Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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 $|A|=|B|$.

share|improve this answer
    
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
1  
@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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.