Cantor's diagonal argument: Prove that $|A|<|A^{\Bbb N}|$

Question

Let $A_1\subseteq A_2\subseteq A_3\subseteq...$ be a raising series of sets such that $\forall n\in \Bbb N \ |A_n|\lt |A_{n+1}|$. We mark $A$ as $A=\bigcup_{n\in\Bbb N}A_n$. Prove that $|A|<|A^{\Bbb N}|$

I've shown that $|A|\le|A^{\Bbb N}|$ since $f:A\rightarrow A^{\Bbb N}$ defined by $f(a)=(a,a,a,...)$ is an injection. I've considered for the sake of contradiction that $|A|=|A^{\Bbb N}|$ and tried to use Cantor's diagonal argument in order to get contradiction, but I got stuck. Thanks

Answer 1

As Asaf points out, Konig's Theorem applies here. It states (Jech, Set Theory, Millenium Edition, Theorem 5.10):

If $\kappa_i < \lambda_i$ for all $i \in I$, then $$ \sum_{i\in I} \kappa_i < \prod_{i\in I} \lambda_i. \tag{Konig} $$

Take $I = \Bbb N$, and for all $n\in I$, $\kappa_n = \lvert A_n\rvert$, $\lambda_n = \lvert A \rvert$. Then by (Konig), $$\begin{align} \lvert A \rvert & = \sum_{i\in I} \kappa_i \\ & < \prod_{i\in I} \lambda_i \tag{by Konig's Theorem} \\ & = \lvert A \rvert^{\lvert \Bbb N \rvert} \\ & = \lvert A ^{\Bbb N} \rvert. \end{align}$$

Note that Cantor's Theorem is an easy consequence of Konig's Theorem (Jech, Corollary 5.11): take $I = \Bbb N$, and $\kappa_i = 1, \lambda_i = 2$ for $i\in \Bbb N$.