If $F$ is strictly increasing with closed image, then $F$ is continuous 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.
 A: 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 $[0,\infty)$.
A: 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.
