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Let $(X, d)$ be a metric space. Denote the open and closed ball as $$B(x_0, r) = \{x \in X \mid d(x, x_0) \lt r\},$$ $$D(x_0, r) = \{x \in X \mid d(x, x_0) \leq r\}.$$ Then $\overline{B(x_0, r)} = D(x_0, r)$.

  • $\overline{B(x_0, r)} \subseteq D(x_0, r)$
    Obviously $B(x_0, r) \subseteq D(x_0, r)$, and since I've already proved that $D(x_0, r)$ is closed, the inclusion follows ($\overline{B(x_0, r)}$ is the smallest closed set that contains $B(x_0, r)$).

  • $D(x_0, r) \subseteq \overline{B(x_0, r)}$
    The points $x$ that satisfy $d(x, x_0) \lt r$ are clearly in both sets. So we consider the points in $$S(x_0, r) = \{x \in X \mid d(x, x_0) = r\}.$$ We have to show that $S(x_0, r) \subseteq \overline{B(x_0, r)}$. Let $x \in S(x_0, r)$, then $$x - x_0 = \lim_{n \to +\infty} \frac{n}{n+1}(x - x_0)$$ and since $$\left\|\frac{n}{n+1}(x - x_0)\right\| = \frac{n}{n+1}r \lt r,$$ it follows that $x \in \overline{B(x_0, r)}$ (from this theorem, which I already proved).

Now, to me the proof seems correct. However, there are some metric spaces in which the equality between the closure of the open ball and the closed ball does not hold. The theorem that I have used is valid in all metric spaces. So where is the error?

rubik
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  • User JDH's answer to this question might be of interest. –  Apr 30 '16 at 07:58
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    Also, reading your question, I don't understand why at some point you switch from metric-space notation to norm. Be careful: norms are a very special and well-behaved case of "distance"! To begin with, they are typically defined on vector spaces and they are homogeneous (whereas general metrics need not have anything of sort). –  Apr 30 '16 at 08:05
  • @G.Sassatelli Yes, that's the problem. Thanks for the clarification. I implicitly assumed that the space had a norm and more specifically that the metric was the one induced by the norm. I realized this after reading Joshua Hunt's answer. I think that I can make the proof correct by adding "X is a normed space" and "$d$ is the metric induced by the norm" to the hypothesis. – rubik Apr 30 '16 at 08:24
  • Indeed, if $X$ is a convex subset of a normed vector space (with the subspace topology) and $d(x,y)=\lVert x-y\rVert$, your proof works. –  Apr 30 '16 at 08:48

2 Answers2

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The standard counterexample is to use the discrete metric on the space $X$: $||x - y|| = \begin{cases}0 ,& x = y \\ 1 ,& x \neq y \end{cases}$. The open unit ball around $x$ is $\{x\}$. The closed unit ball around $x$ is $X$. Your theorem fails when you claim "$x \in \overline{B(x_0,r)}$" because you have assumed that $B(x_0,r)$ actually contains points, $x$, such that $0 < ||x - x_0|| < 1$, which is false in this counterexample.

Eric Towers
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One problem is in the expression $\frac{n}{n+1}(x-x_0)$. Your metric space doesn't necessarily have a vector-space structure! So multiplying by $n/(n+1)$ doesn't have to "make sense" in your space.

Additionally, even if you do have a vector-space structure with a norm, it's not necessarily the case that $\|x-x_0 \| = d(x_0,x)$. For example, if you put the discrete metric on $\mathbb R^d$ then $d(0, y) = 1$ if $y \neq 0$, but $\|y\|$ may well be different. (So the fact that $nr/(n+1) < r$ doesn't imply that $x \in \overline{B(x_0, r)}$.)

Josh Hunt
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  • Oh I see. Then adding the hypothesis "X is a normed space" and "$d$ is the metric induced by the norm" would make the proof correct, wouldn't it? – rubik Apr 30 '16 at 08:11
  • As far as I can see, yes it would - though if your space is finite-dimensional then you're just talking about $\mathbb R^d$ with the Euclidean metric! – Josh Hunt Apr 30 '16 at 09:54