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From the post" 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 to the closed ball. In a Frechet space or a Banach space, the closure of an open 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 to 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 seem that the closure of $B_1(x)$ does not equal to the closed ball $\hat{B}_1(x)$

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

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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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