$\lim 1_{A_n} = 1_{\cup A_n}$, where $(A_n)$ is increasing

Question

Let $X$ be a set and $(A_n)$ be a sequence of increasing subsets of $X$. Show that:

$$\large \lim_{n \to \infty} 1_{A_n} = 1_{\bigcup_{n=1}^{\infty} A_n}$$

Where $1_M$ is the characteristic function of the set $M$.

I don't have any idea about how to tackle this. It seems strange to me. What does it mean to say: $(\lim 1_{A_n})(x)$? Please provide some explanation along with the answer. Thank you.

Answer 1

Let $x\in X$. i) if $x \in \cup_n A_n$, then $1_{\cup_n A_n}(x)=1$. Now in this case there is $k_0\in \mathbb{N}$ such that $x\in A_k$ for every $k>k_0$ (since the sequence is increasing), so the left hand side is a constant sequence with constant value $1$ for $k>k_0$. Now it remains to consider ii) $x\notin \cup_n A_n$.

I suggest you try that case on your own first (?).

Answer 2

Hint: use $A_n=\bigcup_{i=1}^nA_i.$ as $A_n \supseteq A_i \forall i \in\{1,2,3,....,n\}$