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Let $E \subset X, p \in X$ and $(X,d)$ a metric space such that $E := \{q \in X\mid d(p,q) \leq r\}$. By definition, this is a closed ball with center $p$ and radius $r$. I now want to show that this set is closed (i.e. it contains all its limit points).

Proof

Let $k$ be a limit point of $E$. Let $\epsilon >0$ and define $N_{\epsilon}(k)$ as the open ball (neighborhood) with center $k$ and radius $\epsilon$. By definition of limit point, there exists $z \in E \cap N_{\epsilon}(k)$ such that $z \neq k$. Hence we have $d(p,z) \leq r$ and $d(k,z) < \epsilon$, so by triangle inequality, we have $d(p,k) < r + \epsilon$. Because $\epsilon >0$ is chosen arbitrarily, it follows that $d(p,k) \leq r$ (if $d(p,k) > r$, then $d(p,k) - r > 0$, implying that $d(p,k) < d(p,k)$, which is absurd). We deduce that $k \in E$ and hence $E$ is closed. $\quad \square$

Is my proof correct? Is there an easier proof?

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  • $\begingroup$ Related: math.stackexchange.com/questions/661759/… and math.stackexchange.com/questions/1435991/… $\endgroup$ – Henry Aug 20 '17 at 14:34
  • $\begingroup$ I know. I saw this post but that proof uses complements, where as mine is more direct. $\endgroup$ – user370967 Aug 20 '17 at 14:34
  • $\begingroup$ The center of the closed ball $E$ $\endgroup$ – user370967 Aug 20 '17 at 14:42
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    $\begingroup$ Your proof is good. Your proof uses only the def'n of a metric, so it can't really get any easier. E.g. if $X=\mathbb R$ and $d(x,y)=|x-y|,$ prove that $[-1,1]$ contains all its limits...BTW: An intro-course teacher might want you to show details of the use of the triangle inequality......BTW: In general, be careful to note that the $closure$ of the open ball $B_d(x,r)$ may fail to equal $\{y: d(y,x)\leq r\}.$ E.g. if $X$ has more than 1 point and $d(x,y)=1$ when $x\ne y,$ then $B_d(x,1)=Cl(B_d(x,1))=\{x\}$ but $\{y: d(y,x)\leq 1\}=X.$ $\endgroup$ – DanielWainfleet Aug 20 '17 at 17:40
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    $\begingroup$ Thanks for your useful comment. Earlier this day, I conjectured that the closure of an open ball would be a closed ball, but as you showed yourself this isn't true. $\endgroup$ – user370967 Aug 20 '17 at 18:56
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Your proof looks good to me. The key point, that $a<b+\epsilon$ for all $\epsilon>0$ implies $a\le b,$ is used explicitly, which is good expository style.

An alternative proof: the "closed" ball $E$ is the inverse image of the closed set $[0,r]$ under the continuous map $q\mapsto d(p,q).$ Hence it is closed. Or, it is the complement of the inverse image of the open set $(r,\infty)$, and thus closed.

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  • $\begingroup$ Sorry I'm not interested in alternative proofs. I just want to know whether my proof works (In my text, continuity is not yet discussed btw) $\endgroup$ – user370967 Aug 20 '17 at 15:14
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    $\begingroup$ Then why does the last line of your question ask "Is there an easier proof?" $\endgroup$ – kimchi lover Aug 20 '17 at 15:16
  • $\begingroup$ I first want to know whether my proof is correct, then we can talk about alternative proofs. I should have been clearer. Sorry. $\endgroup$ – user370967 Aug 20 '17 at 15:18
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An alternate proof is to show that $X - E$ is open. To show that $x \in X - E$ is an interior point, consider the open ball centered at $x$ with radius $d(x, p) - 1$, and then argue that this is disjoint from $E$ using the triangle inequality.

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Your proof is correct.You proved that the closure of $E$ is equal to $E$ thus $E$ is closed.

To answer also your second question,there is an easier proof for this:

Let $x_n \in E$ such that $x_n \rightarrow x$.

Then $d(x,p) \leq d(x,x_n)+d(x_n,p) \leq d(x,x_n)+r$

Taking limits to both sides of the inequality we have that

$$d(x,p) \leq r \Rightarrow x \in E$$

Thus $E$ is closed.

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