# Infinite set A has the same cardinality as $\cup_{a \in A} B_a$

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!

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 $\bigcup_{a\in A}B_a$.