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From the post" https://math.stackexchange.com/questions/108010/when-is-the-closure-of-an-open-ball-equal-to-the-closed-ball'', we know that the closure of an open ball does not necessarily equal the closed ball. In a Fréchet space or a Banach space, the closure of an open ball is equal to the closed ball. How about the general case?

Let $(X,\tau)$ be a topological vector space that compatible with a metric $d$. i.e., the topology $\tau$ induced by the metric makes vector operations (vector addition and scalar multiplication) continuous.

Does the closure of an open ball $B_{r}(x)=\{y\in X; d(x,y)<r\}$ equal the closed ball $\hat{B}_{r}(x)=\{y\in X; d(x,y)\leq r\}$.

The discrete topology with $\mathbb{R}$ is NOT a topological vector space because scalar multiplication is not continuous. Below is my construction: Let us consider the topology on $\mathbb{R}$ generated by the metric below: $$ d(x,y)=\begin{cases} |x-y|\qquad&\text{if and only if $|x-y|<1$}\\\ 1&\text{otherwise} \end{cases} $$ I'm not quite sure if it is a topological vector space. It seems that the closure of $B_1(x)$ does not equal the closed ball $\hat{B}_1(x)$

WillG
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Yuhang
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  • Looks good! Continuity of the operations holds, since it's effectively a local test, so the absolute value part of $d$ kicks in, as needed. And it's clear that the closure of $B(0,1)$ is equal to $[-1,1]$, whereas $B[0,1]=\mathbb{R}$. – quasi Oct 08 '17 at 05:49
  • I think the claim about Fréchet spaces is incorrect, although I didn't realize this until after posting this question. Counterexample? You gave one! Your metric generates the standard topology on $\mathbb R$, which is a Fréchet space. – WillG Apr 02 '25 at 22:15

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Your counterexample is correct. A quick way to verify it is to notice that your metric induces the same topology as the standard topology on $\mathbb{R}$.

In fact, more generally, if $(M,d)$ is any metric space and $c>0$, then $d'(x,y)=\min(d(x,y),c)$ is another metric which induces the same topology on $M$. It is then easy to see that by choosing $c$ appropriately, you can find a metric on any metrizable space $M$ with more than one point for which there is an open ball whose closure is not the closed ball of the same radius. (Specifically, if $x,y\in M$ are two distinct points, let $c=d(x,y)/2$ and consider the open ball of radius $c$ about $x$.)

Eric Wofsey
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