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.
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 $y$.