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Hi all I am having difficulties in proving the following statement: Suppose $A$ is an infinite set, then there exists a bijection (for me, an injection would good enough) from $\cup_{a\in A} B_a$ to $A$, where each $B_a$ is some finite set. This question arises from the succinct proof of the dimension theorem on Wikipedia: http://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces

In fact, when A is countably infinite, this statement is true because "a countable union of countable sets is countable". However, when A is uncountable, I find it difficult to construct the bijection (or injection) above.

Or perhaps someone has a proof for the dimension theorem as well? Thank you very much!

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Since this fact requires the axiom of choice (as does "countable union of countable sets is countable"), you can't quite write down an injection. Also, the dimension theorem is not true without the axiom of choice so it makes sense. (It follows from a weaker principle, though, so it doesn't imply it.)

On the other hand, using the axiom of choice it is not hard to show that $|A|=|A\times\Bbb N|$, now for each $B_a$ choose a surjection $f_a\colon\Bbb N\to B_a$, and we have that $(a,n)\mapsto f_a(n)$ is a surjection from $A\times\Bbb N$ onto $\bigcup_{a\in A}B_a$.

Finally, using the axiom of choice find an injection reversing this map, and we're done.

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