eventually and frequently for nets

Question

I'm studying nets and there is something in the definition of eventually and frequently that is confusing me.

These are the definitions I have.

Eventually: a net $(x_\lambda)_{\lambda \in \Lambda}$ in a set $X$ is eventually in $Y \subset X$ if $\exists \tilde{\lambda} \in \Lambda$ such that $\forall \lambda \ge \tilde{\lambda}$, $x_\lambda \in Y$.

Frequently: a net $(x_\lambda)_{\lambda \in \Lambda}$ in a set $X$ is frequently in $Y \subset X$ if $\forall \lambda \in \Lambda$, $\exists \mu \in \Lambda, \mu \ge \lambda$ such that $x_\mu \in Y$.

Now, since nets are not necessarily defined on a totally ordered set, I imagine a net might have several independent branches. So the way I intuitively read the definition of eventually is that one of the branches eventually ends up in $Y$ (and other branches might never even go through $Y$), while frequently means that $every$ branch must touch $Y$. Therefore in nets eventually does not imply frequently. Is this correct? Strangely enough I couldn't find an answer to this.

Thanks

Answer 1

It is true that $\Lambda$ is not necessarily a totally ordered set, but it is filtered (or directed), i.e. for any $i,j\in \Lambda$ there exists a $k\in \Lambda$ such that $i\leq k$ and $j\leq k$. Now suppose $(x_{\lambda})_{\lambda}$ lies in $Y$ eventually.

Choose $\lambda\in \Lambda$. We know that there exists an $\tilde{\lambda}\in\Lambda$ such that $x_{i}\in Y$ for all $i\geq \tilde{\lambda}$. Since $\Lambda$ is filtered, there exists a $k\in \Lambda$ such that $\lambda \leq k$ and $\tilde{\lambda}\leq k$. Then $x_k\in Y$. Thus $(x_{\lambda})_{\lambda}$ lies in $Y$ frequently.

This shows that if $(x_{\lambda})_{\lambda}$ eventually lies in $Y$, then it also frequently lies in $Y$.

Answer 2

Even in a linearly ordered index set these notions differ:

e.g. consider the sequence (a special net, indexed by $\mathbb{N})$) $$-1,1,-1,1,-1, 1,-1$$ is not eventually in $(0,\infty)$ but it is frequently in this set..

Eventually does imply frequently becaue of the property that any two elements of the index set have a comon upperbound, so their tails "come together" for a while, i.e. all tail sets $T(i_0 )= \{i: i \ge i_0\}$ intersect (these form a so-called filter base).