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Suppose $(A,\tau)$ is a topological vector space and $B$ is a subset of $A$. Suppose $B$ is metrizable and complete (in this metric). Must $B$ be closed in $\tau$?

Here is my attempt:

First, I will add the assumption that the metric $d$ on $B$ is translation-invariant. Let $\tau_B$ be the induced subspace topology on $B$, so that the metric-topology generated by $d$ on $B$ is equal to $\tau_B$.

Since $d$ is translation invariant, these notes I found here imply that, since $(B,d)$ is complete, so is $(B,\tau_B)$ (in the sense that every Cauchy net in $(B,\tau_B)$ converges to a point in $B$).

Suppose $x$ belongs to the closure of $B$. Then there exists a net $x_\alpha$ of elements from $B$ converging to $x$. I was able to show that any convergent net in a topological vector space is a Cauchy net. Thus $x\in B$ by the completeness of $(B,\tau_B)$, which shows that $B$ is closed.

Remarks: First, I'm not sure if this reasoning is correct, which I am most concerned about. Secondly, I'm not sure what assumptions are necessary, particularly whether the metric must be translation-invariant, and whether this holds more generally in any topological space (not necessarily a topological "vector" space).

user122916
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  • A complete subset of a metric space is closed: see here: https://math.stackexchange.com/questions/1233805/a-complete-subset-of-a-metric-space-is-closed – Math1000 Nov 27 '19 at 20:57
  • The additional structure of a topological vector space is not needed for complete subsets to be closed in a metric space. – Math1000 Nov 27 '19 at 20:59
  • This issue is that $(A,\tau)$ isn't necessarily a metric space, so I do not think I can invoke that result. – user122916 Nov 27 '19 at 21:07
  • If $B$ is complete as a uniform subspace of $A$ and $A$ is Hausdorff, then $B$ is closed (no need for $A$ to be a topological vector space in this). If $A$ isn't Hausdorff, then $B$ need not be closed. If $A$ is a TVS and $B$ is just a subset, not a linear subspace, then it doesn't make sense to say the metric – defined on $B$ – is translation-invariant. Are you interested in linear subspaces or in general subsets? – Daniel Fischer Nov 27 '19 at 22:12
  • @DanielFischer I am interested in the norm-closed unit ball of a Banach algebra in a coarser topology. For example, let $H$ be a separable Hilbert space, let $A$ be the operator algebra of $H$, and let $\tau$ be the strong-operator topology on $A$. Then $(A,\tau)$ is not metrizable, but one can show that the unit ball $B={T\in A: |T|\leq 1}$ is metrizable and complete (as a metric space). $B$ is not a linear subspace. One can also show that $B$ is closed in $\tau$. I was hoping to conclude that $B$ is closed by a more general theory than looking at the specific topology at hand. – user122916 Nov 28 '19 at 00:41
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    I think in that situation it's not just the case that the subspace topology on $B$ is completely metrisable, but the uniform structure that $B$ inherits from $(A,\tau)$ is. In that case, the general "a complete subset of a Hausdorff uniform space is closed" applies. If the metric on $B$ doesn't induce the subspace uniform structure things are hairy. For example every open subset of $\mathbb{R}^n$ is completely metrisable, which shows that we can't have a too general theorem. – Daniel Fischer Nov 28 '19 at 12:54

2 Answers2

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The simple answer is: no. Consider $A=\mathbb{R}$, the topology $\tau=\{\emptyset,A\}$ is anti-discrete and let $B=\{1\}$.

The problem with your reasoning is that $(x_\alpha)$ may be a Cauchy net in $B$ but in $A$ that doesn't even make sense (in order to talk about Cauchy nets we need metric or uniform structure). And even if $A$ was metric the concepts of Cauchy nets in $A$ and $B$ may differ, since metric in $A$ and $B$ are very loosely related.

freakish
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  • Thank you for this example. This topology is not Hausdorff, a condition which I meant to add. Does my reasoning work if I add in the assumptions that $\tau$ is Hausdorff and $(A,\tau)$ is a uniform space? – user122916 Nov 27 '19 at 21:35
  • @user122916 no, see Gae.S. answer. – freakish Nov 27 '19 at 21:39
  • @user122916 $(A,\tau)$ is already uniformizable in a natural way because it is a topological vector space. –  Nov 27 '19 at 21:45
  • This link makes me wonder if this example works, because the anti-discrete topology cannot represent a tvs (except when the space contains one element)? – user122916 Nov 28 '19 at 02:34
  • @user122916 no, the link talks about discrete topology, not anti-discrete. The anti-discrete topology is linear, unlike discrete. – freakish Nov 28 '19 at 05:38
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No: consider $B=\{x\in\Bbb R\,:\, \frac1x\in\Bbb N\}$ in $\Bbb R$. It has the discrete topology (and it is thus completely metrizable with the $0$-$1$ distance), but it is not closed.

  • Thanks to whomever counterdownvoted. I don't understand the downvote in the first place. –  Nov 27 '19 at 21:36
  • Is $\mathbb{R}$ with the discrete topology a topological vector space? I'm thinking about this link here. – user122916 Nov 28 '19 at 02:23
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    @user122916 Gae is not saying that $\mathbb{R}$ has discrete topology. It is considered with the standard topology. And with that the inherited topology on $B$ is discrete. – freakish Nov 28 '19 at 05:40
  • But the inherited topology from $\mathbb{R}$ wouldn't be the discrete topology. – user122916 Nov 28 '19 at 15:27
  • @user122916 It is. –  Nov 28 '19 at 15:41
  • Namely, $${x}=\left{y\in B,:, \left\lvert y-x\right\rvert <\frac{x^2}{x+1}\right}$$ –  Nov 28 '19 at 15:46