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Assume $(x_t)_{t\in\Lambda}$ is a bounded net in $\mathbb{C}$ that doesn't convergent to $0$. Is there a subnet of it and an $\epsilon$ such that is outside of $B(0,\epsilon)=\{ x\in \mathbb{C}: \lVert x\rVert<\epsilon\} $?

This is true for sequences but for net I can't find a counterexample.

Darman
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    I think a place to start would be to consider the sets $\lambda_\epsilon := {t \in \lambda\mid |x_t| \ge \epsilon}$. What happens if all of these sets are bounded in $\lambda$? – Paul Sinclair Nov 23 '19 at 03:04
  • Definition of a subnet:Let $(I, \preceq_I ), (J,\preceq_J )$ be two directed sets and $X$ be the underlying set.${ \eta_j }{j \in J}$ is a subnet of ${ \xi_i }{i \in I}$, if there exists a function $\phi: J \to I$ such that (1) $\eta_j = \xi_{\phi(j)}$ for all $j \in J$. (2) For all $i \in I$ there exists $j $ such that ${ \phi(j') \in I \mid j´\succeq_J j} \subseteq { i' \in I \mid i´\succeq_I i}$ – Darman Nov 23 '19 at 18:32
  • Who says $\Lambda$ is uncountable? Nets can be finite, though those are not particularly useful. But sequences are useful nets, and they are countable. – Paul Sinclair Nov 23 '19 at 18:34
  • As for the rest, I said a place to start, not finish. At least one of the $\Lambda_\epsilon$ is unbounded in $\Lambda$. For that $\epsilon$, you should be able to find a subnet in $\Lambda_n$ – Paul Sinclair Nov 23 '19 at 18:37
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    Excuse me, but $\Lambda$ has pre-order, so yes, boundedness is defined in $\Lambda$. Or more particularly, unboundedness is defined. $A$ is unbounded if for every $\alpha \in \Lambda$, there is a $t \in A$ with $\alpha \le t$. – Paul Sinclair Nov 23 '19 at 18:48
  • Thank you very much. I found out the answer. – Darman Nov 23 '19 at 19:41
  • (Of course, the definition of unboundedness should be: "for every $\alpha\in \Lambda$, there is a $t\in A$ with $\alpha \ngeq t$" is it true?) – Darman Nov 23 '19 at 19:41

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Since $x_t$ doesn't converge to $0$, there is $\epsilon>0$ such that for all $s\in\Lambda$ there is $t\in\Lambda$ such that $t\geq s$ and $|x_{t}|>\epsilon$.

Let's use this to choose the subnet in an orderly manner.

Let $\preceq$ be a well order of $\Lambda$ and $o$ the first element in this order. Choose $t_o\geq o$ with $|x_{t_0}|>\epsilon$.

Assume that we have defined $t_j$ for all $j\preceq i$ with $j\neq i$. If there is already some $t_j\geq i$ defined for some $j\preceq i$, either skip $i$ or define $t_i$ to be one such $t_j$. By transfinite induction we have defined a subset $\Gamma$ of $\Lambda$ of all values of $t_i$ chosen. This set ordered with the original order $\leq$ of $\Lambda$ is directed. In fact, if $t_i,t_j\in\Gamma$ then there are some $i',j'\in\Lambda$ such that $t_i\leq i'$ and $t_j\leq j'$. Since $(\Lambda,\leq)$ is directed, then there is $k\in\Lambda$ with $k\geq i'$ and $k\geq j'$. But then there is $t_k\in\Gamma$ with $t_k\geq k$. This implies that $t_k\geq t_i$ and $t_k\geq t_j$ by transitivity. Therefore $(\Gamma,\leq)$ is also directed.

By construction, for all elements $r\in\Gamma$ we have $|x_r|>\epsilon$.

The restriction of $x$ to $\Gamma$ is a subnet. In fact, if $i\in\Lambda$ then $\{k\in\Gamma:\ k\geq t_i\}\subset\{j\in\Lambda:\ j\geq i\}$, since $k\geq t_i\geq i$ by construction, and this implies $k\geq i$ by transitivity.

conditionalMethod
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