liminf and limsup with characteristic (indicator) function

So first let me state my homework problem:

Let $X$ be a set, let $\{A_k\}$ be a sequence of subsets of $X$, let $B = \bigcup_{n=1}^{+\infty} \bigcap_{k=n}^{+\infty} A_k$, and let $C = \bigcap_{n=1}^{+\infty} \bigcup_{k=n}^{+\infty} A_k$. Show that (a) $\liminf_k\; {\xi_A}_{_k} = \xi_B$, and $(b)$ $\limsup_k \;{\chi_A}_{_k} = \chi_C.$

I know that, in the context I am familiar with, that $$\liminf_{k\rightarrow +\infty}\; X_k = \bigcup_{k=1}^{+\infty} \bigcap_{n=k}^{+\infty} X_n$$ and $$\limsup_{k\rightarrow +\infty}\;X_k = \bigcap_{k=1}^{+\infty} \bigcup_{n=k}^{+\infty} X_n.$$

I also know that the characteristics (indicator) function is defined as $\chi_A(x) = \begin{cases} 1, & x \in A \\ 0, & x \notin A .\end{cases} $

So I wrote out $B$ in some of its `glory': $B= (A_1 \cap A_2 \cap A_3 \cap \cdots) \cup (A_2 \cap A_3 \cap \cdots) \cup (A_3 \cap A_4 \cap \cdots) \cup \cdots$, and as the first argument is the smallest, with increasing size to the right, the last term in the expression for $B$ would be $B$, which would be the largest.

So if I replace the $X_k$'s above with $\chi$'s I still don't see how I can get the correct answer - though it looks pretty clear from the definition of $B$ and that of $\liminf$ being basically the same, except in this case for the $\chi$.

Any direction would be greatly appreciated. By the way, I have checked out limsup and liminf of a sequence of subsets of a set but I was somewhat confused by the topology, the meets/joins, etc.

Thanks much, Nate

First let’s clear up a misconception: there is no last term in the expression for $B$. For any term $A_n \cap A_{n+1} \cap A_{n+2} \cap \dots$ there is a next term $A_{n+1} \cap A_{n+2} \cap A_{n+3} \cap\dots$ that is potentially a larger set.

Now to the problem itself: you must start with the definitions of $\limsup$ and $\liminf$ for sequences of functions. If $\langle f_n \rangle_n$ is a sequence of real-valued functions, $$\limsup_{n\to\infty} f_n(x) = \lim_{n\to\infty}\;\; \sup_{m\ge n}f_m(x),$$ and $$\liminf_{n\to\infty}f_n(x) = \lim_{n\to\infty}\;\;\inf_{m\ge n}f_m(x).$$ In your case $f_n = \chi_{A_n}$, so you’re trying to show that $$\lim_{n\to\infty}\;\;\inf_{m\ge n}\chi_{A_m}(x) = \chi_B(x)$$ and $$\lim_{n\to\infty}\;\; \sup_{m\ge n}\chi_{A_m}(x) = \chi_C(x)\;.$$

Since the only values of a characteristic function are $0$ and $1$, it should be clear that $\inf_{m\ge n}\chi_{A_m}(x)$ is always $0$ or $1$, and it’s not hard to see that it’s $1$ iff $\chi_{A_m}(x)=1$ for every $m\ge n$, i.e., iff $x\in A_m$ for every $m\ge n$. Similarly, $\sup_{m\ge n}\chi_{A_m}(x) = 0$ iff $\chi_{A_m}(x)=0$ for every $m\ge n$, which is the case iff $x \notin A_m$ for every $m\ge n$. From here it’s fairly straightforward to get the desired results.

I assume that $g := \limsup_k\;f_k$ here means the pointwise limit, i.e. $$g(x) = \limsup_k\; f(x) \quad \forall x.$$ (And similarly for $\liminf$.) Since indicator functions only take the values $0$ and $1$, this becomes particularly simple for them: $$(\limsup_k\;\chi_k)(x) = \begin{cases}1 & \text{if }\;\forall i\; \exists j>i: \chi_j(x) = 1, \\ 0 & \text{if }\;\exists i\;\forall j>i: \chi_j(x) = 0.\end{cases}$$ Does this help?