Is $\{ n\in \mathbb N:x_n\in U\}\notin I\iff\{n\in \mathbb N:x_n\notin U\}\in I $ true?

Question

$I$ is an ideal of $\mathbb N$ i.e. a collection of subsets of $\mathbb N$ satisfying

1. $\varnothing\in I$
2. $A\in I, B\subset A \implies B\in I$
3. $A,B\in I \implies A\cup B \in I$

$\{x_n\}_n$ is a sequence of elements of a topological space $X$ and $y$ is fixed point in $X$. Then $y$ is called $I$-cluster point of $\{x_n\}$ if for every open nbd of $y$ $\{n\in \mathbb N: x_n\in U\}\notin I.$ This is the given definition in this paper pg no 2622,Definition $1$.

Now my question is , can I say $\{ n\in \mathbb N:x_n\in U\}\notin I\iff\{n\in \mathbb N:x_n\notin U\}\in I \ ?$ That'll rephrase the definition as $y$ is called $I$-cluster point of $\{x_n\}$ if for every open nbd of $y$ $\{n\in \mathbb N: x_n\notin U\}\in I.$

Thank you.

Answer 1

No, you can't. Unless you have reason to believe that $A\notin I\iff (\Bbb N\setminus A)\in I$