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This is a question about convergence of nets which I don't quite understand yet. In metric spaces convergence of sequences encodes the topology but suppose we want to study convergence of nets even though. When can we pass to countable subnets? In other words,

Given a net $(x_\lambda)_{\lambda\in \Lambda}$ in a separable metric space $X$ that converges to some $x\in X$. Can we find a countable subnet $\Lambda^\prime \subset \Lambda$ such that $(x_\lambda)_{\lambda\in \Lambda^\prime}$ converges to $x$?

Hannes T.
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1 Answers1

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You don’t even need separability.

Let $\langle X,d\rangle$ be a metric space, and suppose that $\langle x_\lambda:\lambda\in\Lambda\rangle$ is a net in $X$ converging to $x\in X$. For each $n\in\Bbb N$ there is a $\lambda_n\in\Lambda$ such that $d(x,x_\lambda)<2^{-n}$ whenever $\lambda\in\Lambda$ and $\lambda_n\preceq\lambda$. Since $\langle\Lambda,\preceq\rangle$ is directed, we may further assume that $\lambda_m\preceq\lambda_n$ whenever $m\le n$. Now let $\Lambda'=\{\lambda_n:n\in\Bbb N\rangle$; then $\langle x_\lambda:\lambda\in\Lambda'\rangle=\langle x_{\lambda_n}:n\in\Bbb N\rangle$ converges to $x$.

(Note, however, that $\langle x_\lambda:\lambda\in\Lambda'\rangle$ is not necessarily a subnet of $\langle x_\lambda:\lambda\in\Lambda\rangle$ by either of the definitions of subnet given in the Wikipedia article or by the better definition used in J.F. Aarnes & P.R. Andenaes, ‘On Nets and Filters’, Mathematica Scandinavica $31$ ($1972$), $285$-$292$.)

Brian M. Scott
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  • Say $X = \mathbb{R}$ with the usual metric, $\Lambda = 2\omega$, $x_n = 2^{-n}$ and $x_{\omega + n} = 0$. Your proof might pick out $\lambda_n = n$. Doesn't this fail the Aarnes-Andenaes definition since you can't pick $N$ such that $x_{\lambda_n} = 0$ for $n \geq N$? (I'm new to these ideas so I may be making some stupid mistake.) – mollyerin Jan 20 '15 at 04:27
  • @mollyerin: You’re right: $\langle x_\lambda:\lambda\in\Lambda'\rangle$ is eventually in every open set that the original net is eventually in, but not necessarily in every arbitrary set that the original net is eventually in. (I had it right originally and then had a mental hiccup and changed it. sigh) Fixed now; thanks! – Brian M. Scott Jan 20 '15 at 04:39
  • Looks like this is the unique problem, though. Thus if your set of $\lambda_n$ isn't cofinal, there's a $\lambda$ larger than all of them, and it follows that $x_\lambda = x$, and then you can just replace your sequence with the constant sequence $\lambda$. This should satisfy Aarnes-Andenaes; the stricter definition on Wikipedia you linked seems hopeless, since some directed sets don't have countable cofinal subsets. – mollyerin Jan 20 '15 at 05:08
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    @mollyerin: Yes, if you’re working in a first countable space you should always be able to get a sequence that’s a subnet in the A-A sense. And you’re certainly right about the notion(s) described in Wikipedia; unfortunately, they’re actually more common, at least in my experience. – Brian M. Scott Jan 20 '15 at 05:13