If memory serves, ten years ago to the week (or so), I taught first semester freshman calculus for the first time. As many calculus instructors do, I decided I should ask some extra credit questions to get students to think more deeply about the material. The first one I asked was this:

1) Recall that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ is said to have a removable discontinuity at a point $x_0 \in \mathbb{R}$ if $\lim_{x \rightarrow x_0} f(x)$ exists but not does not equal $f(x_0)$. Does there exist a function $f$ which has a removable discontinuity at $x_0$ for every $x_0 \in \mathbb{R}$?

Commentary: if so, we could define a new function $\tilde{f}(x_0) = \lim_{x \rightarrow x_0} f(x)$ and it seems at least that $\tilde{f}$ has a fighting chance to be continuous on $\mathbb{R}$. Thus we have successfully "removed the discontinuities" of $f$, but in so doing we have changed the value at every point!

Remark: Lest you think this is too silly to even seriously contemplate, consider the function $f: \mathbb{Q} \rightarrow \mathbb{Q}$ given by $f(0) = 1$ and for a nonzero rational number $\frac{p}{q}$, $f(\frac{p}{q}) = \frac{1}{q}$. It is easy to see that this function has limit $0$ at every (rational) point!

So I mentioned this problem to my students. A week later, the only person who asked me about it at all was my Teaching Assistant, who was an older undergraduate, not even a math major, I think. (I hasten to add that this was not in any sense an honors calculus class, i.e., I was pretty clueless back then.) Thinking about it a bit, I asked him if he knew about uncountable sets, and he said that he didn't. At that point I realized that I didn't have a solution in mind that he would understand (so still less so for the freshman calculus students) and I advised him to forget all about it.

So my actual question is: can you solve this problem using only the concepts in a non-honors freshman calculus textbook? (In particular, without using notions of un/countability?)

[Addendum: Let me say explicitly that I would welcome an answer that proceeds directly in terms of the least upper bound axiom. Most freshman calculus books do include this, albeit somewhere hidden from view of the casual readers, i.e., actual freshman calculus students.]

If you can't figure out how to answer the question at all, I think the following related question helps.

2) Define a function $f: \mathbb{R} \rightarrow \mathbb{R}$ to be precontinuous if the limit exists at every point. For such a function, we can define $\tilde{f}$ as above. Prove/disprove that, as suggested above, $\tilde{f}$ is indeed continuous. [Then think about $f - \tilde{f}$.]

Now that I think about it, there is an entire little area here that I don't know anything about, e.g.

3) The set of discontinuities of an arbitrary function is known -- any $F_{\sigma}$ set inside $\mathbb{R}$ can serve. What can we say about the set of discontinuities of a "precontinuous function"? [Edit: from the link provided in Chandru1's answer, we see that it is countable. What else can we say? Note that taking the above example and extending by $0$ to the irrationals, we see that the set of points of discontinuity of a precontinuous function can be dense.]

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    $\begingroup$ Amazingly enough... math.rutgers.edu/~zeilberg/fb74.html . :) I wonder how to settle "Now try to prove it without saying the word 'uncountable'." $\endgroup$ Commented Sep 1, 2010 at 7:51
  • $\begingroup$ @Pete L. Clark: May i know what makes you think that a proof would exist without using uncountablity of $\mathbb{R}$. $\endgroup$
    – anonymous
    Commented Sep 1, 2010 at 8:42
  • $\begingroup$ @Chandru: I didn't say that I thought that such a proof exists: that's why I asked it as a question! Moreover, the example with the rationals I gave shows that some nontrivial property of the real numbers needs to be used, but it is conceivable to me that there is a proof using, say, the least upper bound axiom. (But, if you are going to make me guess, yes, I'll guess that there is such a proof. The question doesn't mention uncountability, so why should the answer require it?) $\endgroup$ Commented Sep 1, 2010 at 8:49
  • $\begingroup$ But are uncountable sets such an issue for freshmen? $\endgroup$
    – mau
    Commented Sep 1, 2010 at 10:16
  • $\begingroup$ @mau: That's not really the point here. My question is not a pedagogical one. It is a mathematical one -- can you prove $X$ without using $Y$? -- which I have motivated / explained my interest in via a self-effacing pedagogical anecdote, part of the point of which was that the question itself was confusing and off-putting for my non-honors freshman calculus class (at Harvard, but I wouldn't read much into that). $\endgroup$ Commented Sep 1, 2010 at 10:45

3 Answers 3


I think the following works:

Here is a sketch, I will fill in the details later if required.

Let $g(x) = \lim_{t\rightarrow x} f(t)$. Then we can show that $g(x)$ is continuous.

Let $h(x) = f(x) - g(x)$. Then $\lim_{t \rightarrow x} h(t)$ exists and is $0$ everywhere.

We will now show that $h(c) = 0$ for some $c$.

This will imply that $f(x)$ is continuous at $c$ as then we will have $f(c) = g(c) = \lim_{t->c} f(t)$.

Consider any point $x_0$.

By limit of $h$ at $x_0$ being $0$, there is a closed interval $I_0$ (of length > 0) such that $|h(x)| < 1$ for all $x \in I_0$.

This is because, given an $\epsilon > 0$ there is a $\delta > 0$ such that $|h(x)| < \epsilon$ for all $x$ such that $0 < |x - x_{0}| < \delta$. Pick $\epsilon = 1$ and pick $I_{0}$ to be any closed interval of non-zero length in $(x_{0}, x_{0} + \delta)$.

Now pick any point $x_1$ in $I_0$.

By limit of $h$ at $x_1$ being $0$, there is a closed interval $I_1 \subset I_0$ (of length > 0) such that $|h(x)| < 1/2$ for all $x \in I_1$, by argument similar to above.

Continuing this way, we get a sequence of closed intervals $I_n$ such that

$|h(x)| < \frac{1}{n+1}$ for all $x \in I_n$. We also have that $I_{n+1} \subset I_n$ for each $n$, and that length $I_n$ > 0. We could also arrange so that length $I_n \rightarrow 0$.

Now there is a point $c$ (by completeness of $\mathbb{R}$) such that $c \in \bigcap_{n=0}^{\infty}I_{n}$.

Thus we have that $|h(c)| < \frac{1}{n+1}$ for all $n$ and so $h(c) = 0$ and $f(c) = g(c)$.

  • $\begingroup$ @Pete: I see now you already talked about $h(x)$ in your question, so there aren't much details left to be filled in, I suppose. $\endgroup$
    – Aryabhata
    Commented Sep 1, 2010 at 16:36
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    $\begingroup$ @Pierre: I am not arguing by contradiction. It is a constructive proof which directly shows that there is a point where the function equals the limit. Thanks. $\endgroup$
    – Aryabhata
    Commented Sep 2, 2010 at 13:57
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    $\begingroup$ Here is what I understand: Your assumption is that the limit of f(t) for t -> x (t not equal to x) exists everywhere, and your conclusion is that f has a continuity point c. If this is so, why don't you say it? My suggestion is just: tell the reader what you're doing! $\endgroup$ Commented Sep 2, 2010 at 16:04
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    $\begingroup$ Yes, this works, and indeed presents a fairly general technique for converting "uncountable pigeonhole principle" arguments to arguments from completeness. Thanks very much. $\endgroup$ Commented Sep 2, 2010 at 21:45
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    $\begingroup$ @Mathemagician1234: Thanks! I am not too sure if I would try and pursue getting this published. I suppose this site kind of serves the purpose of having it 'out there'. $\endgroup$
    – Aryabhata
    Commented Dec 20, 2011 at 20:57


Please see theorem 5.63 in this link:


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    $\begingroup$ Cruft-less version of the link: books.google.com/books?id=6l_E9OTFaK0C&pg=PA249 $\endgroup$ Commented Sep 1, 2010 at 7:52
  • $\begingroup$ @Chandru1: but this proof uses the notions of un/countable sets (as I alluded to above), so doesn't answer my real question. $\endgroup$ Commented Sep 1, 2010 at 8:04
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    $\begingroup$ Nevertheless I gave +1 because of the progress this makes on the third question above. $\endgroup$ Commented Sep 1, 2010 at 8:20
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    $\begingroup$ @Pete, the edit was intended to be a comment? $\endgroup$ Commented Sep 2, 2010 at 22:46
  • $\begingroup$ @Mariano: for reasons I am now finding difficult to defend, I decided that I wanted my remark to have "higher priority" than a comment. Ceretainly Chandru1 can delete or rollback after he reads the edit. $\endgroup$ Commented Sep 2, 2010 at 23:10

The uncountability of $\mathbb{R}$ must be used, but perhaps we can avoid mentioning it.

Below is a sketch of proof. By restriction of domain and subtraction of its limit from itself, we may consider $f:[0,1]\rightarrow\mathbb{R}$ such that $f$ has zero pointwise limit and removable discontinuity everywhere. If we replace $f$ by $|f|$, we may further assume that $f$ is strictly positive.

Denote the cardinality of a finite set $S$ by $|S|$. Let

$I_0 = \lbrace x: 1\le f(x)\rbrace$ and $I_n = \lbrace x: \frac{1}{n+1}\le f(x)\lt \frac{1}{n}\rbrace$ for $n\ge 1$.

Then each $I_n$ must be finite, or else it has an accumulation point $p\in[0,1]$ and this contradicts the assumption that the limit of $f$ at $p$ is zero. Let

$J_n = \lbrace x\in[0,1]: |x-y|<\frac{1/2}{3^{n+1}|I_n|} \textrm{ for some } y\in I_n\rbrace.$

($J_n$ is empty if $I_n$ is empty.) Hence $[0,1] = \bigcup\limits_{n\ge 0} I_n \subseteq \bigcup\limits_{n\ge 0} J_n$. But the latter has total length $\le \frac{1}{2}$. So we arrive at a contradiction.

[Edit] Alternatively, $\lbrace J_n \rbrace_{n=1,2,...}$ is an open cover of $[0,1]$. So there is a finite subcover. This contradicts that the total length of all $J_n$'s is at most $1/2$.

  • $\begingroup$ Interestingly, the proof you want to give is exactly the same as what I had in mind (reducing to the case where the limit is identically zero), whereas the proof in the link to Chandru's answer is different. Off the top of my head, though, I'm not satisfied with the alternate conclusion: you seem to be using the countable subadditivity of Lebesgue measure, which is out of bounds. $\endgroup$ Commented Sep 1, 2010 at 11:16
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    $\begingroup$ Also, what exactly do you mean when you say that the uncountability of $\mathbb{R}$ must be used? Why couldn't we (conceivably) use the completeness of $\mathbb{R}$ (which implies the uncountability, but we don't need to know that)? $\endgroup$ Commented Sep 1, 2010 at 11:18
  • $\begingroup$ Thanks for your comment. I agree to your point about the completeness of $\mathbb{R}$, but I think you don't need to know anything about Legesgue measure in order to prove the proposition that "if a countable union of open subintervals of $[0,1]$ has total length less than $1$, then the union doesn't cover $[0,1]$". $\endgroup$
    – user1551
    Commented Sep 1, 2010 at 12:29

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