In multiple class examples, and internet examples on discontinuities. I often see that the undefined points are often called "the points at which the function is discontinuous". So If I have say a piecewise function:
$$ f(x) = 1 ; (x > 1) $$
and $$ f(x) = \frac{1}{x} ; x\in[-1, 1] $$
I find examples that would say the function $1/x$ is undefined at x =0, thus it is discontinuous at said point.
Would we say those examples are incorrect, since the function 1/x is not defined at x = 0 ? Even if it is stated in the piecewise description that the interval we are considering does include x=0? Would we say the piecewise function is discontinuous at x =0?
I am just trying to clarify these concepts because many places share the wrong idea that "if a function f is not defined at a point, then it is said that the function is discontinuous at the said point"
Discontinuity of some given function $f:D\to \mathbb{R}$ has nothing to do with being defined/undefined. For example, function $f(x) = \mathrm{sign}\;x = \begin{cases}1,\;x\geq 0\\-1,\;x<0\end{cases}$ is defined at every point $x\in\mathbb{R}$ from its domain, however it has discontinuity at $x=0$. At other points that are not from function's domain $D$ continuity makes no sense.
You can think of continuity in the following way: suppose you have a function $f$ and you want to check if it is continuous at $x\in D$. Then take a look at all points that are, say, to the left of $x$. Start approaching $x$ with these points. What value (of $f$) do you expect to get "by inertia"? Let's take a look at $\mathrm{sign}\;x$. If we take a sequence $x_{n} = -1/n$ that approaches $0$ from the left, we would expect our function to be $-1$ at $0$, because at every point $x_{n}$ we get $\mathrm{sign}\;x_{n} = -1$. But if one approaches $0$ from the right, he would expect function to be equal to $1$. Note that even if both "observers" got the same value, it wouldn't guarantee the continuity of $f$ at $0$. In fact, you should run through all possible sequences, that converge to $0$ (and not equal $0$ at any term). However, this approach doesn't seem to work when $D$ is some "strange" set, or even not a subset of $\mathbb{R}$. To deal with such cases, I would recommend taking topology course.