Theorem. Let $\{ A_k | k \in K \}$ be a collection of sets indexed by the $K$, with $|K| = \kappa$. If $\forall k \in K \ \ |A_k| \leq \lambda$, then
$|\bigcup\limits_{k \in K} A_k| \leq \lambda\kappa$
I wonder how to prove the theorem. One way I can think of is to construct an injection
$f: \bigcup\limits_{k \in K} A_k \to M \times K$ where $|M| = \lambda$
knowing that $\forall k \in K$ we have an injection
$g_k: A_k \to M$
Any hints?
You’re on the right track.
Since $|K|=\kappa$, you might as well replace $K$ by $\kappa$ and index the sets by ordinals less than $\kappa$. For each $\xi\in\kappa$ you have an injection $f_\xi:A_\xi\to\lambda$. Define
$$f:\bigcup_{\xi\in\kappa}A_\xi\to\kappa\times\lambda$$
as follows: for each $x\in\bigcup_{\xi\in\kappa}A_\xi$ let
$$\eta(x)=\min\{\xi\in\kappa:x\in A_\xi\}\;,$$
and let
$$f(x)=\langle\eta(x),f_{\eta(x)}(x)\rangle\;.$$
Now just verify that $f$ is a well-defined injection.