Discontinuous function at an uncountable set with not rationals Does there exists a function $f:[0,1]\to\mathbb{R}$ such that $D(f)$ (its points of discontinuity) is an uncountable set containing no rational number?
First thing I thought of was $\mathbb{R}\setminus\mathbb{Q}$ (uncountable, with no rational), but it won't work, since it's not an $F_\sigma.$ So there is no function whose points of discontinuity is precisely $\mathbb{R}\setminus\mathbb{Q}$.
 A: For example, you can easily construct a Cantor-type set that contains no rational.  I will start with $[a,b]$ where $a$ and $b$ are irrational with $0 < a < b < 1$, and remove at stage $n$ an open interval $U_n$ such that
1) $U_n$ has irrational endpoints
2) The closures of all $U_n$ are disjoint
3) $r_n$ (the $n$'th rational in some enumeration of the rationals in $(a,b)$)
is in $\bigcup_{j=1}^n U_j$.
Take $E = [a,b] \backslash \bigcup_{j=1}^\infty U_j$.
Define $f$ as the indicator function of $E$. 
A: A slick way to show that there is a Cantor set disjoint from the rationals is to recall that $\mathbb{R}\setminus\mathbb{Q}$, as a subspace of $\mathbb{R}$ with the usual topology, is homeomorphic to $\omega^\omega$ with the product topology. (A proof can be found here.) $\omega^\omega$ clearly contains numerous copies of $\{0,1\}^\omega$, which is well-known to be a Cantor set.
A: There are closed subsets of $[0,1]\setminus\mathbb Q$ with positive measure, positive measure implies uncountability, and closed sets are $F_\sigma$s, so yes.  For some other questions that ask about subsets of $\mathbb R\setminus\mathbb Q$ with certain properties, see here, here, and here.  You do not need to consider measure.  See in particular the second link, and note that perfect sets are uncountable.
A: Not any Cantor like set with irrational endpoints will work. For example, if you restrict the length of removed segment to $\epsilon^n$ at each state at the midpoint of the interval, you will run into issues of not reducing the measure to 0 and will have rationals left over (there will be interval near $a$). (even though the method works if you put more restriction).
Just use the Cantor set. Let $C$ be the 1/3 Cantor set (removing 1/3 each time). It's uncountable, compact, and all element are rational or transcendental (I will refer to  wikipedia's Cantor set page. Please look it up with more detail). 
Let set $S = \frac{\sqrt{2}}{2}C$. You can tell that $S \subset [0,1]$, also every element in $C$ that is rational will become irrational, transcendental numbers are not algebric number, so if you multiply by $\frac{\sqrt{2}}{2}$ which is a algebric number, you still get a non-algebric number which can't be rational. It's is rather clear it's uncountable (there is a bijection from $S$ to $C$. (It is also compact). 
