I am stuck in proving two parts of this proof.

Let $U$ be the open set in question. Then, for all $x \in U$, since $U$ is open in $\mathbb{R}^n$, we can build an open ball $B(r,x)$ around $x$ with $r>0$, and construct an $n$-rectangle denoted $R_x = (a_1,b_1)\times \cdots \times (a_n,b_n)$ contained inside $B(r,x)$ where $a_i, b_i \in \mathbb{Q}$ for all $i \in [n]$ since $\mathbb{Q}$ is dense in $\mathbb{R}$.

One part I have problems with is how can I rigorously prove that

$$\bigcup_{x\in U} R_x = U$$

and that there is a countable number of rectangles using the fact that all my points $a_i,b_i$ are rational?

Thank you!

  • $\begingroup$ For all $x$ $R_x \subset U$, then $\cup R_x \subset U$. And there is at most countably many of $R_x$s $\endgroup$ – dEmigOd Jan 25 at 8:54
  • $\begingroup$ I already stated that in my question. My problem is how to prove it? Or is it evident with what I've got already? $\endgroup$ – The Bosco Jan 25 at 9:00
  • $\begingroup$ $\forall x \in U \exists R_x : x \in R_x \Rightarrow \forall x \in U x \in \cup R_x \Rightarrow U \subset \cup R_x$ - this gives the equality. Countability comes from knowing the cardinality of $\mathbb{Q}^n$ $\endgroup$ – dEmigOd Jan 25 at 9:02
  • $\begingroup$ but for example, in $\mathbb{R}$ you could argue this because the subset $U$ is a disjoint countable union of intervals. Therefore, you could argue that each one of them has a unique rational number, but you can't do that in $\mathbb{R}^n$ for $n>1$ since now the rectangles have to overlap (although not completely). $\endgroup$ – The Bosco Jan 25 at 9:06
  • $\begingroup$ nothing in the claim suggests boxes should be disjoint, why bother? $\endgroup$ – dEmigOd Jan 25 at 9:14

As $R_x\subseteq U$ for all $x \in U,$ we have $\bigcup_{x\in U} R_x \subseteq U.$ Now if $x$ is any point in $U,$ then $x\in R_x \subseteq \bigcup_{x\in U} R_x.$ Thus $$ \bigcup_{x\in U} R_x = U. $$

There is only a countable number of rectangles with coordinates in $\mathbb{Q}^n$ as $\mathbb{Q}^n$ is countable, as is normally shown using Cantor's diagonal argument.


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