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https://en.m.wikipedia.org/wiki/Classification_of_discontinuities https://en.m.wikipedia.org/wiki/Classification_of_discontinu... is responsive and quite access

https://en.m.wikipedia.org/wiki/Classification_of_discontinuities https://en.m.wikipedia.org/wiki/Classification_of_discontinu... is responsive and quite accessible. It notes that there doesn't have to be an undefined point for a function to be discontinuous (and that terminology often conflates the two), and matches what I recall of determining that if the limit of the derivative from both sides of the discontinuity exists and is equal, the derivative exists.

bikenaga 2y ago
> ... there doesn't have to be an undefined point for a function to be discontinuous. That's right. In the example f(x) = (x² - 1)/(x - 1) for x ≠ 1, if we further define f(1) = 0, the function is now defined at x = 1, but discontinuous there. > ... if the limit of the derivative from both sides of the discontinuity exists and is equal, the derivative exists. (You probably mean "both sides of the point", since if there's a discontinuity there the derivative can't exist.) Your point that, if the left and right-hand limits both exist and are equal, then the derivative exists (and equals their common value) is true for all limits. Also, there's a difference between the use of the word "continuous" in calc courses and in topology. In calc courses where functions tend to take real numbers to real numbers, a function may be said to be "not continuous" at a point where it isn't defined. So f(x) = 1/(x - 2) is "not continuous at 2". But in topology, you only consider continuity for points in the domain of the function. So since the (natural) domain of f(x) = 1/(x - 2) is x ≠ 2, the function is continuous everywhere (that it's defined).
- dwattttt 2y ago
  I was actually aiming for the situation where a function is defined on all reals but still discontinuous (e.g. the piecewise function in the wiki article for the removable discontinuity). So there's a discontinuity (x=1), however the function is defined everywhere.

bikenaga 2y ago

> ... there doesn't have to be an undefined point for a function to be discontinuous. That's right. In the example f(x) = (x² - 1)/(x - 1) for x ≠ 1, if we further define f(1) = 0, the function is now defined at x = 1, but discontinuous there. > ... if the limit of the derivative from both sides of the discontinuity exists and is equal, the derivative exists. (You probably mean "both sides of the point", since if there's a discontinuity there the derivative can't exist.) Your point that, if the left and right-hand limits both exist and are equal, then the derivative exists (and equals their common value) is true for all limits. Also, there's a difference between the use of the word "continuous" in calc courses and in topology. In calc courses where functions tend to take real numbers to real numbers, a function may be said to be "not continuous" at a point where it isn't defined. So f(x) = 1/(x - 2) is "not continuous at 2". But in topology, you only consider continuity for points in the domain of the function. So since the (natural) domain of f(x) = 1/(x - 2) is x ≠ 2, the function is continuous everywhere (that it's defined).

dwattttt 2y ago
I was actually aiming for the situation where a function is defined on all reals but still discontinuous (e.g. the piecewise function in the wiki article for the removable discontinuity). So there's a discontinuity (x=1), however the function is defined everywhere.