Prove that if $1 < \kappa_i \leq \lambda_i$ for all $i \in I$ then $\sum_{i \in I} \kappa_i \leq \prod_{i \in I} \lambda_i$ Prove that if $1 < \kappa_i \leq \lambda_i$ for all $i \in I$ then $\sum_{i \in I} \kappa_i \leq \prod_{i \in I} \lambda_i$ where $\kappa_i$ and $\lambda_i$ (for all $i \in I$) are cardinal numbers.
I am trying to prove this, which is similar to König's Theorem, but the technique to prove that does not work here because it could be that $\kappa_i = \lambda_i$ for any or all $i \in I$.
Suppose $\langle A_i \,|\, i \in I \rangle$ are mutually disjoint sets where $|A_i| = \kappa_i$ for all $i \in I$.
Likewise suppose that $\langle B_i \,|\, i \in I \rangle$ are sets such that $|B_i| = \lambda_i$ for all $i \in I$.
Since $\kappa_i \leq \lambda_i$ we can assume that $A_i \subseteq B_i$ for all $i \in I$.
We can then show the result by constructing an injection from $\bigcup_{i \in I} A_i$ to $\prod_{i \in I} B_i$ by the definition of cardinal sums and products.
I have found the determination of a general rule for such a mapping such that it is injective to be very difficult. I believe that I am able to come up with a mapping but it must treat the cases where $|I| = 0$ (so $I = \emptyset$, which is the trivial case), $0 < |I| < 3$, and $|I| \geq 3$ separately, which is messy, and the rules for mapping within the cases is also pretty messy.
Is there a simple way to construct this injective mapping that I am missing? Presumably it will depend on the fact that both $A_i$ and $B_i$ have at least two distinct elements for every $i \in I$.
NOTE: I found this question: Cardinality of the Union is less than the cardinality of the Cartesian product, and I actually came up with the mapping that Asaf gives in his answer. However, it seems that this mapping is not actually injective if $|I| = 2$, and I note that his mapping does not depend on the fact that $\kappa_i$ is strictly less than $\lambda_i$ (not here but in the actual König's Theorem), which is needed for a correct proof of König's Theorem.
To see how this fails suppose that $I = 2 =\{0, 1\}$ and
$$
A_0 = B_0 = \{0, 1\} \\
A_1 = B_1 = \{0', 1'\} \,.
$$
In his notation, if we let $\langle b_i \,|\, i \in I \rangle = \langle 0, 0' \rangle$, then it has to be that $\langle c_i \,|\, i \in I \rangle = \langle 1, 1' \rangle$.
His mapping $f$ then becomes
$$
f(0) = \langle 0, 1' \rangle \\
f(1) = \langle 1, 0' \rangle \\
f(0') = \langle 1, 0' \rangle \\
f(1') = \langle 0, 1' \rangle \,.
$$
Unless I am misunderstanding his mapping, this is clearly not injective, and I don't think that we have violated any of the assumption used to build the mapping.
I was able to come up with a slightly different mapping that works when $|I| = 2$ but, again, having to treat these cases separately is messy.
 A: After working through the details, I was mistaken about needing to treat the cases on $|I|$ separately. We only need two cases.
First, if $I = \varnothing$, then we have the trivial case of
$$
\sum_{i \in I} \kappa_i = 0 \leq 1 = \prod_{i \in I} \lambda_i \,.
$$
If $I \neq \varnothing$, then there is an $i_0 \in I$.
We also have distinct $\alpha_i$ and $\beta_i$ in $B_i$ (for any $i \in I$) since we have $1 < \kappa_i \leq \lambda_i = |B_i|$ so that $2 \leq |B_i|$.
Note that choosing these elements requires the Axiom of Choice.
Then, for an $x \in \bigcup_{i \in I} A_i$ there is an $i_x \in I$ where $x \in A_{i_x}$.
So, for any $i \in I$, define
$$
a_i = \begin{cases}
        x & i = i_x \\
        \alpha_i & i \neq i_x \text{ and } i = i_0 \text{ and } x = \alpha_{i_x} \\
        \beta_i & i \neq i_x \text{ and } i = i_0 \text{ and } x \neq \alpha_{i_x} \\
        \beta_i & i \neq i_x \text{ and } i \neq i_0 \text{ and } x = \alpha_{i_x} \\
        \alpha_i & i \neq i_x \text{ and } i \neq i_0 \text{ and } x \neq \alpha_{i_x} \,.
      \end{cases}
$$
Then define $f(x) = \langle a_i \,|\, i \in I \rangle$.
It is then easy but tedious (involving a lot of messy case work) to show that $f$ is an injective mapping from $\bigcup_{i \in I} A_i$ to $\prod_{i \in I} B_i$, which shows the result.
