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If $X$ is a metric space and $f: X \to \mathbb{R}$ a map such that $E_r = \{x \in X \mid f(x) < r \}$ is open for all $r \in \mathbb{Q}$, then show that $E_r$ is open for all $r \in \mathbb{R}$.

If $E_r$ is open for all $r \in \mathbb{Q}$ it means that for all $x \in E_r$ there is an $\varepsilon >0$ such that $B(x, \varepsilon) \subset E_r$. However I also have that $\mathbb{R} \subset \mathbb{Q}$ doesn't this immediately imply that $E_r$ is open for all $r \in \mathbb{R}$?

Tomer Amstah
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  • I think the inclusion you want to use is false, any irrational number isn't rational! – blamethelag Mar 19 '21 at 15:57
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    You certainly don't have $\mathbb{R} \subset \mathbb{Q}$, in fact the inclusion is the other way around. I would suggest trying a density argument. – Gary Moon Mar 19 '21 at 15:57

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For every $r\in\mathbb{R}$ there exists a sequence $q_n,\in\mathbb{Q}$ with $lim_nq_n=r, q_n<r$. We have $\cup_nE_{q_n}=E_r$. Firstly remark that since $q_n<r, E_{q_n}\subset E_r$,

Let $x\in E_r, f(x)<r$, there exists $q_n$ with $f(x)<q_n<r$ implies that $x\in E_{q_n}$ and $E_r\subset \cup_n E_{q_n}$.

We deduce that $E_r$ is open since it is the union of open sets.

Tsemo Aristide
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