# Some definitions without full choice

Assume DC($\aleph_1$).

Can we define the following:

1. Basis for a vector space $V$ over a field $K$ such that $\operatorname{card}(K) \leq \aleph_1$ and we happen to find a generating set of $V$ of cardinality $\leq \aleph_1$.
2. Linear dimension making the same assumption as above.
3. Transcendence degree of a field $K$ over $\mathbb{Q}$ if we happen to know that $K$ is of cardinality $\leq \aleph_1$. etc...
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## 1 Answer

Without any appeal to the axiom of choice, if you have a vector space which is well-ordered then it has a basis. Moreover, if you have a generating set which is well-ordered then the vector space has a basis. The latter one allows you to have that the field itself is not necessarily well-ordered (as in your first point, where it is).

The same point applies to all three points. Once an object is well-ordered, we can use transfinite induction to do anything without any appeal to the axiom of choice.

(Do note, however, that in models like Solovay's model the cardinality of $\Bbb C$ is not $\aleph_1$, and it may or may not have a transcendence basis over $\Bbb Q$.)

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Thank you. But what is the cardinality of $\mathbb{C}$ in Solovay's model then? – user38200 Aug 8 '13 at 11:51
I read in Jech's axiom of choice that the generalized continuum hypothesis implies the axiom of choice. What for the continuum hypothesis itself? – user38200 Aug 8 '13 at 11:57
In the Solovay model, the cardinality of $\Bbb C$ cannot be described by an ordinal. To see more about the continuum hypothesis itself, math.stackexchange.com/questions/404807/… – Asaf Karagila Aug 8 '13 at 12:01
And the same thing for Shelah's model, right? – user38200 Aug 8 '13 at 12:03
And also, is $\aleph_1$ the same thing as $\omega_1$ in these models? – user38200 Aug 8 '13 at 12:06