# Union of a countable collection of open balls

Show that a subset of $\mathbb{R^n}$ is open if and only if it is the union of a countable collection of open balls.

Attempt:

I know a set G is open if there exists a $\epsilon >0$ such that every point $x\in A$ satisfies $||x-y||<\epsilon$. Then I must show that $x$ is in some ball in $A$ which will make every point an interior point. What I don't understand in this question is the use of the words "countable collection" why does it have to be countable? Is it because an infinite collection of open sets is not open?

• What is A here? – user60469 Feb 10 '13 at 7:17
• Are you asking the question for $\mathbb{R}$ only, or for general $\mathbb{R}^n$? For the former, it would be more common to say "open intervals" and not "open balls", which is more common in higher dimensions. – Herng Yi Feb 10 '13 at 7:19
• Intuitively, when unioning an uncountable collection of open balls, we can achieve the same result by using a union of a countable subcollection. It's something like $\bigcup_{x \in \mathbb{R}^+}(x, \infty) = \bigcup_{y \in \mathbb{Q}^+}(y, \infty)$. It has something to do with the completeness property of the real numbers, which is used to construct the real numbers (uncountable) from the rationals (countable). – Herng Yi Feb 10 '13 at 7:27
• @HerngYi Yes, I meant to say $\mathbb{R^n}$. I will edit – Lays Feb 10 '13 at 7:38

($\rightarrow$) Let $Y \subset \mathbb{R}^n$ be open, then for every $x \in Y$, there exists an open $U_x \subset Y$ such that $x \in U$. Clearly the set $\left\{ \displaystyle\bigcup U_x \colon x \in Y \right\} = Y$. However the collection $\mathscr{F} := \left\{ U_x \colon x \in Y \right\}$ may not be countable. Since $\mathbb{Q}^n$ is dense in $\mathbb{R}^n$, it is also dense in $Y$. Restrict $\mathscr{F}$ by considering then $\mathscr{G} := \left\{ U_x \in \mathscr{F} \colon x \in Y \cap \mathbb{Q}^n \right\}$, which is countable and covers $Y$ since $\mathbb{Q}^n \cap Y$ is dense in $Y$.

($\leftarrow$) Suppose that $Y \subset \mathbb{R}^n$ is the union of a countable collection $\mathscr{C}$ of open sets. By hypothesis, for every $x \in Y$, there exists some open set $U_x \in \mathscr{C}$ such that $x \in U_x$ and $U_x \subset Y$. Therefore by definition, $Y$ is open.

• Tom, what do you mean by dense? – Lays Feb 10 '13 at 7:43
• I made some typos, let me correct them! By $A$ is a dense subset of $B$ I mean if you pick any element $x \in B$ and choose any $\epsilon > 0$, then there is some $y \in A$ such that $x$ is in the ball of radius epsilon of $x$. – tomcuchta Feb 10 '13 at 7:44
• It's not really possible in general to describe a closed set as the union of open sets of a given topological space unless you use to the subspace topology on your set, but that could be argued as cheating :). – tomcuchta Feb 10 '13 at 8:33
• @tomcuchta: It's great to see you on here! +1 by the way! – Clayton May 10 '13 at 16:55
• I know this is old, but for future reference I'm not sure the first part of the proof is complete. $U_x$ for $x\in Y\cap\mathbb{Q}$ could be chosen such that the radius of each open ball is $|x-\pi|$, in which case your set wouldn't contain $\pi$. I think you should union over all balls with rational radius which are contained in $Y$. – JLA Jun 23 '13 at 20:05

see: Math 104. Metric Spaces, Proposition 29.

This question shouldn't be read to imply that an uncountable union of open balls/sets is not necessarily open. Indeed, every union of open sets is open, a fact that can easily be proved using the definition of openness implicit in your question (and something that you have likely seen already). This question is giving a finer analysis of open subsets of $\mathbb{R}^n$: Clearly every open subset of $\mathbb{R}^n$ is a union of open balls, but in fact you only need countably many of them to "make" any given open set.

This property is something closely related to the metric structure of $\mathbb{R}^n$ as well as the following topological properties (if these terms are unfamiliar, don't worry about them too much right now; you'll see them if you continue your mathematical studies):

• $\mathbb{R}^n$ is separable (meaning there is a countable dense subset; think of the collection of all points in $\mathbb{R}^n$ with all coordinates rations);
• $\mathbb{R}^n$ is second-countable (meaning that there is a countable collection of open sets such that every open set is a union of sets from this collections; think of the collection open balls of rational radius centred at points with all coordinates rational).
• $\mathbb{R}^n$ is Lindelöf (meaning for every collection of open sets covering $\mathbb{R}^n$ there is a countable subcollection which also covers $\mathbb{R}^n$).

But be warned that not all metric spaces are as nicely behaved as $\mathbb{R}^n$. Indeed, there are metric spaces where not every open set can be expressed as a countable union of open balls. Here are two examples, one of which is somewhat more basic than the other:

• Given any set $X$ the function $d : X \times X \to [ 0 , + \infty )$ defined by $$d ( x , y ) = \begin{cases}0, &\text{if }x = y \\ 1, &\text{if }x \neq y\end{cases}$$defines a metric on $X$ (called the discrete metric). In this space every subset of $X$ is open. If $X$ is uncountable, then any uncountable proper subset $U \subsetneq X$ is an open set which cannot be expressed as the union of countably many open balls.
• A (slightly) more advanced example (which I'm tied to, I guess) would be a (metrizable) hedgehog space of uncountable "spininess".

Again, these examples (especially the latter one) might be somewhat beyond your current level, but they are offered as a warning against thinking that open sets in all metric spaces are countable unions of open balls.

• What about the discrete metric on your favorite uncountable set? That is a simple example for a metric space where "most" open sets are not countable unions of open balls. – Asaf Karagila Feb 10 '13 at 8:37
• @Asaf: Ever get to a point where the simplest examples escape you? Thanks, I'll edit the above presently. – user642796 Feb 10 '13 at 8:55
• Arthur, ever? I'm the point at infinity of the points which you describe! :-) – Asaf Karagila Feb 10 '13 at 10:43
• Many thanks for this! – Lays Feb 11 '13 at 1:27

In every topological space, every union of open sets is open. This is one of the axioms of topology.

However countable unions are easier to handle, compared to to uncountable unions, and they still have more expressive power than finite unions (e.g. the unit "square" $D=\{(x_1,\ldots,x_n)\mid-1<x_i<1,i=1,\ldots n\}$ is not a finite union of balls, and vice versa).

So we are particularly interested in spaces that have the property that some nicely described collection is sufficient to generate all the topology very "quickly", i.e. every open set is expressible as a countable union of these sets. In our case open balls, and we can reduce this even more: open balls around points in $\mathbb Q^n$ whose radii is rational.

• Many thanks to you! – Lays Feb 11 '13 at 1:30