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.
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.
Best Answer
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 $E.$