Let $A=\bigcup_{i\in I_A}A_i$ where $I_A$ is the index set of $A$ and such that $A_i \cap A_j=\emptyset$ for any $i,j\in I_A$. Similarly, let $B=\bigcup_{j\in I_B}B_j$ where $I_B$ is the index set of $B$ and such that $B_i \cap B_j=\emptyset$ for any $i,j\in I_B$. Suppose:
- For every $i\in I_A$ and every $j\in I_B$, $|A_i|>|B_j|$.
- $|I_A|>|I_B|$.
Can we conclude that $|A|>|B|$? I think we can, but I don't know how to prove it unfortunately. There is definitely an injection from $B$ to $A$. I assumed that there is a bijection from $B$ to $A$ for reductio. But after this, I'm not sure how to proceed.
No. For instance, suppose $|I_A|=\aleph_1$ and $|A_i|=\aleph_\omega$ for each $i$, while $|I_B|=\aleph_0$ and for each $n\in\mathbb{N}$, there is one $j$ such that $|B_j|=\aleph_n$. Then $|A|=\aleph_1\cdot\aleph_\omega=\aleph_\omega$ but $|B|=\aleph_\omega$ as well since $|B|\geq\aleph_n$ for each $n$.
In general, we will have $|B|\leq|I_B|\cdot\kappa$ where $\kappa=\sup_j |B_j|$, and $|A|\geq |I_A|\cdot \kappa$ since $|A_i|\geq\kappa$ for each $i$. However, if $\kappa\geq |I_A|$ and no $|A_i|$ is strictly greater than $\kappa$, then $|A|$ and $|B|$ will both be equal to $\kappa$ as long as $\kappa$ is infinite, as in the example above.