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It is perhaps well known that the sign function is discontinuous, if defined for $f:\mathbb{R}\rightarrow \mathbb{R}$. However, if we were to define the sign function for $f:\mathbb{R} \setminus \left \{ 0 \right \}\rightarrow \mathbb{R}$, would the sign function still remain discontinuous?

My belief is yes simply because at any given $x_{0}>0$ or $x_{0}<0$ the function will still not be continuous by virtue of the epsilon-delta proof (there exist a value of $\varepsilon$ where $\left |f(x)-f(x_{0}) \right | < \varepsilon $ is not satisfied.) Is my reasoning correct?

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Where is the delta in your epsilon-delta proof?? – Rahul Nov 20 '12 at 21:11
up vote 8 down vote accepted

Any function becomes continuous if you remove its points of discontinuity from the domain.

In your case, the only discontinuity is at $0$, so by removing $0$ from the domain you make the function continuous.

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Consider, within the set R\{0}, the sequence (a_n) defined by a_n = 1/n. Then the sequence is Cauchy but does not converge, showing that R\{0} is incomplete. Is it the property of incompleteness of R\{0} which allows the horrible continuity above to happen? – Adam Rubinson Nov 21 '12 at 12:16
No, it's the fact that the set has two connected components. The function is continous on $[-5,-4]\cup[1,2]$, which is complete. – Martin Argerami Nov 21 '12 at 13:39
I see. Thanks Martin – Adam Rubinson Nov 21 '12 at 13:49

If we expell $0$ from the domain, the sign function becomes continuous (in fact, it is even locally constant). For $x\ne 0$ (and $\epsilon>0$) pick $\delta=|x|$. Then you have $f(y)=f(x)$ for all $y$ with $|y-x|<\delta$.

More trivia: Contrary to popular belief, the function given by $f(x)=\frac1x$ is continuous (on its domain of definition)! It is simply not defined where it would be discontinuous.

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If we restrict the domain to exclude the point $0$, the function becomes continuous. This is easily verified using the definition of continuity. Let $\epsilon>0$ be given. For any $x\neq 0$, simply take $\delta=|x|$. Then any $y$ with $|y-x|<\delta$ will be such that $f(y)=f(x)$ so that $|f(y)-f(x)|=0<\epsilon$.

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Good job, Will Hunting! – amWhy Nov 21 '12 at 0:02

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