A closed ball in a metric space is closed Prove that any closed ball in a metric space is closed.


(Note that this is not a duplicate as this is a proof verification question and have a different proof, in my opinion, as compared to other proofs on this site)

My Attempted Proof:   Let $(X, d)$ be a metric space. Pick a closed ball $\Phi = \overline{B(x, r)} = \{y \in X \ | \ d(x, y) \leq r\}$. Note that at this moment in time $\overline{B(x, r)}$ is just a notation, we don't know if $\Phi$ is actually closed.

We now show that $\Phi$ is closed. To do this we show that $X \setminus \Phi$ is open. Observe that $X \setminus \Phi = \{y \in X \ | \ d(x, y) &gt; r\}$. Pick $y \in X \setminus \Phi$. For this $y$ we have $d(x, y) &gt; r$ which implies that $d(x, y) = r + \epsilon$ for some $\epsilon &gt; 0$. 

We claim that $B(y, \epsilon) \subseteq X \setminus \Phi$. To prove this claim, pick $z \in B(y, \epsilon)$. We now show that $z \in X \setminus \Phi$ by showing that $d(x, z) &gt; r$. To that end observe that the triangle inequality gives us 

\begin{align*} 
  &amp; d(x, y) \leq d(x, z) + d(z, y) \\ 
  &amp; \implies \epsilon + r \leq d(x, z) + d(y, z) \\
  &amp; \implies \epsilon + r &lt; d(x, z) + \epsilon \ \ \ \ \ \  \ \ \  \text{since $d(y, z) &lt; \epsilon$} \\
  &amp; \implies r &lt; d(x, z)
  \end{align*}

as desired. Hence $z \in X \setminus \Phi$ and we have $B(y, \epsilon) \subseteq X \setminus \Phi$, and since $y$ was arbitrary we have that $X \setminus \Phi$ is an open set in $(X, d)$ and thus $\Phi$ is a closed set. $\square$



Is this a rigorous and satisfactory proof? Can it be improved in any way?
Your proof is flawless.

The only recommendation that I have is to change the notation on your closed ball. 

$\overline{B(x, r)}$ is used for the closure of the open ball B(x, r) not the closed ball.

$\overline{B}(x,r)$ is a better notation for a closed ball.

The closure of an open ball is not necessarily the closed ball depending on the topology.
We know that a closed ball in a metric space is closed. If we replace
the triangular inequality of the metric d by:
d1
(x, y) ≤ s[d1
(x, z) + d1
(z, y)],
where s &gt; 1. Then check if still the closed ball in a metric space with
the metric d1 is closed?explain