$X$ is a topological space. and $I$ is an ideal of $\mathbb N.$ $x=\{x_n\}_{n\in \mathbb N}$ is a sequence of elements in $X.$
The following two definitions:
$L(x):x\in L(x) \text{ if for every neighbourhood }W \text{ of }z,\{n\in \mathbb N: x_n\in W\}\text{ is infinite. }$
$I(C_x)=\left\{z\in X:\text{ for every neighbourhood }U \text{ of }z,\{n\in \mathbb N:x_n\in U\}\notin I\right\}$
Now my question is what is the inclusive relationship between them? That is whether $I(C_x)\subset L(x)$ or $L(x)\subset I(C_x)?$
If I take any point from $W$, say $y$ and a nbd $U$ of $y$ , then the set $\{n\in \mathbb N: x_n\in W\}$ is infinite but that does not tell whether it is in $I$ or not for both $I$ and $F(I)$ can contain infinite sets.So $L(x)\nsubseteq I(C_x).$
Then it must be the other way round.Take $z\in I(C_x)$ and any nbd $U$ of $z.$ Then $\{n\in \mathbb N:x_n\in U\}\notin I.$ Had they said $I$ is admissible i.e. containing all the singletons then we could conclude from here that since all finite sets are in $I$, this 'necessarily outside of $I$' set must be infinite and hence $I(C_x)\subset L(x)$ but not such thing is said in this paper Some further results on ideal convergence in topological spaces by Pratulananda Das.
But in the proof of theorem $5$ something of this sort has been used and I do not understand exactly what he has done. Please help me understand this. Thank a lot. Or is there some typo in the proof and some modification required, like a couple of cases before?
Let us consider a sequence $(x_n)$ as a function $x \colon \mathbb N \to X$. This can simplify notation a bit, since we can write simply $x^{-1}(U)$ instead of the longer notation $\{n\in\mathbb N; x_n\in U\}$.
Let us denote by $\newcommand{\Fin}{\mathrm{Fin}}\Fin$ the ideal of finite sets.
So we now have \begin{align*} z\in I(C_x) &\Leftrightarrow \text{for each neighborhood $U$ of $z$ we have } x^{-1}(U) \notin I\\ z\in L(x) &\Leftrightarrow \text{for each neighborhood $U$ of $z$ we have } x^{-1}(U) \notin \Fin \end{align*}
If $I$ is admissible ideal, then $\Fin\subseteq I$ and therefore $$x^{-1}(U) \notin I \implies x^{-1}(U) \notin \Fin.$$
So we get that $I(C_x)\subseteq L(x)$.
If the ideal $I$ is not admissible, the same claim is not true. Simply choose some integer $n_0$ and take $I=\{A\subseteq\mathbb N; n_0\notin A\}$. Then the only $I$-cluster point (and at the same time $I$-limit) of a sequence $x$ is equal to $x_{n_0}$. And it is easy to find a sequence such that $L(x)\ne\{x_{n_0}\}$.