvMoers the semi-norms are precisely those that allow there to be non-zero elements with semi-norm zero – Martín Vacas Vignolo Nov 04 '21 at 20:51

2 Answers2

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These definitions are equivalent, unlike the case for limits, as in the question you link. Limits do not inherently refer to $f(a)$, so allowing $|x-a|$ to vanish adds something in that case but not in this one.

Kevin Carlson
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Definition $0$ describes $\lim_{x\to a} f(x)=f(a)$ if $a$ is a limit point. Even if $a$ is not a limit point, this definition still works because we can choose $\delta>0$ such that $0<|x-a|<\delta$ never occurs for any $x\in Dom(f)$. Hence, the conclusion $|f(x)-f(a)|<\epsilon$ is vacuously true for "any such $x$."

Definition $1$ is equivalent because if $f(a)$ exists, it has to be true that $0=|a-a|<\delta \implies 0=|f(a)-f(a)|<\epsilon$.

ProfOak
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