# Proving that $U_{n}$ is dense.

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.Now following the sequence of steps required to complete the solution of the problem regarding to the second part as suggested here

Prove that there doesn't exist any function $f:\mathbb R\to \mathbb R$ that is continuous only at the rational points.

I have to do the following: for $$f:\mathbb R\to \mathbb R$$

1. Suppose $$f$$ is continuous at the rationals. Show $$U_n$$ is also dense. Hence, $$U_n^c$$ is closed and nowhere dense.

2. 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: by the definition of dense set in a metric space I have to show that every nonempty open subset of $$\mathbb{R}$$ contains a point of $$U_n$$ for every $$n,$$ but I am unable to prove this. Could anyone help me in doing so please?

Every point of continuity of $$f$$ belongs to $$U_n$$. [For this use: $$|f(y)-f(z)| \leq |f(y)-f(x)|+|f(x)-f(z)|$$ and use definition of continuity]. Since $$f$$ is assumed to be continuous at all rational points it follows that $$U_n$$ contains $$\mathbb Q$$. Hence $$U_n$$ is dense.
• Does this follows from the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$?
• And why the closure of $U_n^C$ will be hollow?
• @Mathstupid It follows by definition of a dense set. No open set can be contained in $U_n^{c}$. Mar 18, 2020 at 10:34