# How to cover an open subset of $\mathbb{R}^n$ with balls?

I am trying to solve the following exercise:

If $U$ is an open subset of $\mathbb{R}^n$, show that there exists an increasing sequence $\{A_k\}^\infty_1$ of compact contented sets such that $U=\bigcup^\infty_{k=1}\ \mathrm{int}\ A_k$.

Hint: Each point of $U$ is contained in some closed ball which lies in $U$. Pick the sequence in such a way that $A_k$ is the union of $k$ closed balls.

It seems "increasing sequence" means that $A_k \subset A_{k+1}$ for all $k$.

Now, I think the hint says that $A_{k+1}$ should be $A_k \cup B_{k+1}$ for some suitable ball $B_{k+1}$, starting with $A_1 := B_1$. But I don't see a way to choose those balls so that eventually all points of $U$ are covered?

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What does "contented" mean? –  t.b. Jan 5 '12 at 13:10
I think he means "connected". –  Paul Jan 5 '12 at 13:19
Can you maybe take all rational points (points with rational coordinates) in $U$, and take balls around those? –  Thomas Rot Jan 5 '12 at 13:29
"Contented" (as posted) is correct... it means, more or less, a set to which a volume can be assigned. The exact definition, however, is irrelevant, since balls (and their unions) are contented (thus, if the hint is used, this condition is satisfied). –  koletenbert Jan 5 '12 at 13:30
Apparently contented is to be read as having CONtent, not as being conTENT. –  Brian M. Scott Jan 5 '12 at 13:31
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HINT: Every open set in $\mathbb{R}^n$ is a union of open (or closed) balls whose centres have rational coordinates and whose radii are rational; how many such balls are there?

In case the first statement isn’t obvious, suppose that $B$ is the ball of radius $r$ about a point $x\in\mathbb{R}^n$. If $x$ has all rational coordinates, there’s nothing to be done. Otherwise there is a point $y$ with all coordinates rational inside the ball of radius $r/2$ centred at $x$. Let $d$ be the distance between $x$ and $y$, let $q$ be a rational number such that $d<q<r/2$, and let $B'$ be the ball of radius $q$ centred at $y$; then $x\in B'\subseteq B$.

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To extend Brian's hint:

First cover $U$ as suggested by Brian by "rational" open balls whose closures are contained in $U$ . You may enumerate this covering: $$U=\bigcup_{i=1}^\infty F_i.$$ where each $F_i$ is an open ball with $\overline F_i$ contained in $U$.

Now consider the sequence: $A_1=\overline{F_1}\$, $A_2=\overline{F_1}\cup\overline{ F_2}\$, $A_3= \overline{F_1}\cup \overline{F_2}\cup\overline{ F_3}\$, $\ldots$.

Note that: finite unions of compact sets are compact

and

$F_i\subset \text{int} A_i$.

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