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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?

user525966
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2 Answers2

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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 $\,x \in \mathbb{Z}\,$.

dxiv
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  • "The domain can not include 'undefined points', by definition." It always seemed nicer to me to consider $1/x$ to be a partial function from $\mathbb R \rightarrow \mathbb R$, allowing it to have an undefined point at 0. – eyeballfrog Jan 29 '18 at 07:47
  • @eyeballfrog The "domain" of partial function $1/x$ is usually taken to be $,\mathbb{R} \setminus {0},$, and it's usually made clear in the context when used otherwise. I somehow don't think that's what the OP meant here. – dxiv Jan 29 '18 at 07:52
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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 $0$...