I am trying to answer this question:
Let $f$ be a real-valued function on a metric space $X.$ Show that the set of points at which $f$ is continuous is the intersection of a countable collection of open sets. Conclude that there is no real-valued function on $\mathbb{R}$ that is continuous at the rational numbers only.
Here is my answer for the first part:
For $n\in\mathbb{N}$, consider the sets
$$
U_n:=\{x\in X:\exists\delta>0,\forall y,z \in X, \, y,z\in B(x,\delta)\implies |f(y) – f(z)|<1/n\}.
$$
I managed to prove that those $U_{n}$ are open and that the set of continuities of $f$ is a $G_{\delta}$ set and that $U_{n}$ is dense. Now following the sequence of steps required to complete the solution of the problem regarding to the second part as suggested here
I have to do the following: for $f:\mathbb R\to \mathbb R$
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Suppose $f$ is continuous at the rationals. Show $U_n$ is also dense. Hence, $U_n^c$ is closed and nowhere dense.
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Using the previous statement and the fact that the rationals are countable, write $\mathbb{R}$ as a countable union of nowhere dense sets, contradicting the Baire category theorem.
Taking complements can be used to restate the Baire category theorem in the following equivalent way: a countable intersection of dense open subsets of $\mathbb{R}$ is dense. \
My question is:
I know that $\mathbb{Q}$ is in $U_{n}$ (not in $U_n^c$ which are nowhere dense sets) how this can lead me to write $\mathbb{R}$ as a countable union of nowhere dense sets.
Best Answer
As explained in the answer you linked, the set of points at which $f$ is continuous (i.e. $\mathbb{Q}$) is the intersection of all the $U_n$. Thus: \begin{align*}\mathbb {Q}&=\bigcap_{n\in\mathbb{N}} U_n\\ \mathbb{R}-\mathbb{Q}&=\bigcup_{n\in\mathbb{N}} U_n^c\\ \mathbb{R}&=\bigcup_{n\in\mathbb{N}} U_n^c\bigcup_{q\in\mathbb{Q}}\{q\}\end{align*}
Since $\mathbb{Q}$ is countable and $\{q\}$ is nowhere dense we obtain $\mathbb{R}$ as a countable union of nowhere dense sets