# Basic topology - can we find such a cover always

$A \subset \mathbb R^n$ is open, and $B \subset A$ is a null set, meaning:

$\forall \epsilon > 0$ there are open boxes $Q_1, Q_2, \dots \subset \mathbb R^n$ such that $$B \subset \bigcup_{j=1}^{\infty}Q_j$$

and $$\sum_{j=1}^{\infty}v(Q_j) < \epsilon$$

I want to show that not only do such boxes exist, but we can even find such boxes that also have the property that $Q_j \subset A$ for all $j$, and it's quite difficult for me.

Since $A$ is open, it can be represented as a countable union of closed boxes $A = \bigcup_{j=1}^{\infty}\overline{Q_j}$

So $B \subset \bigcup_{j=1}^{\infty}\overline{Q_j}$, but we don't know that the sum of the volumes of these boxes is less than epsilon, and these boxes are not open.

My intuition tells me that there is a very very small $\epsilon$ such that if the volumes of the covers of $B$ is less than $\epsilon$, then they are all in $A$, but I can't prove it.

Suppose $A$ is not open. Would that make a difference?

In $\mathbb{R}^n$ every open set can be expressed as a countable union of open boxes. You can get the required sequence of boxes the folowing way:

1) Take $Q_j$ such that $B\subset\bigcup\limits_j Q_j$ and $\sum\limits_j \nu(Q_j)<\varepsilon$.

2) Express $Q_j \bigcap A$ as a countable union $\bigcup\limits_i \tilde{Q}_j^i$.

3) Since a countable union of countable sets is countable you can reindex all $\tilde{Q}_j^i$ as $\tilde{Q}_k$.

• I know that every open set is a countable union of closed boxes. Is it also true for open boxes? Commented Apr 19, 2018 at 11:23
• Yes, because the open boxes with rational sides form a countable base of the topology in $\mathbb{R}^n$. Commented Apr 19, 2018 at 11:27