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

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up vote 3 down vote accepted

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 $E$.

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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.

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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.

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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).

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