Do discontinuities only exist strictly inside the domain of a function?

Question

For example the graph $f(x) = 1/x$ approaches $\infty$ at $x=0$ but we would not say this is an infinite discontinuity, just an asymptote, correct? Unless we specifically said the domain of the function included $x=0$? Or must domains by definition exclude discontinuities and undefined points?

Do discontinuities only exist if we can "split" the domain into two new non-empty intervals that each exclude the discontinuity?

Answer 1

Continuity is only defined at points within the domain of the function . Since a discontinuity is usually defined as a point where the function is not continuous, that must be within the domain as well. It would be meaningless to say that $\,\sqrt{x}\,$ is discontinuous at $\,x=-1\,$, for example.

Or must domains by definition exclude discontinuities and undefined points?

• The domain can not include "undefined points", by definition.

• The domain can include discontinuity points, and often does, for example the integer part function $\,f(x) = \lfloor x \rfloor\,$ is defined on $\,\mathbb{R}\,$, which includes all the discontinuity points $\,x \in \mathbb{Z}\,$.

Answer 2

Consider $f:\mathbb R\to \mathbb R$ where $f(x)=\begin{cases}1, \text{ if } x\in \mathbb Q\\ 0 ,\text{ if } x\notin \mathbb Q \end{cases} $. This function is discontinuous everywhere.

But, in your example, $f(x)=\frac1x$ isn't defined at $x=0$. You could define $g$ to be the same as $f(x) $ away from zero, and give it some value, say $1$, at $0$. Then $g$ would not be continuous at $0$...