How does $K\notin I$ imply $\lim_{n\rightarrow \infty} \phi(k\backslash [1,n])\neq 0$

Question

Theorem $3$ of 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.

$X$ is a topological space such that Every closed subset of $X$ is separable. $I$ is an analytic $P$-ideal of $X$. $\phi$ is a lower semi-continuous submeasure. Then it says that

If $K\notin I$ then $\lim_{n\rightarrow \infty} \phi(K\backslash [1,n])=\beta\neq 0$. And from the lower semicontinuity of $\phi$ there are pairwise disjoint sets $C_j,j\in \mathbb N$ with $C_j\subset K$ and $\lim_{n\rightarrow \infty }\phi(C_j)=\beta$

I cannot understand how these follow. Please explain to me anyone.

Answer 1

A word about notation: If $n\in\omega$ then $n=\{0,1,\dots,n-1\}=[0,n-1]$. This is a well-known fact and it is commonly used in some areas of set theory, since it often simplifies the notation. If you study some papers about analytic P-ideals, you are very likely to encounter it very often. I will use this notation below.

Let us recall what we know. We know that $\phi$ is a submeasure. This means that:

• $\phi(\emptyset)=0$,
• $A\subseteq B$ $\Rightarrow$ $\phi(A)\leq\phi(B)$,
• $\phi(A\cup B)\leq \phi(A)+\phi(B)$,
• $\phi(\{n\})<+\infty$ for $n\in\omega$.

Moreover, $\phi$ is lsc, i.e., $$\phi(A)=\lim\limits_{n\to\infty} \phi(A\cap n)$$

We also know that $$\mathcal I=\operatorname{Exh}(\phi)=\{A\subseteq\omega; \|A\|_\phi = 0\}$$ where $$\|A\|_\phi=\limsup_{n\to\infty} \phi(A\setminus n)=\lim\limits_{n\to\infty} \phi(A\setminus n).$$

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For any $K\subseteq\omega$ the limit $\lim_{n\rightarrow \infty} \phi(K\setminus n)$ exists, simply because $\phi(K\setminus n)$ is a decreasing sequence.

If $K\notin\mathcal I$, then this limit is non-zero. So we denote it $\lim_{n\rightarrow \infty} \phi(K\setminus n)=\beta\ne0$.

Now let us try to show (inductively) that there are sets $C_j$ with the required properties.

• Let $\varepsilon_j=\frac1{2^j}$. (Or any sequence which decreases to zero.)
• We will inductively construct $n_j$ and $C_j$ such that $n_j<n_{j+1}$, $C_j\subseteq [n_j,n_{j+1})$ and $\beta<\varphi(C_j)<\beta+\varepsilon_j$.
• We put $n_{-1}=0$. (Just to be able to write base of induction and inductive step at the same time.)
• In the $j$-th inductive step, we will choose $C_j$ and $n_j$.
  • Since $\varphi(K\setminus n)\searrow\beta$, there is $n_j>n_{j-1}$ such that $\beta<\varphi(K\setminus n_j)<\beta+\varepsilon_j$.
  • Since $\varphi$ is lsc, we have that $$\varphi(K\setminus n_j) = \lim\limits_{k\to\infty} \varphi((K\setminus n_{j})\cap k) = \lim\limits_{k\to\infty} \varphi(K\cap [n_j,k)).$$
  • Since the above limit is between $\beta$ and $\beta+\varepsilon_j$, we get that there exists $k$ such that $$\beta < \varphi(K\cap [n_j,k)) < \beta+\varepsilon_j.$$
  • Now we can simply put $C_j:=K\cap [n_j,k)$ and $n_{j+1}:=k.$

Notice that the conditions $n_j<n_{j+1}$ and $C_j\subseteq [n_j,n_{j+1})$ imply that $C_j$'s are pairwise disjoint. The condition $\beta<\varphi(C_j)<\beta+\varepsilon_j$ together with $\varepsilon_j\searrow0$ means that $\varphi(C_j)\to\beta$.