From the paper Some further results on ideal convergence in topological spaces by Pratulananda Das (Topology and Its Applications, vol.159, pp.2621-2626):
Theorem 3: . Let $X$ be a space with $\operatorname{hcld}(X) = \omega$. Then for each $F_\sigma$ -set $A$ in $X$ there exists a sequence $x = (x_n)_{n∈\mathbb{N}}$ in $X$ such that $A = \mathcal{I}(L_x)$ provided $\mathcal{I}$ is an analytic P-ideal.
Proof:. Let $A =\bigcup_{i=1}^{\infty} A_i$ where each $A_i$ is a closed subset of $X$. By Lemma 1, for each $i$, we can find a sequence $(y_{i,j})_{j∈\mathbb{N}} \subset A_i$... such that $A_i = L((y_{i,j}))$.
Before we proceed, we first observe that if $K \notin \mathcal{I}$ then $\lim_{n\to\infty} \varphi (K\setminus [1,n]) = \beta$ (say) $\neq 0$. Then $\beta > 0$ (possibly $\beta = \infty$). From the lower semicontinuity of $\varphi$, there are finite pairwise disjoint sets $C_j$ , $j \in \mathbb{N}$ with $C_j \subset K$ and $\lim_{j\to\infty} \varphi(C_j) = \beta$. Let $\mathbb{N} =\bigcup_{i=1}^{\infty}D_i$ be a decomposition of $\mathbb N$ into pairwise disjoint subsets of $\mathbb N$. Put $K_1 = (K\setminus \bigcup_j C_j) \cup ( \bigcup_{j\in D_1} C_j )$ and for $i > 1$ $K_i=\bigcup_{j∈D_i}C_j$ . Then one can check that the sets $K_i$’s are pairwise disjoint subsets of $K$, $K =\bigcup_{i=1}^{\infty}K_i$ . Further it follows that $$\lim_{n\to\infty} \varphi(K_i\setminus[1,n])= \beta \quad \forall i \in \mathbb{N}.$$ Clearly then $K_i \notin \mathcal{I}$ $\forall i$.
We now come back to the main proof. First decompose $\mathbb{N}$ into pairwise disjoint sets $\mathbb N =\bigcup_{i∈\mathbb N} M_i$ where $M_i \notin \mathcal{I}$ for any $i$. Further for each $i \in \mathbb N$, decompose $M_i=\bigcup_{j∈\mathbb N} B_{i,j}$ where $$\lim_{n\to\infty} \varphi(B_{i,j}\setminus [1,n])=\lim_{n\to\infty} \varphi(M_i\setminus [1,n])=β_i \text{ (say) } \forall j \in \mathbb{N}$$ and $B_{i,j} \notin \mathcal{I}$ $\forall j \in \mathbb N$. Define a sequence $x = (x_n)_{n\in\mathbb N}$ as follows: $$x_n = y_{i,j} \quad\text{for each } n ∈ B_{i,j}.$$ We shall show that $A = \mathcal{I}(L_x)$. The following two cases arise.
And so do my problems again
I believe at this point he is proving the equality by inclusion-reverse inclusion method. (i), below, will show $L(x)\subset A$ by contrapositivity and (ii) will show $A\subset L(x)$ and thus arriving at the desired conclusion.
(i) First suppose that there is a subsequence $(x_{n_k})_{k\in \mathbb N}$ of $x$ converging to $y$ and $y\notin A$.Then for every $i$,$\bigcup_{j=1}^i M_j$ contains only finite number of terms $n_1,n_2,\ldots ,n_s$ but $\{n_1,n_2, \ldots ,n_s\} \in \mathcal I$ (since $\mathcal I$ is admissible). So $y \notin \mathcal{I}(L_x)$.
I cannot understand how we know that $\bigcup_{j=1}^i M_j$ is finite here. $y\notin A \Rightarrow y\notin A_i$ for any $i$. If possible let infinitely many $n_k$'s are in $\bigcup_{j=1}^i M_j$.Wh ere is the contradiction? $y$ is a limit point of that sequence.
(ii) Now take $y \in A$. Then $y \in A_i$ for some $i \in \mathbb N$. Then there is a subsequence $(x_{i,j_k})_{k\in \mathbb N}$ of $(x_{i,j})_{j\in\mathbb N}$ which converges to $y$.
Where did this sequence $(x_{i,j})_{j \in \mathbb N}$ come from? And why would we assume that $y\in A_i$ implies there is a subsequence of $x_{i,j}$ converging to $y$?
Choose an $\epsilon > 0$ such that $\epsilon < \beta_i$ . Following the proof in Theorem 1 we can find a sequence of positive integers $t_1 < t_2 < t_3 < \cdots$ such that $$\varphi (B_{i,j_k} \cap (t_k,t_{k+1}])\ge \beta_i−\epsilon.$$ We now define $B=\bigcup_{k\in \mathbb N}\{B_{i,j_k} \cap (t_k,t_{k+1}]\}$. Then it readily follows that $$\lim_{n\to \infty} \varphi (B\setminus [1,n])\ge \beta_i−\epsilon.$$
I do not know how.
Therefore $B\notin \mathcal{I}$ and it is easy to observe that $\lim_{n\to \infty,n\in B} x_n = y$.
I cannot understand this one either.
This shows that $y \in \mathcal{I}(L_x)$.
I managed the earlier parts thanks to Martin Sleziak. Please explain these fourthings to me. I wish I could speak to the author personally.