If $A_i$ is a sequence of sets, define$$\liminf_i A_i = \bigcup_{j = 1}^\infty \bigcap_{i = j}^\infty A_i, \quad \limsup_i A_i = \bigcap_{j = 1}^\infty \bigcup_{i = j} A_i.$$Given a set $D$ define the function $\chi_D$ by $\chi_D(x) = 1$ if $x \in D$ and $\chi_D(x) = 0$ if $x \notin D$. Does it follow that$$\chi_{(\liminf A_i)}(x) = \liminf_i \chi_{A_i}(x), \quad \chi_{(\limsup A_i)}(x) = \limsup_i \chi_{A_i}(x)$$for each $x$?
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, &amp; x \in A \\ 0, &amp; 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  but I was somewhat confused by the topology, the meets/joins, etc.  Thanks much, Nate
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