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5

You can't construct such example. The fact that it is only consistent and not provable is exactly the issue here. Not only you can't construct one, you can't prove that one exists when the axiom of choice fails. These remarks, at the introductory level, are meant to guide the reader and suggest that a choice-free proof cannot be found. This is used to both ...


5

The axiom of choice is necessary for producing a Hamel basis for $\Bbb R$ over $\Bbb Q$, meaning we can prove that without the axiom of choice we cannot prove the existence of a Hamel basis; and of course it is not equivalent to the existence of such basis. Let me start with the easy part. The second part. There is no statement which is limited to a ...


3

No. To show the existence of such a basis one uses the Axiom of Choice. And anything that uses this axiom explicitly and necessarily defies explicit description. (If one could give an explicit description, one would make choices that can be described in finitely many words). For a deeper insight into this, check out Asaf's answer that will occur here in a ...


3

Both statements are provably false without the axiom of choice. The first one fails because of Hartogs theorem: each set $X$ has an ordinal $\alpha$ such that there is no injection from $\alpha$ into $X$. In particular $\mathcal P(\omega)$ has such $\alpha$, and the least such $\alpha$ is an $\aleph$ number by definition (whose index can be anything larger ...


2

All three points you suspect choice are using choice. Although in the third one you can perhaps remove it. There might be many finite subcovers at each step, and you have to choose one each time. How do you choose it? There might be many candidates at each time. How do you choose them? Ideally, you have a finite enumeration of $Q_n$, and you choose from ...


2

No, you can't prove this without the axiom of choice. Given any field, $F$, we can extend the universe of set theory, such that over $F$ there is a vector space which is not finitely generated, but every subspace is finitely generated. In particular, this means that there are no linear functionals from this vector space to the field (because a linear ...


1

You are looking for the Hamel Basis. There is no way to write this basis explicitly, since its existence is dependent on the Axiom of Choice.


1

In their paper, Howard and Tachtsis discuss these questions. The paper was published just last year, so I suspect that there has been any significant progress since then. Paul Howard and Eleftherios Tachtsis, On vector spaces over specific fields without choice, MLQ Math. Log. Q. 59 (2013), no. 3, 128--146. In general, it seems that the axiom of ...


1

Let me answer your underlying question. You don't need the axiom of choice. The proof in Jech "Set Theory" (3rd edition) of this claim, which is Lemma 14.28 is a choice free proof. As is the following theorem which extends that to arbitrary statements in the language of forcing (not just atomic ones). It's really just by induction on the maximal rank of the ...


1

$|A\cup B| \leq |A| + |B| \leq |B| + |B| = |B|$ since $|B|$ is of infinite cardinality and $2\cdot \aleph_k = \aleph _k$ for whichever cardinal $|B|$ happens to be. Then also $|B|\leq |A\cup B|$ by subadditivity. Hence $|A\cup B| = |B|$ when $B$ is an infinite set and $|A|\leq |B|$ I don't think you need to use the axiom of choice for the proof, though ...



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