In my notes I have a proof that if $B_1$ and $B_2$ are basis of a vector space, then $|B_1|=|B_2|$. The proof assumes w.l.o.g $|B_1|\ge|B_2|$ and has two parts; the second part consider the case $|B_1|$ infinite.
Let be $B_1=\{v_j\mid j\in J\}$ and $B_2=\{w_k\mid k\in K\}$. Then $w_k=\sum_{j\in S_k}x^jv_j$, for some finite $S_k\subset J$.
The I have the following conclusion that use some set theory result:
Since $|B_1|>|B_2|$, respectively $|J|>|K|$, and the $S_k$'s are finite subsets of $J$, we have also $|J|>|\bigcup_{k\in K} S_k|$.
What is this result? (*) The union of finite subsets $S_k$ of an infinite set $J$ with $k\in K$ with $|K|<|J|$ is strictly smaller than $J$.
If I try to construct a counterexample in $\mathbb{Z}$, it is impossible, since $|K|<|\mathbb{Z}|$ means that $K$ is finite. On one site this make the result plausible, but in $\mathbb{Z}$ there is only one type of infinity.
Does the result (*) has some name? Some reference? Some idea for the proof?
EDIT I think that the idea behind is that if the $S_k$ are finite, the cardinality of $|\bigcup_{k\in K} S_k|$ is equal to the cardinality of $K$.
If $K$ is finite, then clearly $\bigcup_{k\in K}S_k$ is also finite and therefore $\left|\bigcup_{k\in K}S_k\right|<|J|$. If $K$ is infinite, then
$$|K|\le\left|\bigcup_{k\in K}S_k\right|\le|K|\cdot\aleph_0=|K|\,,$$
so $\left|\bigcup_{k\in K}S_k\right|=|K|<|J|$.