Can all open sets in R^n be expressed as countable union of open cubes?

[CantorSet]

Hi everyone,

I came across a problem that requires knowing this fact.

But can any open set in $$R^n$$ be expressed as the countable union of "cubes". That is subsets of the form $$(a_1,b_1) \times ... \times (a_n, b_n)$$.

 

[micromass]

Hi CantorSet!

The answer is yes! For notational purposes, I'll do this is one dimension, but multiple dimensions is quite analogous.

So, take G open, then around every $x\in G$, we can find an open interval such that

$$x\in ]a,b[\subseteq G$$

By shrinking the interval, we can assume the endpoints to be rational. Thus

$$G=\bigcup\{]a,b[~\vert~a,b\in \mathbb{Q},~]a,b[\subseteq G\}$$

We can even take the intervals/cubes to be disjoint, but that's somewhat more difficult...

 

[CantorSet]

Thanks for the response, micromass!

I see that we can express any open set as the union of open intervals. But can we express it as a countable union of open intervals. What if the number of points in G is uncountably infinite?

You mention that we can even use disjoint intervals. Would this guarantee the union is over countably many?

 

[micromass]

What if the number of points in G is uncountably infinite?

Doesn't the same proof apply regardless on the number of elements in G?

 

[AKG]

CantorSet, what micromass didn't mention is that when you write an open set as a union of open cubes with rational endpoints, then the result is a union of countably many sets. This is because there are only countably many cubes with rational endpoints (something you can prove for yourself probably). In his procedure, you have your open set $G$, a point $x \in G$, and then you choose an open interval, $x \in ]a_x, b_x[ \in G$. Then you shrink $]a_x, b_x[$ to an interval with rational endpoints, say $]p_x, q_x[$. If $G$ is uncountable, and you do this for every $x \in G$, then since there are only countably many intervals with rational endpoints, it means that for many different $x \neq y \in G$, you'll have ended up choosing the same interval: $p_x = p_y,\ q_x = q_y$.

 

[CantorSet]

Got it. Thanks for the help, guys.