Question: Suppose I have two collections of sets, $\{A\_i\}\_{i \in I}$ and $\{B\_i\}\_{i \in I}$. It is given in the problem that $$\mathrm{card}(A\_i) < \mathrm{card}(B\_i) \text{ for all }i \in I $$ 

I want to show that 
$$ \operatorname{card}\left(\bigcup\_{i \in I} A\_i \right) < \operatorname{card}\left( \prod\_{i \in I} B\_i \right) $$ 

There are no stipulations given about the cardinalities of $A\_i, B\_i$ or $I$. 

This question came up when I was looking through a book on Naive Set Theory, and it has proved intractable so far. Any help would be much appreciated!

Answer: 
This is known as **König's Theorem** (there is also a lemma called the son of this one). We are going to assume the axiom of choice through this answer, otherwise one cannot prove this theorem (more on that at the bottom).

Michael Greinecker gave a concise argument for the fact there is no surjective function from the union to the product. Which under the axiom of choice is enough to conclude there is a surjection in the other direction, and an injection from the union to the product, as wanted. Here is a more explicit construction of an injection:

If any of the $A\_i$ is empty let us drop it from the list, and remove the corresponding $B\_i$ as well. I hope it's clear why this is not going to harm the result. We may also assume that the $A\_i$'s are pairwise disjoint sets.

The fact that $|A\_i|<|B\_i|$ means that there is an injection $f\_i\colon A\_i\to B\_i$. And there is no injection in the other direction, else by Cantor-Bernstein we would have equality rather than a sharp inequality. For simplicity we can also assume that $A\_i\subsetneqq B\_i$ for all $i\in I$.

Fix $\langle b\_i\mid i\in I\rangle$ some sequence in the product. Let us define an injection now. Let $a\in A\_j$, if $a\neq b\_j$ then send $a$ to the sequence $\langle \overline b\_i\mid i\in I\rangle$ such that $\overline b\_i=b\_i$ for $i\neq j$ and $a$ otherwise.

If $a=b\_j$, we know that $B\_i$ has at least two elements for every $i\in I$ then we can fix $\langle c\_i\mid i\in I\rangle$ in the product, such that $c\_i\neq b\_i$ for every $i\in I$. So in the case $a=b\_j$ we send $a$ to the sequence of $\langle \overline c\_i\mid i\in I\rangle$ such that $\overline c\_i=c\_i$ for $i\neq j$, and $a$ otherwise.

It is not hard to see that this is an injective function. Therefore the union has cardinality less or equal than that of the product.

---

A word about the axiom of choice. It is consistent that without the axiom of choice there is a countable set of disjoint *pairs* that has no choice function. Namely $\{P\_n\mid n\in\omega\}$ such that $\prod P\_n=\varnothing$. It is easy to see that even if we take $A\_n=\{0\}$ for all $n$, then $|A\_n|=1<2=|P\_n|$, and $$\left|\bigcup A\_n\right| = |\{0\}| = 1 > 0 = |\varnothing| =\left|\prod P\_n\right|$$

I will remark, however, that if we assume that $A\_i$ and $B\_i$ are all sets of ordinals then the lemma holds regardless to the axiom of choice.
This is called [König's theorem](http://en.wikipedia.org/wiki/K%C3%B6nig%27s_theorem_%28set_theory%29). The proof I know is reasonably simple but not something I'd expect anyone to think up as an exercise. See the bottom of page 15 [here](http://tartarus.org/gareth/maths/notes/iii/Set_Theory_2012.pdf) (page 17 in the PDF).