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Say we have a function

$ f = \dfrac{1}{\arctan|x|^3} $

If we add to that definition with

$ f(0) = +\infty $

Can $ f$ now be considered continuous? I'm assuming you can't just say that function equals infinity at one point.

If we can't do that, is there any way to add to the definition of the function to make it continuous in $0$?

amWhy
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Luka Horvat
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2 Answers2

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Thanks to the absolute value, $\lim_{x\to 0} f(x) = \infty$ (as $x$ approaches $0$ from either side). Yes, you can make your function go from $\mathbb{R}$ to the "extended real numbers" $\{-\infty\} \cup \mathbf{R}\cup \{\infty\}$, a topological space that is homeomorphic to $[0,1]$, using a topology that should be pretty obvious. Then if you define $f(0) = \infty$, your function is continuous at $0$.

Stefan Smith
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  • The function in question is actually defined as $f:R{0}->R$ Would in that case the above still be possible? – Luka Horvat Apr 07 '13 at 22:57
  • Without the absolute value in the expression for $f$, we would have $\lim_{x\to 0^-}f(x)=-\infty$, so this would only work if we use the one-point compactification of $\mathbb R$, i.e. identify $+\infty$ and $-\infty$, making essentially a circle instead of an interval. – Hagen von Eitzen Apr 07 '13 at 22:57
  • @Darwin : What I propose is extending the domain of $f$ to $\mathbf{R}$. You also have to extend the target set. There's no other way to make $f$ continuous at $0$, because $f(x) \to \infty$ as $x \to 0$. And you don't actually have to add $-\infty$ to the target space, just $\infty$. – Stefan Smith Apr 07 '13 at 23:09
  • I understand that. What I'm trying to figure out is if the extension of the target set would be permitted in the problem I'm trying to solve. It obviously can be extended, but I'm not sure if the problem setter had that in mind. I'd just assume the answer is no, but then there's a follow up question in case the first answer was yes. Kind of confused. – Luka Horvat Apr 07 '13 at 23:14
  • @Darwin : I don't know who asked you the question and what they had in mind. Can you tell me? And what is the followup question? If someone asked me this question in the middle of a freshman calculus class, I would definitely not answer the question the same way - I would just say no, you can't make $f$ continuous at $0$. Allowing functions to take the value $\infty$ and mentioning topology would just confuse the students. If a genuinely curious and talented student who was acing the class asked me this question in my office, I might give him/her an answer like the one I gave you. – Stefan Smith Apr 07 '13 at 23:35
  • Yeah. Very sorry for giving so little info. I understand what you mean. I'm a student. First year of mathematics (minor? I don't really know how our education translates to other countries). The follow up is "Is that expansion of class $ C^1(R) $?". – Luka Horvat Apr 07 '13 at 23:39
  • @Darwin : that followup question is pretty hard. I think that for $f$ to be in $C^1(\mathbf{R})$, $f'(x)$ has to exist and be a finite real number for all real $x$, and $f'$ has to be continuous. I'm not sure whether the definition of $C^1(\mathbf{R})$ even allows $f$ to take infinite values. But you try to evaluate $f'(0)$ (with $f(0)$ defined to be $\infty$) using the definition, you get $\infty$ on one side and $-\infty$ on the other side. So $f'(0)$ is undefined (even if you allow infinite values). So $f$ is definitely not in $C^1(\mathbf{R})$. – Stefan Smith Apr 07 '13 at 23:47
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It is better to specify the function domain, thus, for reals we will have non removable discontinuity in $x=0$ when $f(x)=∞$. So, the function will be semi-continuous, but by playing with a domain of the function we can achive continuity, for example by extending reals with infinity itself as it was mentioned above.

rook
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