# In $\mathbb{R}^n$ with the usual metric, is the closure of an open ball equal to the closed ball?

I read here : When is the closure of an open ball equal to the closed ball? and here : closed ball in euclidean space that in $(\mathbb{R}^n, d)$, which is Euclidean Space with the standard metric, the closure of an open ball is equal to the closed ball.

But pick a point $x \in \mathbb{R}^n$, and consider the open ball around $x$ of radius $0$, that is $\Phi = B_{(\mathbb{R}^n, d)}(x, 0) = \{y \in \mathbb{R}^n \ | \ d(x, y) < 0\} = \emptyset$. Now compare that to the closed ball around $x$ of radius $0$, which is $\Gamma = B_{(\mathbb{R}^n, d)}(x, 0) = \{y \in \mathbb{R}^n \ | \ d(x, y) \leq 0\} = \{x\}$.

Based on the answers in the questions above, we should have $\overline{\Phi} = \Gamma$ however we have $\overline{\Phi} =\Phi = \emptyset \neq \Gamma = \{x\}$, contradicting the answers above, so what am I missing here?

• Is an open ball of radius zero really an open ball? Commented Dec 19, 2017 at 6:20
• Where in those posts do you see the explicit statement that a "radius" can be zero? Commented Dec 19, 2017 at 6:20
• Hold up: the metric $d$ satisfies $d(x,y) \geq 0$ for all $x$, $y$. Also, you would not have $\overline{\Phi} = \Gamma$; instead, you'd have $\overline{\Phi} = \Phi = \varnothing$. Commented Dec 19, 2017 at 6:24
• @MarkTwain Correct me if I'm wrong, but that's the same as what I said in my last paragraph Commented Dec 19, 2017 at 6:49
• @LordSharktheUnknown. I suppose you could say so, as it's the empty set, which is open. Commented Dec 19, 2017 at 8:07

You should exclude $0$ as a radius. Defining $B(x,r) =\{y: d(x,y) < r\}$ and $D(x,r) = \{y: d(x,y) \le r\}$ for $r>0$, in any metric space, we can easily prove that for all $r>0$:
$$\overline{B(x,r)} \subseteq D(x,r)$$
with equality holding in e.g. all metrics derived from norms in a vector space, of which the Euclidean distance is an example. In general metric spaces the inclusion can be proper, and if we do allow $r=0$, it's always proper for that "radius".