# Is it possible to prove that the metric space is an open set without choice?

Suppose that $(X, \rho)$ is a metric space, $|X| > 1$. Is it possible to prove that $X$ is an open set without assuming the axiom of choice?

As I understand it, the challenge is to find a way to assign an open ball to each point of the space (assign a radius, basically). The way I found is to select some point $y \in X$ and assign to each point $x \in X\setminus\!\{y\}$ the ball $B_{2\rho(x,y)}(x)$. Then it is easy to show that the union of these balls is $X$, and so $X$ is open.

Is there another way to prove that $X$ is open without relying on the axiom of choice? Or am I wrong and the axiom of choice is not required for the above proof?

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for a top space $(X,\mathcal{T}\ )$ we always have $\emptyset, X\in\mathcal{T}$. Or if you want to show $X$ is a union of balls, just take the ball of radius one around each point. – yoyo Mar 2 '11 at 16:53
The proof that you sketched doesn't use the axiom of choice in the first place: you explicitly defined the radius of all the balls in terms of a single point $y$. You don't need the axiom of choice to know that a nonempty space has a point in it. Any space with $|X|>1$ has at least two points. – Carl Mummert Mar 2 '11 at 16:56
@Carl: Can you explain further, please? I don't understand how it can be used to prove that $X$ is an open set. – Alexei Averchenko Mar 2 '11 at 16:57
@yoyo: But why is $X$ open? Is it just a convention or must it necessarily be open in order for topological space to generalize metric space? – Alexei Averchenko Mar 2 '11 at 16:58
Alexei: Yoyo answered that, and Chris Eagle below: just take the ball of radius one around every single point of the space. The space is certainly the union of these balls. But your original proof didn't use the axiom of choice anyway. What definition are you using for "open set"? – Carl Mummert Mar 2 '11 at 17:00

## 4 Answers

You can pick the radius of the open ball centred at $x$ however you like, and the union of all the balls will be $X$. For example, you could just let all the radii be 1.

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What if the metric is $$\rho(x,y) = \begin{cases} 0 & \text{if } x = y\\ 2 & \text{otherwise}\end{cases}$$ I don't think that it is true that for any metric there exists a ball smaller than some fixed $M > 0$. – Alexei Averchenko Mar 2 '11 at 16:54
What if it is? In that case, the open ball of radius 1 around $x$ is just $\{ x \}$, so their union is $\bigcup_{x \in X} \{ x \}=X$. – Chris Eagle Mar 2 '11 at 16:55
@Alexei: The union of all open balls of radius 1 around all points is still equal to the entire space. So the entire space is still a union of open balls. – Arturo Magidin Mar 2 '11 at 16:56
Do we consider $\{x\}$ to be a ball? – Alexei Averchenko Mar 2 '11 at 17:00
For any point $x$ and any radius $r > 0$ there is an open ball of radius $r$, by definition. The ball may only contain $x$, but that's not important here. If $x$ is an isolated point then $\{x\}$ will always be a ball. – Carl Mummert Mar 2 '11 at 17:01

There is no choice issue here at all, and no need to pick a particular radius, since in the definition of open set, there is no obligation for us to assign to each point a unique radius. Rather, we can assign many radii to each point, and this will also verify opennes. We just need to know that every point has at least one ball around it contained in the set; perhaps there are many balls around that point contained in the set (and indeed, of course any smaller ball also works). So let us simply take all balls around any point of any possible radius to see that the whole space is open.

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$X$ is an open set because every topology (=collection of open sets) on $X$ must contain $X$, see the definition. Nothing metric here.

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That's exactly my point - do we define $X$ as open by convention or is it provably open at least for a metric topology? – Alexei Averchenko Mar 2 '11 at 16:52
@Alexei Being open means belonging to the topology. A topology satisfies some axioms, one of them is that $X$ belongs to the topology. Maybe the fact that the emptyset is closed is more natural to you? And this is equivalent to $X$ being open... – Did Mar 2 '11 at 16:56
Forget about topology for a second. There is a definition of an open set for a metric space, that's what I'm concerned with for now. – Alexei Averchenko Mar 2 '11 at 17:01
@Alexei Not sure I understand your point. Consider for example Chris's (excellent) solution: this uses (rightly) the fact that any union of open balls is an open set. But why is that so? Because open balls are, well... open sets and any union of open sets is an open set, right? Oops... here is precisely one of the axioms which define a topology. – Did Mar 2 '11 at 17:34
Do you think axioms of topology are God-given or something? I'm looking into why they are the way they are. – Alexei Averchenko Mar 2 '11 at 17:37

As an aside, note that you don't need Choice to validate the openness of an arbitrary open subset $U\subseteq X$. That is, we can assign an open ball around each $u\in U$ without Choice: let $\delta_u = \frac12 \inf\{\rho(u,x) : x\notin U\}$ and set $F_u = B_{\delta_u}(u)$.

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