How can $\lim \phi(C_j)$ be equal to $\lim \sum_{j\in D_i} \phi(C_j) \ ?$

Question

Proof of theorem $3$ in the paper Pratulananda Das, Some further results on ideal convergence in topological spaces, Topology and its Applications 159(2012):2621–2626. DOI: 10.1016/j.topol.2012.04.007.

Theorem $3$. Let $X$ be a space with $hcld(X) = \omega$. Then for each $F_σ$ -set $A$ in $X$ there exists a sequence $x = (x_n)_{n∈\mathbb N}$ in $X$ such that $A = I(L_x)$ provided $I$ is an analytic $P$ -ideal.

proof: $\ \ $Let $A =\bigcup^{∞}_{i=1} A_i$ where each $A_i$ is a closed subset of $X$. By Lemma $1$, for each $i$, we can find a sequence $(y_{i,j})_{j∈\mathbb N} ⊂ A_i$ such that $A_i = L((y_{i,j}))$.

Before we proceed, we first observe that $$K \notin I \implies \lim_{n→∞} ϕ(K\backslash [1,n]) = β \text{ (say) }\neq 0.$$ Then $β > 0$ (possibly $β = ∞$).

From the lower semicontinuity of $ϕ$, there are finite pairwise disjoint sets $C_j$, $j ∈ \mathbb N$ with $C_j ⊂ K$ and $\lim_{j→∞} ϕ(C_j) = β$. Let $\mathbb N =\bigcup^{∞}_{i=1} D_i$ be a decomposition of $\mathbb N$ into pairwise disjoint subsets of $\mathbb N$.

Put $$K_1 = (K\backslash \cup_j C_j) \bigcup (\cup_{j∈D_1} C_j)\\ K_i=\bigcup_{j∈D_i}C_j;\ \ i\ge 1.$$ Then one can check that the sets $K_i$ ’s are pairwise disjoint subsets of $K, K=\bigcup^{∞}_{i=1}K_i$.

(Upto this point, things are clear to me, thanks to Martin Sleziak) Then another problem occurs: Further it follows that $\lim_{n→∞} ϕ(K_i\backslash [1,n])=β ∀i ∈ \mathbb N$.

I do not get how for,$$\phi(K_i\backslash [1,n])\\=\phi((\bigcup_{j\in D_i}C_j)\backslash [1,n])\\=\phi (\bigcup_{j\in D_i}(C_j\backslash [1,n]))\\ \le \sum_{j\in D_j}\phi(C_j\backslash [1,n])$$ How can I say this tends to $\beta$?

Answer 1

First, notice that since $K_i\subseteq K$, we get from the monotonicity $\varphi(K_i\setminus[1,n])\le\varphi(K\setminus[1,n])$ and $$\lim_{n\to\infty} \varphi(K_i\setminus[1,n]) \le \lim_{n\to\infty} \varphi(K\setminus[1,n]) = \beta.$$

Now let us fix some $n$. The set $K_i\setminus[1,n]$ contains all but finitely many sets from $\{C_j; j\in D_i\}$. (Since only finitely many of them can intersect $[1,n]$.) So there is some integer $j_n$ such that $C_j\subseteq K_i\setminus[1,n]$ and $$\varphi(C_j) \le \varphi(K_i\setminus[1,n])$$ for each $j\in D$, $j\ge j_n$.

This also implies $$\beta = \lim_{\substack{j\to\infty\\j\in D_i}} \varphi(C_j) \le \varphi(K_i\setminus[1,n])$$ for each $n$.

So together we get $$\beta\le\lim_{n\to\infty} \varphi(K_i\setminus[1,n]) \le \lim_{n\to\infty} \varphi(K\setminus[1,n]) = \beta.$$