1
$\begingroup$

Definition

If $x_\lambda$ is a net from a directed set $\Lambda$ into $X$ and if $Y$ is a subset of $X$ then we say that $x_\lambda$ is redisually in $Y$ if there exsit $\lambda_0\in\Lambda$ such that $X_\lambda\in Y$ for any $\lambda\ge\lambda_0$

Definition

If $x_\lambda$ is a net from a directed set $\Lambda$ into $X$ and if $Y$ is a subset of $X$ then we say that $x_\lambda$ is frequently in $Y$ if for any $\lambda\in\Lambda$ there exist $\lambda_0\ge\lambda$ such that $x_{\lambda_0}\in y$

What shown belove is a reference from "General Topology" by Stephen Willard

enter image description here

So I want to discuss the claim for which if an ultranet is frequently in $E$ then it is residually in $X-E$.

Cleraly if $x_\lambda$ is a net residualliy in $Y$ then for any $\overline{\lambda}\in\Lambda$ there exist $\lambda_0$ such that if $\lambda\ge\lambda_0$ then $x_\lambda\in Y$ and so if we pick $\overline{\lambda}_0\in\Lambda$ such that $\overline{\lambda},\lambda_0\le\overline{\lambda}_0$ (we can do this since $\Lambda$ is a directed set) then it follows that $x_{\overline{\lambda}_0}\in Y$ and $\overline{\lambda}_0\ge\overline{\lambda}$ so that $x_\lambda$ is frequentely in $Y$.

So clearly any ultranet is a net and so for what we have proved above if an ultranet is residually in $E$ then it is frequentely too.

However I can't prove the inverse implication so I ask to do it. Could someone help me, pease?

$\endgroup$
2
$\begingroup$

Suppose that $\nu=\langle x_d:d\in D\rangle$ is an ultranet that is frequently in some set $E$. Then for each $d_0\in D$ there is a $d\in D$ such that $d_0\le d$ and $x_d\in E$, so $\nu$ cannot be residually in $X\setminus E$. But $\nu$ is an ultranet, so by definition it is either residually in $E$ or in $X\setminus E$, and since it is not residually in $X\setminus E$, it must be residually in $E$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.