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 $X \setminus \Phi = \{y \in X \ | \ d(x, y) > r\}$. Pick $y \in X \setminus \Phi$. For this $y$ we have $d(x, y) > r$ which implies that $d(x, y) = r + \epsilon$ for some $\epsilon > 0$.
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 $d(x, z) > r$. To that end observe that the triangle inequality gives us
\begin{align*}
& d(x, y) \leq d(x, z) + d(z, y) \\
& \implies \epsilon + r \leq d(x, z) + d(y, z) \\
& \implies \epsilon + r < d(x, z) + \epsilon \ \ \ \ \ \ \ \ \ \text{since $d(y, z) < \epsilon$} \\
& \implies r < 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?
Best Answer
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.