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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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    @Yuval: Well put. That is what I had in mind with my first proof, but you have better conveyed the intuition.2010-11-05

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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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    @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 $f$.2010-11-05
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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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    @Chandru1: Ah, I should have guessed that was a typo. Thank you.2010-11-05