Prove that $L(C_x)=L(C_y)$ when $I(C_x)=I(C_y)$

Question

this Some further results on ideal convergence in topological spaces by Pratulananda Das

Definition: $y ∈ X$ is called an $I$-limit point of $x$ if there exists a set $M = \{m_1 < m_2 < m_3 < ···\} ⊂ \mathbb N$ such that $M \not∈ I$ and $\lim_{k→∞} x_{m_k} = y$

Theorem $4.$ The case for the $I$-limit points. The thing I cannot prove it. Here is my attempt:

$z\in L(C_x)$. Let $U$ be a nbd of $z\ $. Then $x_{m_k}\in U \forall k\ge n_0$ Now I want to show that $z\in L(C_y)$ as well. For this we have to prove that for some $m_0\in \mathbb N$ $y_{m_k}\in U\forall k\ge m_0$. If possible let this does not hold i.e. for every $n,\exists n_1`\ge n$ s.t $y_{n_k}\notin U$

I believe that after this step somehow I should use the fact that $L(C_x)\subset I(C_x)$ but cannot decide how. Please just give me some hints as to how to begin the next step, hopefully I can manage the rest. Thank you.

Answer 1

The way this theorem is formulated in the linked paper is definitely incorrect. Here is the exact quote, to stress the incorrect part I will put it as a centered formula.

Theorem 4. (incorrect version) Let $x=(x_n)_{n\in\mathbb N}$ and $y=(y_n)_{n\in\mathbb N}$ are two sequences in $X$ such that $$\{n\in\mathbb N; x_n\ne y_n\}\notin\mathcal I$$ then $\mathcal I(C_x)=\mathcal I(C_y)$ and $\mathcal I(L_x)=\mathcal I(L_y)$.

(Here $\mathcal I(C_x)$ denotes the set of $\mathcal I$-cluster points of the sequence $x$ and $\mathcal I(L_x)$ is the set if $\mathcal I$-limit points.)

To find a counterexample, simply take $x$ and $y$ as two different constant sequences.

Looking at the proof, the author probably intended to write this:

Theorem 4. (corrected version) Let $x=(x_n)_{n\in\mathbb N}$ and $y=(y_n)_{n\in\mathbb N}$ are two sequences in $X$ such that $$\{n\in\mathbb N; x_n\ne y_n\}\in\mathcal I$$ then $\mathcal I(C_x)=\mathcal I(C_y)$ and $\mathcal I(L_x)=\mathcal I(L_y)$.

The intuition is that if we change values of some sequence on a "small set" (set from ideal), then this does not influence $\mathcal I$-limits and other similar notions which basically say how sequence behaves "on large sets" or "modulo this ideal".

After this correction the proof of $\mathcal I(C_x)=\mathcal I(C_y)$ from the paper works.

About limit points: Let $\mathcal F$ be a dual filter of the ideal $\mathcal I$. Let $a$ be a limit point of $x$, i.e. there exists a set $M\in \mathcal F$ such that the subsequence $x|_M$ converges to $a$ in the usual sense. Since $$A=\{n\in\mathbb N; x_n\ne y_n\}\in\mathcal I$$ we get that $M'=M\setminus A\in\mathcal F$. For this set we have that $y|_{M'}=x|_{M'}$. This implies that $y_{M'}$ is a subsequence of $x|_M$ and therefor it also converges to $a$. Thus $a$ is $\mathcal I$-limit point of the sequence $y$.