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Consider the sequence $f: \mathbb{N} \rightarrow \mathbb{N}$ defined by $n \mapsto n$. Then this sequence should have a universal subnet. I’ve seen an exercise that says that a sequence is a universal net if and only it is eventually constant. So, it cannot be the case that a universal subset of the sequence $f$ is a subsequence.

This might be an incredibly difficult thing to do here but is it possible to construct an explicit universal subset of $f$? I’ve seen a proof that every net yields a universal subnet but that proof relies on Hausdorff’s maximum principle to demonstrate existence of such a subnet.

Feel free to include other tags that may be relevant. I’m very new to the definition of a net and that of a subnet and so I thought of this while trying to wrap my mind around the relevant definitions.

user328442
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No, there is no way to explicitly construct a universal subnet of $f$. One way of making this precise is that it is impossible to prove such a subnet exists in ZF (i.e., without the axiom of choice). Indeed, if $g:I\to \mathbb{N}$ is a universal subnet of $f$, then $\{A\subseteq\mathbb{N}:g(i)\in A\text{ for all sufficiently large }i\}$ is an ultrafilter on $\mathbb{N}$, and is nonprincipal since $g$ is a subnet of $f$. But you cannot prove there is a nonprincipal ultrafilter on $\mathbb{N}$ in ZF; see The non-existence of non-principal ultrafilters in ZF.

Eric Wofsey
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