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: 
Let $f:\bigcup\_i A\_i\to\prod\_i B\_i$ be any function. We show it can not be surjective. Let $\pi\_i:\prod\_i B\_i\to B\_i$ be the projection onto the $i^\text{th}$ factor for all $i$. Let $X\_i=B\_i\backslash \pi\_i\big(f(A\_i)\big)$. By assumption, $X\_i\neq\emptyset$ for all $i$. So $\prod\_i X\_i\neq\emptyset$. By construction, an element of $\prod\_i X\_i$ is not in the range of $f$.

Remark: This is an old result from König. The proof here is taken from [this book](http://rads.stackoverflow.com/amzn/click/0387984127).
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).