Proving a natural map is open

Question

This actually have been asked and answered here. I still have question for proving that the natural map $p$ is open and i can't comment since my reputation is not enough yet, so i make a new question.

Henno suggests to prove that $p^{-1}(p(O)) = [0,1]$ so that since [0,1] is open in $I$, then $p^{-1}(p(O))$ is open in $I$, implying $p(O)$ is open in $I/\sim$. I have tried to prove it by showing that $p^{-1}(p(O)) \subseteq [0,1]$ and $[0,1] \subseteq p^{-1}(p(O))$. The former is obvious, so i focused on the latter. We have that

$$ \begin{equation} x \in p^{-1}(p(O)) \Leftrightarrow p(b) \in p(O) \Leftrightarrow \exists y \in O \ni p(x) = p(y) \Leftrightarrow \exists y \in O \ni [x] = [y]. \end{equation} $$

Let $a \in [0,1]$. Since $O$ is open in $I$, then it contains rational and irrational numbers. If $a$ is rational, then i can choose any rational $b$ in $O$ so that $[a] = [b]$, since all rational in [0,1] have the same equivalence class. This proves that every rational number in [0,1] is also in $p^{-1}(p(O))$.

I have been struggling when $a$ is irrational. Here is my work.

If $a$ is irrational, then $a$ is either in $O$ or not. It is obvious that if $a \in O$, then $a \in p^{-1}(p(O))$. I am stuck when proving for $a \notin O$. Is my approach correct or is there a better way to prove this?

Note: i use the same book as the asker in the question i linked.

Answer 1

Let $O$ be a nonempty open subset of $[0,1]$, our goal is to prove that $p^{-1}(p(O)) = [0,1]$.

The quickest way is to show that for every $x \in [0,1]$, the equivalence class of $x$, i.e. $p ^{-1}(p(x))$ or $\{x+\mathbb{Q}\}\cap[0,1]$, is dense in $[0,1]$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, $x+\mathbb{Q}$ is also dense in $\mathbb{R}$. This is actually a bit tricky, but I suppose you know that any open subset of $\mathbb{R}$ is the disjoint union of countably many open intervals, or that the open intervals form a basis for the topology of $\mathbb R$.

Let $(a,b)$ be any nonempty open interval in $\mathbb{R}$, then $a-x<b-x$, and there exists some $q \in \mathbb{Q}$ such that $a-x<q<b-x\implies a<x+q<b$. Therefore, $x+\mathbb{Q}$ is dense in $\mathbb{R}$.

Now since a continuous function maps dense subsets to dense subsets, $\{x+\mathbb{Q}\}\cap[0,1]$ is dense in $[0,1]$. We conclude that $p ^{-1}(p(x))$ intersects every open subset $O$ of $[0,1]$, thus there exists some $y \in O$ satisfying $p(x)=p(y)$.

Answer 2

For each $x \in [0,1]$ we have $p^{-1}(p(x)) = (x + \mathbb Q) \cap [0,1]$.

Consider any $a \in [0,1]$.

Pick any $a' \in O' = O \setminus \{0,1\}$ (note that $O'$ must be non-empty because otherwise $O \subset \{0,1\}$ which is impossible for an open set).

There exists $\epsilon > 0$ such that $(a'- \epsilon, a' + \epsilon) \subset O$. There exists a rational number $d$ such that $a - a' - \epsilon < d < a - a' + \epsilon$. Then $a' - \epsilon = a - (a -a' + \epsilon) < a - d < a - (a - a' - \epsilon) = a' + \epsilon$, hence $a - d \in (a'- \epsilon, a' + \epsilon) \subset O$. Therefore $a \in ((a-d) + \mathbb Q) \cap [0,1]) = p^{-1}(p(a-d)) \subset p^{-1}(p(O))$.