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A function is, (almost) by definition, continuous at its isolated points.

What is the rationale behind this?

Intuitively it would seem to me that isolated points "shouldn't" be considered points at which the function is continuous. Indeed, often in books/courses, it is noted that somewhat surprisingly, a function is continuous at its isolated points and this is often assigned as an exercise (example).

I am guessing that there are advantages to doing so but I don't know what they are.


Often, we first define $\lim_{x\rightarrow a} f(x)$ if and only if $a$ is a limit point of the domain of $f$.

In which case, in the usual definition of continuity that includes isolated points, we are unable to shorten the definition of continuity to simply: $\lim_{x\rightarrow a} f(x) = f(a)$, because that would fail to include isolated points. So it would seem to me that including isolated points actually complicates the definition of continuity.

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    $\begingroup$ Simply put: it makes it easier to define continuous at a point without special cases. $\endgroup$ – Thomas Andrews Oct 30 '18 at 1:45
  • $\begingroup$ @ThomasAndrews: Please see my edited post $\endgroup$ – dtcm840 Oct 30 '18 at 1:49
  • $\begingroup$ Well, my favorite definitions of continuity make it clear that the function is continuous at isolated point: the direct $\varepsilon,\delta$ definition, good for any two metric spaces; and the requirement that if $U$ is open in the codomain, then $f^{-1}(U)$ is open in the domain. This second one is good for all topological spaces. $\endgroup$ – Lubin Oct 30 '18 at 1:52
  • $\begingroup$ The more general notion of continuity is not defined in terms of limits, but in terms of “open set.” This is the“point set topology” approach. A point is isolated if the singleton containing that point is an open set. $\endgroup$ – Thomas Andrews Oct 30 '18 at 1:54
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    $\begingroup$ Actually, the requirement that your point be nonisolated is a way of sneaking into the story a conclusion that the limit at a point is unique, if it exists. I prefer not to require nonisolatedness, and bite the bullet to recognize that in this case, there is no uniqueness. $\endgroup$ – Lubin Oct 30 '18 at 1:59

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