Why is $f(x)=\inf \{ r \in \mathbb{Q} ;x \in U(r)\}$?

Question

Why is $f(x)=\inf \{ r \in \mathbb{Q} ;x \in U(r)\}$ where $f$ is a real valued function from a metric space $(S,\rho)$ to $\mathbb{R}$. It is given that $$U(r)=\{x \in S;f(x) \le r \}$$ I could show that $f(x) \le \inf \{ r \in \mathbb{Q} ;x \in U(r)\}$ since given an $x \in U(r)$, it follows by the definition of $U(r)$ that $f(x) \le r$ and since this holds for all $r \in \mathbb{Q}$ such that $x \in U(r)$ we can apply infimum to the inequality $$ f(x) \le r_n $$ where $r_n \in \{r \in Q;x\in U(r)\}$. I attempted to show the other direction by trying to prove by contradiction but it didnt work. Any hints would be appreciated.

Answer 1

In general for a constant $c\in\mathbb{R}$: $$\inf\left\{ r\in\mathbb{Q}\mid c\leq r\right\} =\inf\left\{ r\in\mathbb{R}\mid c\leq r\right\} =c\tag1$$ This because $\mathbb{Q}$ is dense in $\mathbb{R}$.

We have: $$x\in U\left(r\right)\iff f\left(x\right)\leq r\tag2$$

So applying $(1)$ and $(2)$ we find: $$\inf\left(\left\{ r\in\mathbb{Q}\mid x\in U(r)\right\} \right)=\inf\left(\left\{ r\in\mathbb{Q}\mid f\left(x\right)\leq r\right\} \right)=f\left(x\right)\text{ for every }x\in S$$