# “Every open ball is closed" and "every closed ball is open", does one imply the other?

In a metric space $$(X,d)$$, by a closed ball I mean a set of the form $$\{y: d(y,x)\le r\}$$, where $$r > 0$$.

A common example where every open ball is a closed set and every closed ball is an open set is the ultrametric spaces: the metric spaces where $$d(x,y)\le \max\{d(x,z),d(y,z)\}$$. I would like to have an example where every open ball is a closed set but not every closed ball is an open set, and an example vice versa.

Definitions. An open ball is a set of the form $$B_r(a) = \{x \mid d(x,a) < r\}$$ where $$r > 0$$. A closed ball is a set of the form $$\overline{B}_r(a) = \{x \mid d(x,a) \le r\}$$ where $$r > 0$$.

• Nice question.. May 21, 2022 at 14:47
• @GEdgar Thanks! May 22, 2022 at 14:55
• I came here today intending to offer a bounty. But it has already been done. May 26, 2022 at 14:29
• @GEdgar Since you are intending the same as me, I will offer the bounty to the answer from Jason DeVito :) Thanks again for your contribution and concern for this question! May 27, 2022 at 12:44

Example 1. In the line, consider the set $$X = \{0, 1, 1/2, 1/3, 1/4,\dots\}$$ with the usual metric. I claim: every closed ball in $$X$$ is open in $$X$$; but the open ball $${B}_1(1)$$ is not closed in $$X$$.

Here's an example going the other direction: all open balls are closed, but there a closed ball which is not open. But let me note that I'm going to use the notation $$CB_r(p)$$ for the closed ball, rather than $$\overline{B}_r(p)$$, because the latter notation makes me think of the closure of an open ball. Then one has $$\overline{B}_r(p)\subseteq CB_r(p)$$, but the inclusion can be strict. E.g., in a discrete metric space $$\overline{B}_1(p) = p$$, while $$CB_1(p)$$ is the whole space.

Consider the Hilbert cube $$H:=[0,1]^\mathbb{N}$$ with metric $$d( (x_1,x_2,...), (y_1,y_2,..)) = \sum_i \frac{|x_i - y_i|}{2^i}$$.

For each $$i\in \mathbb{N}$$, let $$e_i$$ denote the element of $$H$$ which is all $$0$$s except for a $$1$$ in the $$i$$th slot. So, $$e_1 = (1,0,0,..)$$, $$e_2 = (0,1,0,...)$$, etc. I'll also write $$0$$ for the all-zeroes element $$(0,0,...)$$

Let $$Y\subseteq H$$ be given by $$Y = \{e_i, 0\}$$. Then $$Y$$ will be the desired example. That is, I claim that 1) every open ball in $$Y$$ is closed but 2) that there is a closed ball in $$Y$$ which is not open.

As a preliminary observation, note that for $$i\neq j$$, that $$d(e_i,e_j) = 1/2^i + 1/2^j > \max\{1/2^i, 1/2^j\}$$, and that $$d(e_i,0) = 1/2^i$$. It follows that each $$e_i$$ is isolated, with, e.g., $$B_{1/2^i}(e_i) = \{e_i\}$$ being an open ball containing just the one point $$e_i$$. Moreover, for each $$e_i$$, $$0$$ is the unique closest point to it.

Here's the proof of 1) To begin with, note first that every open ball centered at $$0$$ is closed because the complement can only possible contain some of the $$\{e_i\}$$ which are all isolated.

So, consider $$B = B_{r}(e_i)$$, an open ball centered at $$e_i$$ of radius $$r > 0$$. Since all the $$e_i$$ are isolated, the only case we need to consider is if $$0\notin B$$. But $$0$$ is the closest point to $$e_i$$, so if $$0$$ is not in $$B$$, then neither are any $$e_j$$ with $$i\neq j$$. Thus, any ball around $$0$$ which doesn't contain $$e_i$$ witnesses the fact that the complement of $$B$$ is open. E.g., one can take $$B_{1/2^i}(0)$$. This concludes the proof of 1).

We now prove 2), that there is a closed ball which is not open. To that end, consider $$C:=CB_{1/2}(e_1)$$. This set clearly contains $$e_1$$ and $$0$$; since $$0$$ is the unique closest point to $$e_1$$, it follows that $$C = \{e_1,0\}$$. So, to show $$C$$ is not open, it's enough to show that every ball around $$0$$ contains an $$e_i$$ other than $$e_1$$.

But $$d(e_i,0) = 1/2^i\rightarrow 0$$ as $$i\rightarrow \infty$$, so any open ball around $$0$$ contains all but finitely many of the $$e_i$$.