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I want to prove the theorem that two transcendence bases for a transcendental field extension have the same cardinality. I've come with this situation : if $A$ and $B$ are two transcendence bases for $E/F$ a field extension and $A$ is infinite, for each $S \subseteq A$ finite, there exists $B_S \subseteq B$ finite with $|B_S| \le |S|$ and $$ B = \bigcup_{S \subseteq A} B_S. $$ How does that imply that $|A| = |B|$? The proof I've read stops here and concludes the result must be true. I believe only set-theoretical arguments are used at this point. I've tried to come up with the fact that $$ |B| = \left| \bigcup_{S \subseteq A} B_S \right| \le \left| \sum_{S \subseteq A} B_S \right| \le \left| \sum_{S \subseteq A} S \right| $$ where the sum symbol stands for disjoint union (to get the last inequality, I simply use the fact that $|B_S| \le |S|$ to get an injection from $B_S$ to $S$ for each $S$ and I merge all those injections together). But I am not sure that the set of all finite subsets of $A$ has the same cardinality as their disjoint union.

To prove that thing that I am not sure of, I thought of writing $$ \sum_{S \subseteq A} S = \sum_{1 \le \ell < \aleph_0} \underset{|S| = \ell}{\sum_{S \subseteq A}} S $$ and then assume something like "countable additivity" to see that $$ \left| \sum_{1 \le \ell < \aleph_0} \underset{|S| = \ell}{\sum_{S \subseteq A}} S \right| = \sum_{1 \le \ell < \aleph_0} \left| \underset{|S|=\ell}{\sum_{S \subseteq A}} S \right| = \sum_{1 \le \ell < \aleph_0} \ell |A|^{\ell} = \sum_{1 \le \ell < \aleph_0} |A| = \aleph_0 |A| = |A|. $$ I'm worried about pretty much all the steps I'm using since I haven't done much set theory with cardinals before... anybody either knows how to prove the statement or knows if this is correct?

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up vote 2 down vote accepted

Your argument is a bit more complicated than necessary, but it’s basically correct. Let $[A]^n$ be the family of subsets of $A$ of cardinality $n$, and let $[A]^{<\omega}$ be the family of finite subsets of $A$. Then $\left|[A]^0\right|=1$, and for $1\le n<\omega$ there is an obvious injection from $[A]^n$ to $A^n$, so $$\left|[A]^n\right|\le\left|A^n\right|=|A|^n=|A|\;.$$ Thus, $\left|[A]^{<\omega}\right|\le\omega\cdot|A|=|A|$. On the other hand, $a\mapsto\{a\}$ is obviously an injection of $A$ into $[A]^{<\omega}$, so $\left|[A]^{<\omega}\right|=|A|$. (Note that this does use the axiom of choice.)

Added: To finish up, simply note that $$|B|\le\omega\cdot\left|[A]^{<\omega}\right|=\omega\cdot|A|=|A|\;.$$ Now apply the same argument with the rôles of $A$ and $B$ interchanged to get $|A|\le|B|$ and hence that $|A|=|B|$.

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Um, you seem to show that $$ |[A]^{<\omega}| = |A|, $$ but I seem to need the fact that $$ \left| \sum_{S \in [A]^{<\omega}} S \right| \le |A|. $$ The difference is that instead of counting the number of finite subsets of $A$, I take their disjoint union, which is how I got my bound for $|B|$... – Patrick Da Silva Mar 15 '13 at 22:29
@Patrick: That last step is trivial: Clearly $|B|\le\omega\cdot\left|[A]^{<\omega}\right|=\omega\cdot|A|=|A|$. (I probably should add it anyway, though.) – Brian M. Scott Mar 15 '13 at 22:34
I don't really care much about not using choice at this point, since the existence of a transcendence base is given by AC... haha =) – Patrick Da Silva Mar 15 '13 at 22:39
How do you get $|[A]^{<\omega}| \le |B|$? It seems wrong. (Not false, but just... weird?) In my proof I begin by assuming that $|A| \le |B|$ so it follows, but you seem to obtain it from the construction and I don't see how. – Patrick Da Silva Mar 15 '13 at 22:41
@Patrick: You’re welcome. (It’s amazing what you can prove by misreading, though!) I used choice in a couple of places, but yes, that’s one of them. – Brian M. Scott Mar 15 '13 at 22:55

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