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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?

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  • $\begingroup$ yes, its correct. I assume, from the proof, that an open set was defined as a set where there exists some open ball centered at every point that is contained in the set. $\endgroup$ – Masacroso Mar 20 '18 at 19:52
  • $\begingroup$ Note that the notation of $\overline{B(x,r)}$ is reserved for the closure of the open ball, rather than the closed ball. $\endgroup$ – TrostAft Mar 20 '18 at 19:55
  • $\begingroup$ why $d(y,z)<\varepsilon$ with the same $varepsilon$ of $r-\varepsilon$? $\endgroup$ – gbox Jul 23 at 14:52
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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.

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Yes, it is rigorous and satisfactory. It can be shortened, though:$$d(x,z)\geqslant d(x,y)-d(z,y)>r+\varepsilon-\varepsilon=r.$$

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