The metric $d = |x-y|$ is given.
How can we say that closure of an open ball equals to closed ball with the same radius on the Euclidean space?
A general example of the equality of closure of open ball and closed ball is the Euclidean space.
Why?
[Math] How to say that closure of an open ball equals to closed ball with the same radius on the Euclidean space
general-topology
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Best Answer
WLOG we can assume that the ball is $B(0,1)$ (translations and dilations are homeomorphisms). Let $B$ be the open ball, $B_c$ the closed ball, $\bar{B}$ the topological closure of $B$. It's easy to show that $B_c$ is closed, because $f : x \mapsto d(0,x)$ is continuous (it's Lipshitz) and $B_c = f^{-1}[0,1]$. It contains $B$, therefore $\bar{B} \subset B_c$. Therefore we just need to prove that points on the "boundary" of $B_c$ are in $\bar{B}$ (ie. the points at distance exactly 1 from 0, we don't know it's the actual boundary yet). Let $x \in E$ st. $|x| = 1$, and let $x_n = (1-{1 \over n})x$, then $x_n \in B$ and $x_n \rightarrow x$, therefore $x \in \bar{B}$. Therefore $\bar{B} = B_c$.
Note that in this proof I only used the properties of a normed vector space, not necessarily euclidean; but it is false in general metric spaces. Take for example with the discrete metric on $\mathbb{N}$ ($d(x,y) = 1$ for $x \not= y$), then $B = B(0,1) = \{ x \in \mathbb{B} : d(0,x) < 1 \}$ is just $\{0\}$ and $\bar{B} = B$, but $B_c = \{ x \in \mathbb{N} : d(0,x) \leq 1 \} = \mathbb{N}$.