Union of a countable collection of open balls

Question

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?

Answer 1

($\rightarrow$) 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.

Answer 2

see: Math 104. Metric Spaces, Proposition 29.

Answer 3

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.

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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.

Answer 4

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.