Cardinality of infinite transcendence bases

Question

I'm reading a proof that any two transcendence bases of a field extension have the same cardinality here. It's the proof of Lemma 9.26.3 on that page, in the second paragraph of the proof, it says:

"By construction, $B^*$ is a subset of $B$, but we claim that in fact the two sets are equal. To see this, suppose that they are not equal, say there is an element $\beta\in B\text\B^*$ . We know $\beta$ is algebraic over $E(B′)$ which is algebraic over $E(B^*)$. Therefore $\beta$ is algebraic over $E(B^*)$, a contradiction. "

I understand that, but then it goes on to say that thus the cardinality of $B$ is smaller or equal to the cardinality of $B^*$... But didn't we just show that since $B^*$ is a subset of $B$, and there is no element in $B \text\B^* \Rightarrow |B|=|B^*|$?

Obviously when $|B|=|B^*|$, $|B|$ is also smaller or equal to $|B^*|$, but why don't they just say $|B|=|B^*|$ right away at that point? Is there something else going on?

Additionally, a bit earlier in the proof, we define $B_{\alpha}$ to be a subset of B, and that's why $B^*$ is a subset of $B$, right?

Thanks so much for your help!

Answer 1

For your first question: yes, $|B|=|B^*|$ holds, but I suppose they just want to emphasise that $|B|\leq|B^*|$ as they're trying to show that $|B|\leq|B'|$. But in any case I agree, it's confusingly written.

For the second question: yes, this is simply because $B_\alpha\subset B$. We can pick such $B_\alpha$ as $B$ is a transcendence basis of $E/F$, so we know that every $\alpha\in B'\subseteq E$ is algebraic over $F(B)$ and thus also $F(B_\alpha)$ for some finite $B_\alpha\subseteq B$.

[Small comment: it seems like there are some typos in the proof in which they write $E$ in places where I would guess $F$ should be used, but as I'm no expert in these matters I'm hesitant to claim that it definately is an error.]