# 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 $\mathbb{R}\setminus\mathbb{Q}$.

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 $E$.

• This works great, thanks. Just a question: why is it necessary that $a$, $b$, and the end points of all the $U_n$ be irrational? What "goes wrong" if they're rational? – FPP Dec 7 '11 at 1:59
• Any such point will end up in $E$. – Robert Israel Dec 7 '11 at 17:36
• Why, though? If such a point were rational, wouldn't it get removed by some other $U_n$? – FPP Dec 8 '11 at 0:25

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.

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.

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).