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I have an elementary question on nets because I'm not familiar with this concept. Here are two basic facts:

  • Every subsequence of a sequence is a subnet;
  • Not every subnet of a sequence is a subsequence.

For the second fact, I have seen the following example:

Given a sequence $(x_n)=(x_1,x_2,x_3,x_4,...)$, the net $$(x_\alpha)=(x_1,x_2,x_2,x_3,x_3,...,x_{1+[\frac{n}{2}]},...)$$ is a subnet of $(x_n)$ that is not a subsequence of $(x_n)$.

In this example, $(x_\alpha)$ has a subnet which is a subsequence of $(x_n)$, namely, the sequence $(x_n)$. Could someone give me an example where this doesn't happen?

Explicitly: I'd like an example of sequence $(x_n)$ and a subnet $(x_\alpha)$ of $(x_n)$ such that no subnet of $(x_\alpha)$ is a subsequence of $(x_n)$.

Motivation for the question: I have a bounded sequence in the dual of a normed space. If the space was separable, then I could pass to a weak-* convergent subsequence. However the space is not separable. So, all I have is a subnet weak-* convergent. Presumably, I can't pass to a subsequence. As I said, I'm not familiar with the concept of net and thus I'd like to see an example where the existence of the subsequence fails.

Martin Sleziak
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Pedro
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  • I think you cannot do that. If we have a subnet of a sequence, then there exists an increasing cofinal $\phi : \Omega\to\mathbb{N}$ where $\Omega$ is a directed set. Because $\phi$ is cofinal, you can find, for every $n\in\mathbb{N}$, $\lambda_n\in\Omega$ such that $\phi(\lambda_n)<\phi(\lambda_{n+1})$. Then you can construct a subsequence from here, I think. – JonSK Jan 10 '16 at 17:08
  • I think http://math.stackexchange.com/a/1209702/4280 (the ultrafilter subnet of a sequence ) is an example as you seek. – Henno Brandsma Jan 10 '16 at 17:27
  • There are a few, subtly different, definitions of subnet in use. As @JonSK points out, according to at least one there is no such net. What's yours? – BrianO Jan 10 '16 at 17:33
  • @JonSK So, every sequence in a compact space has a convergent subsequence? – Pedro Jan 10 '16 at 17:41
  • @BrianO I'm interested in the definition according to which the following fact is true: A space $X$ is compact if and only if every net in $X$ has a subnet with a limit in $X$. – Pedro Jan 10 '16 at 17:41
  • @Pedro this (compact iff convergent subnet) holds in all definitions of subnet. The subsequence of the subnet need not converge; otherwise every sequence would have a convergent subsequence in compact spaces, which is false. – Henno Brandsma Jan 10 '16 at 18:05
  • @HennoBrandsma But every subsequence is a subnet. And a net converges if and only if every subnet converges. So, shouldn't the subsequence of the subnet necessarily converge? (The facts on nets and subnets that I'm mentioning are taken from wikipedia.) – Pedro Jan 10 '16 at 18:17
  • @Pedro a subsequence of a net (!) need not be a subnet of the net (there need not even be a cofinal sequence). A subsequence of a sequence (!) is a subnet. So take a sequence $(x_n)$ in a compact space without a convergent subsequence. Then this sequence has a convergent subnet. But this subnet has no convergent subsequence. – Henno Brandsma Jan 10 '16 at 18:20
  • @HennoBrandsma This also happens in the example of your link: we have a sequence ${\delta_n}$ with no convergent subsequence and with a convergent subnet $\nu$. So, $\nu$ has no subnet which is a subsequence of ${\delta_n}$ (otherwise ${\delta_n}$ would have a convergent subsequence), right? – Pedro Jan 10 '16 at 18:32
  • @Pedro indeed. This is a more or less concrete example of this phenomenon – Henno Brandsma Jan 10 '16 at 18:33
  • @HennoBrandsma I suggest you post your comments as an answer because they clarified the point. – Pedro Jan 10 '16 at 18:35

1 Answers1

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This question and answer give a concrete example of a sequence $(\delta_n)$ in a compact space that has no convergent subsequence (so this compact space is not sequentially compact). The answer gives a "concrete" (if you believe ultrafilters are concrete) subnet $(x_d), d \in D$ that converges to some $f_\mathcal{U}$. This subnet cannot have a subsequence (that is also a subnet!) that converges, because this would be a subsequence of the original sequence that would converge (and this cannot be). It is possible to find $d_n$ that are increasing in the index set $D$ of the subnet, but this is not a subnet of the subnet (as they will not be cofinal).

A subsequence of a sequence is a subnet as well, but a subnet need not have a cofinal subsequence.

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Henno Brandsma
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