Is there a bounded function discontinuous on a countable dense subset?

Question

Is there a function that is continuous everywhere except on a countable dense subset, but is bounded?

Bounded in the sense that supremum of $f$ is a finite number.

Answer 1

The function $f(p/q)=1/q$ and zero at irrationals seems to fit the bill.

Answer 2

Yes. The typical example is K.J. Thomae's function $f$ from 1875 mentioned in another answer: Define $f(x)=0$ if $x$ is irrational. If $x$ is rational, write $x=p/q$ where $q>0$ and $p,q$ are relatively prime, and set $f(x)=1/q$. One easily checks that the function is bounded (having range contained in $[0,1]$), continuous at all irrationals, and discontinuous at all rationals.

More generally, given any real-valued function (with domain an interval), its discontinuities form an $F_\sigma$ set (a countable union of closed sets), and any $F_\sigma$ set can be obtained this way. This is a result of W.H. Young from 1903. The example above corresponds to taking as $F_\sigma$ set the set of rationals. Note that the rationals are countable union of singletons, and every singleton is closed.