Real analysis: continuity of functions Is it possible for an increasing function $[0,1]$ to be discontinuous at every irrational number? Can you help me with this problem?
 A: Monotone functions can have only jump discontinuities, and only countably many of those. Hence, the answer is no, since the irrationals in $[0,1]$ are uncountable.
More generally, the set of discontinuities of a function must be a countable union of closed sets, which the set of irrationals in $[0,1]$ is not, so in fact there is no function $f:[0,1]\to\Bbb R$--increasing or otherwise--whose set of discontinuities is the irrationals in $[0,1]$.
Let me give you an idea of how to prove the former claim (I will use the fact that real-valued functions on real intervals can have only three possible types of discontinuity, as well as the fact that the rational are countable and dense in the reals.)
Proof(ish): Suppose that $f$ is an increasing real-valued function defined on an interval $I$ (open, closed, or half-open; bounded above, below, both, or neither), and that $f$ is discontinuous at some point $x_0\in I$.
Now, if $x\in I$ with $x<x_0$, then $f(x)\le f(x_0)$, so $\lim_{x\to x_0^-}f(x)$ exists (why?) so long as $x_0$ isn't the lower endpoint of $I$, and is necessarily no greater than $f(x_0)$. Similarly, if $x$ isn't the upper endpoint of $I$, then $\lim_{x\to x_0^+}f(x)$ exists and is no less than $f(x_0)$. This rules out an essential discontinuity or removable discontinuity (why?), so $f$ has a jump discontinuity at $x_0$. Since $x_0$ was an arbitrary point of discontinuity and $f$ an arbitrary increasing real-valued function on an arbitrary interval, then any increasing real-valued function on any interval can only have jump discontinuities.
Now, let $$Y=\{y\in\Bbb R:\exists x_1,x_2\in I\text{ s.t. }f(x_1)<y<f(x_2),\text{ and }\exists a<y<b\text{ s.t. }(a,b)\cap\text{range}(f)=\emptyset\}.$$ Note that $Y$ is a union of pairwise disjoint open intervals, corresponding precisely to the "jumps" taken at any jump discontinuities of $f$. (Why?) Letting $\{q_n\}_{n\in\Bbb N}$ be any enumeration of the rationals, note that for each component interval $J$ of $Y$, there is some least $n$ such that $q_n\in J$--call this $n_J$ for each such $J$. Different components of $Y$ correspond to different $n_J$ (why?), so the function $J\mapsto n_J$ is a one-to-one function from the set of component intervals of $Y$ into the naturals, meaning that $Y$ has at most countably-many component intervals. Since each component interval corresponds to a jump discontinuity of $f$, then $f$ has at most countably-many discontinuities. $\Box$
