I came across this statement in the first chapter or Rudin Real and Complex Analysis.

Rudin states that every open set in the plane is a countable union of rectangles.

Looking for a proof I stumbled upon this question that is very similar, but I could not understand the proof provided by the accepted answerer.

How do i prove that every open set in $\mathbb{R}^2$ is a union of at most countable open rectangles?

If I construct open k-cells containing $x \ \forall{x} \in E $ that are contained in the open ball around x then would I not be missing a part of the open balls that is a part of the set $E$ ?

So the countable union would not be $E$ because I am missing a part of the balls that form $E$.

Maybe a fresh proof of the statement in Rudin will clear my mind.


The kicker to that proof is that, regardless of which $x\in E$ is chosen, we can construct such a $k$-cell. Note, then, that each such $k$-cell is a subset of $E,$ and that each $x\in E$ is contained in such a $k$-cell. While it's true that each $k$-cell misses a part of $E,$ no part of $E$ is missed by all such $k$-cells (that is, every part of $E$ is hit by at least one of the $k$-cells). Consequently, the union of all such $k$-cells contained in $E$ is precisely the set $E$!

Since there are only countably-many $k$-cells with vertices having only rational coordinates, then for each open $E,$ there are at most countably-many such $k$-cells contained in $E.$

  • $\begingroup$ Thanks got it, now I am unsure to whom I should give the accepted answer. $\endgroup$ – Monolite Nov 5 '14 at 15:13
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    $\begingroup$ That's up to you. Some people accept the (chronologically) first useful answer. Some people accept the answer that made it easiest to understand, regardless of chronology. $\endgroup$ – Cameron Buie Nov 5 '14 at 15:16
  • $\begingroup$ Would this work even if we took contained closed rectangles? $\endgroup$ – Monolite Nov 5 '14 at 16:22
  • $\begingroup$ Indeed! You can show that every $x\in E$ is contained in some closed $k$-cell with vertices having only rational coordinates, and such that the $k$-cell is contained in $E.$ Hence, $E$ is a union of countably-many such closed $k$-cells by similar reasoning. Note that this gives us an example in which an infinite union of closed sets need not be closed. $\endgroup$ – Cameron Buie Nov 5 '14 at 17:06

For every $x \in E$, we can get an open rectangle $R_x$ containing x within E such that $R_x$ has rational vertices (as the linked question describes). So, it is certainly the case that $\cup_{x \in E} R_x = E$ ($E \subset \cup_{x \in E} R_x$ since for each $x\in E, x \in R_x$). Also, the sets $R_x$ each come from the set $\{(a_1,b_1)\times(a_2,b_2) : a_1,a_2,b_1,b_2\in \mathbb{Q}\}$ which is countable. So, there can only be countably many distinct $R_x$.


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