When is a discontinuity of a real valued function removable?

Question

This is a follow-up to a question asked by a calculus student here: Is the function $\frac{x^2-x-2}{x-2}$ continuous?

It got me thinking about a more interesting related question in classical analysis: When in general is a discontinuity for a real valued map removable? A rational function with a factorable polynomial in the numerator with a factor in the denominator is an almost trivial case - but when is it true in general? In complex analysis, for holomorphic functions, there's a very well-defined theory based on a theorem of Riemann. But what about real valued maps?

While I'm on the subject, I should note that "removable discontinuity" in the sense used in this problem is something of a misnomer - the point really should be called a removable singularity. The technical distinctions are nicely summed up here:

http://en.wikipedia.org/wiki/Classification_of_discontinuities

Addendum: I just found this older,related post at this board: Is there a function with a removable discontinuity at every point? If this is a valid algorithm, it may provide a starting point for a theoretical basis for the solution to my question!

When the left and right limits of the function at the singularity exist and agree? — user7530, Nov 26 '11 at 08:33
You can't make a general statement for continuity without prooving the limits of the function in the discontinuity-point. The general statement is the existence of the limit. Or didn't I get your point? — ulead86, Nov 26 '11 at 08:42
@Mathemagician1234: Your comment was off-topic. Please either calmly ask "Why the downvote?", or don't say anything about it. — Zev Chonoles, Nov 26 '11 at 08:57
@user7530, Daniel: I'm not even sure the question makes sense if the left and right hand limits around the singularity don't agree. The 2 sided limit exists or whether or not we can remove the singularity is irrelevant-we CAN'T make it continuous at a point where the one sided limits don't agree,yes? — Mathemagician1234, Nov 26 '11 at 10:07
I wasn't the first downvote, but I just added a second downvote. Given your self-reported background, this question is really silly and trivial. A version of it appears as an exercise in every textbook I've ever used to teach freshman calculus. — Adam Smith, Nov 26 '11 at 21:21
@Mathemagician1234: Your comment does not seem to me to make a lot of sense; the two-sided limit exists if and only if both one-sided limits exist and are equal. So "the 2-sidedd limit exists" is exactly the same as "the one sided limits exist and agree". How can the former be irrelevant, but the latter be the key, if they are equivalent? — Arturo Magidin, Nov 26 '11 at 21:33
@Arturo You misunderstood my response. I was agreeing with Daniel's assessment. It's important to make sure there's agreement on the obvious points-you never know which will be important in the solution of a problem. — Mathemagician1234, Nov 27 '11 at 03:26
@AdamSmith - The content of a question is what's important, not who asked it. If Terry Tao came in and asked a question you deemed below his station, would you downvote it? I cannot imagine any situation in which people should avoid asking a question when they don't know the answer - even if you think it "ought" to be within their ability, all you should worry about is giving a good answer for the benefit of the online math community as a whole. — Zev Chonoles, Nov 29 '11 at 16:01
I think it is clear that, while your comments on Mathemagician1234's posts in most cases do constitute valid criticisms, you have consciously focused on criticizing him in particular, and it has gotten to the point where the adversarial nature of your comments is doing more harm than good. At this time, I will ask you to let other users interact with Mathemagician1234 on occasions in which criticisms might be given to him, and only post yourself if some time has passed and no one has yet made the point you want to make (though, if that is the case, how important could it really be?) — Zev Chonoles, Nov 29 '11 at 16:24
But if necessary, the next step will be a "forced separation" (no interaction on the site, on pain of suspension). — Zev Chonoles, Nov 29 '11 at 16:26
I don't understand why this question got 6 downvotes. I would not upvote it but now I will to compensate for inappropriate downvotes. +1 — Rudy the Reindeer, Oct 12 '12 at 09:58

score 7 · Answer 1 · answered Nov 26 '11 at 11:53

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Let $f: \mathbb{R}\to\mathbb{R}$ be a function that is not continuous (or undefined) at $x_0 \in \mathbb{R}$. We can try to remove the discontinuity by replacing $f$ with the "patched" function $g$ given by

$$g(x) = \begin{cases} f(x), &x \neq x_0\\C, &x = x_0\end{cases}$$

for some real number $C$. Almost immediately from the definition of continuity, it follows that $g$ is now continuous at $x_0$ if and only if $$\lim_{x\to x_0^+} f(x) = \lim_{x\to x_0^-} f(x) = C.$$ In other words, the singularity is removable if and only if the left and right limits of $f$ near the singularity exist, and are equal.

answered Nov 26 '11 at 11:53

user7530

49,280

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+1. Simple and elegant solution. It can also clearly be extended to jump discontinuities,that's the impressive thing about it. The next obvious question is whether it can be generalized even further and how one would do so. – Mathemagician1234 Nov 26 '11 at 20:13
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This is exactly the obvious answer that Daniel gave above and you criticized. – Adam Smith Nov 26 '11 at 21:18
@Adam No,it isn't-he didn't give this response. Daniel commented that both sided limits must exist to be able to define the question sensibly-and I was agreeing with him. And my question involved the general case.I've never seen a rigorous statement of user7530's response as an exercise in any calculus text (save perhaps Spivak ,of course). I did not take a rigorous calculus course as a freshman.In any event,it does not give the general solution to my question,merely gives an important special case. And clearly I wasn't the only one unaware of it from the upvoting. – Mathemagician1234 Nov 27 '11 at 03:23
@Adam(cont.) You obviously were so fast to insult my perceived lack of mathematical knowledge, you didn't carefully read the exchange. – Mathemagician1234 Nov 27 '11 at 03:31
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@Mathemagician1234 : I have no idea what kind of point you are pressing here. It's just the definition of a limit -- nothing subtle is happening here. And I'm not talking about rigorous calculus courses, just ordinary calculus courses that discuss limits (without properly defining them). I don't keep copies of calculus books on hand when I'm not teaching it, so I can't give a precise reference, but this kind of thing is definitely discussed in, e.g., Stewart's book. – Adam Smith Nov 27 '11 at 04:54
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@Adam The point I'm pressing is that I agreed with Daniel and that this simple example may be the first step towards a general solution. And if it's in Stewart's book, I never covered it. Then again, that wouldn't be a great shock considering the pathetic excuse for a calculus course most students have these days, is it? I'm self taught in undergraduate analysis-there's bound to be holes in my training and sometimes, it manifests itself in me being unaware of what you consider basic examples.I'm not ashamed to admit that. – Mathemagician1234 Nov 27 '11 at 05:45
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@Adam BTW,I checked my beat up old copy of the 3rd edition of Stewart-you're wrong,he does NOT discuss the patching function in section 1.5, where continuity and discontinuity are discussed. He DOES define removable discontinuities,but gives no rigorous discussion of when this is possible beyond the 2 sided limit existing and "redefining" the function. – Mathemagician1234 Nov 27 '11 at 09:31
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@Mathemagician1234 : I don't see any material difference between what you describe in Stewart and what is contained in the above answer. – Adam Smith Nov 27 '11 at 17:36
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@Adam Whatever. I don't know how else to respond, you don't see a difference between Stewart's handwaving definition and the definition given above. I actually get exhausted debating you. – Mathemagician1234 Nov 27 '11 at 19:08
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@Mathemagician1234: I have removed an irrelevant remark from the above comment. And as I am sure you are aware, each user is only able to vote once on any given post, so it does not make sense to blame a single user for the total score of a post. – Zev Chonoles Nov 28 '11 at 01:21
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@Zev Alright. I'm trying not to get myself thrown off here. I think I was very civil about it. I'm trying very hard to be. – Mathemagician1234 Nov 28 '11 at 02:18

When is a discontinuity of a real valued function removable?

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