I'll do this with an example, although I don't know how to do the big fancy piecewise bracket curly brace thing:

Consider function $f$ where $$f(x)=\begin{cases} -1 & \text{if }x \leq 0,\\ 1&\text{if }x > 0. \end{cases}$$

And a slight modification of the same function:

$$g(x)=\begin{cases} -1 & \text{if }x < 0,\\ 1&\text{if }x > 0. \end{cases}$$The only difference between $f$ and $g$ is that $0$ is excluded from the domain of $g$. Here is a picture to show both graphs in case that is easier:

enter image description here

My question:

For $f(x)$, since $f(0)$ is defined, we can say $0$ is part of $f$'s domain. However, the two-sided limit does not exist at $x=0$, which means we can say $f$ is discontinuous at $x=0$.

For $g(x)$, we have $x=0$ excluded from the domain. Does this, somewhat counter-intuitively, make $g$ a continuous function? Since the domain is split into something that looks like $(-\infty, 0) \cup (0, \infty)$, each of those two intervals are continuous on their own, and a function is continuous iff it is continuous at all points in its domain.

Is that right? $f$ is discontinuous, but $g$ is continuous?

  • $\begingroup$ somewhat counter-intuitively Both $\,f\,$ and $\,g\,$ are continuous, in fact identical, on $\,\mathbb{R} \setminus \{0\}\,$. Then you extend the domain to include one extra point, in this case $\,0\,$. Not sure why you'd find that it would be intuitively expected that continuity were preserved on the newly extended domain. $\endgroup$ – dxiv Feb 9 '18 at 7:58

Yes, $f$ is discontinuous but $g$ is continuous. But the domain of $g$ doesn't just “look like” $(-\infty,0)\cup(0,+\infty)$ (you wrote $\cap$ here, but I suppose that that's a typo); it is that set. By removing from the domain of $f$ the only point at which it was discontinuous, you turned it into a continuous function.

  • $\begingroup$ Fixed my typo, thanks for the catch $\endgroup$ – user525966 Feb 9 '18 at 0:18

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