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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.

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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

2 Answers 2

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.

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@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

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 $[0,\infty)$.

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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
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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

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