# How can we say that closure of an open ball equals to closed ball with the same radius on the Euclidean space?

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?

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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 $B_c = \{ x \in \mathbb{N} : d(0,x) \leq 1 \} = \mathbb{N}$.

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The open ball is contained in the closed ball. The closed ball is closed, being the inverse image of the closed set $[0,r]\subseteq\mathbb R$ by a continuous function $f:x \mapsto d(x,x_0)$. Hence the closure of the open ball is contained in the closed ball because the closure is the intersection of all closed sets that contain the open ball. Finally, the closure contains the closed ball because for each $z$ in the border of the closed ball there is a sequence of points in the open ball that converges to it. This means that the closure is the closed ball.

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To show that set $B$ is the closure of set $A$, you need to show that $$B=A\cup\partial A$$ where $\partial A$ is the set of all limit points of $A$.
Now take any point $x$ outside of the closed ball $B$. The distance between $x$ and the center $z$ of the ball is greater than $r$, therefore, there is a small open ball of radius $\le(d(x,z)-r)$ around $x$ that has no points in $A$.
On the other hand, the distance between every point $x$ in $B$ and $z$ is $\le r$. And for every positive $\epsilon$ the ball of radius $\epsilon$ centered at $x$ has some points in $A$ (must be proved: just pick a point).
Therefore, no point outside of $B$ is a limit point, and every point in $B$ is a limit point.