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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.

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  • $\begingroup$ 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. $\endgroup$ – JonSK Jan 10 '16 at 17:08
  • $\begingroup$ I think math.stackexchange.com/a/1209702/4280 (the ultrafilter subnet of a sequence ) is an example as you seek. $\endgroup$ – Henno Brandsma Jan 10 '16 at 17:27
  • $\begingroup$ 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? $\endgroup$ – BrianO Jan 10 '16 at 17:33
  • $\begingroup$ @JonSK So, every sequence in a compact space has a convergent subsequence? $\endgroup$ – Pedro Jan 10 '16 at 17:41
  • $\begingroup$ @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 $X$. $\endgroup$ – Pedro Jan 10 '16 at 17:41
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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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