Examples of nowhere continuous functions Do you know examples of nowhere continuous functions, besides the Dirichlet function and its modifications?
 A: The examples given so far seem to be rather "tamely" discontinuous, in that their graphs are included in the union of the graphs of two continuous functions. A somewhat more dramatic example would be a function $f:\mathbb R\to\mathbb R$ whose graph is dense in the plane $\mathbb R^2$.  To produce such an $f$, notice first that the set of all rectangles $(a,b)\times (c,d)$ with $a,b,c,d\in\mathbb Q$ is a countable set, so enumerate it as $\{A_n:n\in\mathbb N\}$.  Then choose, by induction on $n$, points $(x_n,y_n)\in A_n$ in such a way that each $x_n$ is different from all previously chosen $x_j$'s ($j<n)$.  Then define $f(x)$ to be $y_n$ if $x=x_n$, and define $f(x)$ to be $0$ if $x\notin\{x_n:n\in\mathbb N\}$.  The graph of $f$ contains all the pairs $(x_n,y_n)$, hence meets every $A_n$, and hence is dense in $\mathbb R^2$.   
(Despite the apparent wildness of $f$, the construction doesn't need the axiom of choice.  Fix an enumeration of the rationals and then, at each step, take $x_n$ and $y_n$ to be the first, in the enumeration, rational values that satisfy the required conditions.)
A: One of my favorites is on $(0,1) \to (0,1)$ but you can extend it to $\Bbb {R \to R}$ with your favorite bijection.  For $x \in (0,1)$, write it in ternary, using the terminating version if there is a choice.  If there are an infinite number of $2$'s in the expansion, or the tail is all $0$'s or all $1$'s, set $f(x)=x$  If the last $2$ in the expansion is in the $3^{-m}$ place, multiply by $3^m$ and take the fractional part (erase everything through the last $2$).  Now read it as a binary expansion.  This function takes all values in every interval, so (if extended to $\Bbb {R \to R}$ with a bijection) is dense in the plane.  The point is that on any interval $(a,b)$ you can find an interval of the form $(\frac {3k+2}{3^m},\frac {3k+3}{3^m})$ and you get all the binary expansions in $(0,1)$ within that interval.
