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Is the following function considered piecewise-continuous??

I'm reading conflcting definitions in different places: some highlight that that the function need not be defined at the (jump/removable) discontinuities, others explicitly state that the end-points of the pieces must be well-defined. Hmm!

enter image description here

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All functions are defined everywhere in their domain. The domain is the set where the function is defined. – Baby Dragon Oct 12 '12 at 3:44
@BabyDragon Good point. Edited my question for clarity. – Ryan Oct 12 '12 at 3:48
up vote 2 down vote accepted

I think you may be confusing a "piecewise-continuous function" with a continuous "piecewise function" (that is, a function defined piecewise). Piecewise functions must be defined on the endpoints of their subdomains if they are to be continuous on a larger interval, but piecewise-continuous functions need only approach finite limits as they approach the endpoints of their subdomains. Both the Wikipedia article and the lecture notes seem to be in agreement on this, though the Wikipedia article spends more time on when a piecewise-defined function is continuous than on when a function is piecewise-continuous.

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Oh ok, thanks. The wikipedia citation was a bad example. My course notes states this: "A function is said to be piecewise continuous if its graph consists of a finite number of continuous pieces on any part of its domain. The endpoints of the pieces must be well-defined." Several other places also explain it like this. So, I'm still undecided. – Ryan Oct 12 '12 at 7:06
I see. Perhaps "the endpoints of the pieces must be well-defined" is indicating that the one-sided limits have to exist for the function to be piecewise continuous? But if that's not how it's used in your notes I would chalk it up to a genuine disagreement; functions like the one in the picture are generally considered piecewise continuous. – Owen Biesel Oct 13 '12 at 5:02
Oh okay. Thanks for your input; I'll go with your definition. Cheers. – Ryan Oct 13 '12 at 15:41

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