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If $(x_t)_{t\in \Lambda}$ is a net in complex numbers that converges to $0$; Is it true that $(x_t)_{t\in\Lambda}$ has a countable subnet?

Definition of a subnet is the first definition in the:Different definitions of subnet

I say:

$x_t$ is convergent to $0$, So $$\forall n\in\mathbb{N} \ \ \exists t_n \ \ s.t\ \ \ t\ge t_n \ \Rightarrow \ \ \lVert x_{t_n} \rVert \le \frac{1}{n} .$$ But we can't say that the suquence $(x_{t_n})_{n\in \mathbb{N}}$ is a subnet.

On the other hand in this page:Does every net have a countable subnet? it is said that there is a countable subnet. But I can't prove it.

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  • $\begingroup$ Do you just want any countable subnet, or a countable subnet that also converges to $0$? $\endgroup$ – Jack M Nov 24 '19 at 9:30
  • $\begingroup$ Why cannot you say it is a countable subnet? What's missing in your opinion? $\endgroup$ – Henno Brandsma Nov 24 '19 at 9:32
  • $\begingroup$ If there is a countable subnet then this subnet is converged to $0$ automatically. $\endgroup$ – Darman Nov 24 '19 at 9:50
  • $\begingroup$ We can't say that $x_{t_n}$ is a subnet, because we should say for every $t\in \Lambda $ there is an $N$ such that $t_n\ge t$ for all $n\ge N$ $\endgroup$ – Darman Nov 24 '19 at 9:52
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Consider the directed set $\Lambda=(0,1]\times\omega_1$ where $\omega_1$ is the first uncountable ordinal and the order in $\Lambda$ is given as $(a_1,b_1)>(a_2,b_2)$ if $a_1<a_2$ and $b_1>b_2$.

This is a directed set since for $(a_1,b_1),(a_2,b_2)\in\Lambda$ we have that $(\min(a_1,a_2),\max(b_1,b_2))$ is larger than both.

Let $\theta:\omega_1\to[0,2\pi)$ be a bijection.

Define $x:\Lambda\to\mathbb{C}$ by $x(a,b)=ae^{i\theta(b)}$.

This net converges to $0$. In fact, for all neighborhood of $0$, there is a ball $\{|z|<r\}$ contained in it. Then, taking $s\in(0,1]$ with $s<r$ and $b\in\omega_1$, we have that $\{x(a,b)\in\Lambda:\ (a,b)>(s,b_0)\}\subset\{|z|<r\}$

Now, there is no countable (Kelley) subnet of $x$. In fact, if $y$ were a subnet on a countable directed set $(J,\preceq)$ with $\phi:J\to\Lambda$ the function showing that it is a subnet, then let $B\subset\omega_1$ be the set of all the second components of $\phi(J)\subset\Lambda$. Then $B$ is countable and there must exist $\alpha\in\omega_1$ such that $\alpha$ strictly larger than all elements of $B$, for example $\bigcup_{b\in B}b + 1$. Then $\{i\in\Lambda:\ i>(1,\alpha)\}$ doesn't contain any of the (images by $\phi$ of the) indexes of $y$.

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  • $\begingroup$ I was thinking along those lines (using $\omega_1$) but stopped short of this. Nice. $\endgroup$ – Henno Brandsma Nov 24 '19 at 14:32
  • $\begingroup$ Maybe something similar can be done for this question of the OP's? $\endgroup$ – Henno Brandsma Nov 24 '19 at 14:35
  • $\begingroup$ @HennoBrandsma I went from first thinking that the given net would have to be countable (as it happens with sequences) if it was convergent, to thinking about the net of circles with center $0$ in which points on the same circle are comparable but are smaller than any of the points in any circle with smaller radius. This example has tails that are too fat. This makes it allow for countable subnets. So, I tried to allow fewer comparisons to the points in smaller circles by putting the angles in well-order. $\endgroup$ – conditionalMethod Nov 24 '19 at 15:40
  • $\begingroup$ @HennoBrandsma I think that in that other problem the answer is positive. $\endgroup$ – conditionalMethod Nov 24 '19 at 16:05

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