Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be a strictly increasing function on $S$, a subset of the real line. If you know that $F(S)$ is closed, prove that $F$ is continuous.

share|cite|improve this question
Intuitively, since $F(S)$ is closed and $F$ is increasing, there are no jumps: if there's a jump from $a$ to $b$ then one of the endpoints would be a limit point of $F(S)$ not in $F(S)$; since $F$ is increasing, $F$ will never equal $a$ again. – Yuval Filmus Nov 5 '10 at 4:04
@Yuval: Well put. That is what I had in mind with my first proof, but you have better conveyed the intuition. – Jonas Meyer Nov 5 '10 at 4:11

Let $f$ be any strictly increasing (not necessarily strictly) function on $S$. To show that $f$ is continuous on $S$, it is enough to show that it is continuous at $x$ for every $x \in S$. If $x$ is an isolated point of $S$, every function is continuous at $x$, so assume otherwise.

The key here is that monotone functions can only be discontinuous in a very particular, and simple, way. Namely, the one-sided limits $f(x-)$ and $f(x+)$ always exist (or rather, the first exists when $x$ is not left-isolated and the second exists when $x$ is not right-isolated): it is easy to see for instance that

$f(x-) = \sup_{y < x, \ y \in S} f(y)$.

Therefore a discontinuity occurs when $f(x-) \neq f(x)$ or $f(x+) \neq f(x)$. In the first case we have that for all $y < x$, $f(y) < f(x-)$ and for all $y \geq x$, $f(y) > f(x-)$. Therefore $f(x-)$ is not in $f(S)$. But by the above expression for $f(x-)$, it is certainly a limit point of $f(S)$. So $f(S)$ is not closed. The other case is similar.

Other nice, related properties of monotone functions include: a monotone function has at most countably many points of discontinuity and a monotone function is a regulated function in the sense of Dieudonné. In particular the theoretical aspects of integration are especially simple for such functions.

Added: As Myke notes in the comments below, the conclusion need not be true if $f$ is merely increasing (i.e., $x_1 \leq x_2$ implies $f(x_1) \leq f(x_2)$). A counterexample is given by the characteristic function of $[0,\infty)$.

share|cite|improve this answer
Remark: Having looked more carefully at Jonas Meyer's answer, I cannot claim any essential difference in mine: really it is the same answer presented in two moderately different ways. I think that multiple explanations are in the spirit of the site, and I hope mine will be helpful to readers as well. – Pete L. Clark Nov 5 '10 at 8:08
I think your explanation is different enough :) (and good, +1). You also made explicit that "strictly" is superfluous. In my second proof, "strictly" is used to prove that $x=z$, which is sufficient but not necessary; all that is really needed is $f(x)=f(z)$, which follows without "strictly". – Jonas Meyer Nov 5 '10 at 15:28
I am not sure if "strictly increasing" is superfluous or not.The function $f(x) = 0$ for $x <0$ and $f(x)=1$ for $x \ge 0$ is not continuous. But the image of $f$ is closed. – Digital Gal Nov 5 '10 at 16:45
@Myke: Ah, you are right! I hadn't thought much about whether it was necessary when I used it, but on noticing that I didn't need the full strength of some of my statements, I hastily leapt to the wrong conclusion. That misconception then led me to misinterpret Pete's clear answer, which uses strictness at "$f(y)\lt f(x-)$". Thanks for the correction. – Jonas Meyer Nov 5 '10 at 17:21
@Myke: you are absolutely right, and I have modified my answer accordingly. We need $f$ to be strictly increasing through $x$ so that $f(x-)$ is not a value of $f$. – Pete L. Clark Nov 5 '10 at 22:20

Here's an approach by contraposition. Let $f$ be a strictly increasing function discontinuous at $x\in S$. Then $f(x)\lt\lim_{y\to x+}f(y)$ or $f(x)\gt\lim_{y\to x-}f(y)$ (or both). Suppose $f(x)\lt\lim_{y\to x+}f(y)$. Then you can show that $\lim_{y\to x+}f(y)$ is in $\overline{f(S)}\setminus f(S)$, so $f(S)$ is not closed. To see that the limit is in the closure of $f(S)$ is a straightforward unwinding of definitions. It's not in $f(S)$ because for every $z\lt x$, $f(z)\lt f(x)\lt\lim_{y\to x+}f(y)$, and for every $z\gt x$, $\lim_{y\to x+}f(y)\lt f(z)$. (Similarly on the other side. It may help to keep in mind that $\lim_{y\to x-}f(y)=\sup_{y\lt x}f(y)$ and $\lim_{y\to x+}f(y)=\inf_{y\gt x}f(y)$.)

Here's a way that doesn't use contraposition (although there is a bit of contradiction). Let $x$ be an element of $S$, and let $x_1,x_2,\ldots$ be an increasing sequence in $S$ converging to $x$. Then $f(x_1),f(x_2),\ldots$ is an increasing sequence bounded above by $f(x)$, and hence it converges. Since $f(S)$ is closed, there is a $z\in S$ such that $f(x_n)\to f(z)$ as $n\to \infty$. I claim that $z=x$. If $z$ were bigger than $x$, then we'd have $f(x_n)\leq f(x)\lt f(z)$ for all $n$, making the convergence impossible. If $z$ were smaller than $x$, we'd have $z$ smaller than $x_n$ for some $n$, so $f(z)\lt f(x_n)\leq f(x_{n+1})\leq\cdots$, again making the convergence impossible. So $z=x$ as claimed. This implies that the left-hand limit of $f$ at $x$ exists and equals $f(x)$. Similarly on the right, so $f$ is continuous.

share|cite|improve this answer
@Jonas: I am feeling disappointed that i couldn't solve it in the given amount of time. – anonymous Nov 5 '10 at 3:43
@Jonas: I was thinking to prove that $f^{-1}(\mathbb{R}\setminus S)$ is open => $f$ is continuous. – anonymous Nov 5 '10 at 3:43
@Jonas: You solved it in say perhaps 15mins, i worked on this for over 15mins, but couldn't do it!... – anonymous Nov 5 '10 at 3:44
@Chandru1: (Sorry I deleted a comment; probably should have edited instead) There's no reason to feel disappointed, and no time limit for working on more solutions. I don't understand how $f^{-1}(\mathbb{R}\setminus S)$ could be useful. – Jonas Meyer Nov 5 '10 at 3:46
@Jonas: Proving $f^{-1}(\mathbb{R} \setminus S)$ because continuous mapping takes inverse images of open sets to open sets – anonymous Nov 5 '10 at 3:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.