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One of the classical uses of the existence of bases of vector spaces (which is equivalent to the axiom of choice) is the following theorem:

If $V$ is an infinite vector space over a field $F$, and $|V|>|F|$, then $|V|=\dim V$.

As a corollary, we immediately get that any two uncountable fields of the same characteristics are isomorphic as additive groups if and only if they have the same cardinality. Proof? They have the same dimension over their prime field. So for example $(\Bbb R,+)\cong(\Bbb C,+)$.

Without the axiom of choice, things are not that nice. If every set is Lebesgue measure, or every set has the Baire property, then $(\Bbb R,+)$ and $(\Bbb C,+)$ are not isomorphic. Therefore there are two $\Bbb Q$-vector spaces of the same uncountable cardinality, which are not isomorphic.

Question. Suppose that whenever $W,W'$ are uncountable $F$-vector spaces and $|W|=|W'|>|F|$, we have that $W\cong W'$. Does the axiom of choice holds?

One of the classical uses of the existence of bases of vector spaces (which is equivalent to the axiom of choice) is the following theorem:

If $V$ is an infinite vector space over a field $F$, and $|V|>|F|$, then $|V|=\dim V$.

As a corollary, we immediately get that any two uncountable fields of the same characteristics are isomorphic as additive groups if and only if they have the same cardinality. Proof? They have the same dimension over their prime field. So for example $(\Bbb R,+)\cong(\Bbb C,+)$.

Without the axiom of choice, things are not that nice. If every set is Lebesgue measure, or every set has the Baire property, then $(\Bbb R,+)$ and $(\Bbb C,+)$ are not isomorphic. Therefore there are two $\Bbb Q$-vector spaces of the same uncountable cardinality, which are not isomorphic.

Question. Suppose that whenever $W,W'$ are uncountable $F$-vector spaces and $|W|=|W'|>|F|$, we have that $W\cong W'$. Does the axiom of choice holds? Or at least imply that $W$ has a basis over $F$?

(Looking through Rubin–Rubin II, there is a lot of focus on bases of vector spaces, but not so much on their cardinality. At most, there was a mention of the cardinality of a basis, but here I am not requiring that $W$ or $W'$ have a basis.)

One of the classical uses of the existence of bases of vector spaces (which is equivalent to the axiom of choice) is the following theorem:

If $V$ is an infinite vector space over a field $F$, and $|V|>|F|$, then $|V|=\dim V$.

As a corollary, we immediately get that any two uncountable fields are isomorphic as additive groups if and only if they have the same cardinality. Proof? They have the same dimension over their prime field. So for example $(\Bbb R,+)\cong(\Bbb C,+)$.

Without the axiom of choice, things are not that nice. If every set is Lebesgue measure, or every set has the Baire property, then $(\Bbb R,+)$ and $(\Bbb C,+)$ are not isomorphic. Therefore there are two $\Bbb Q$-vector spaces of the same uncountable cardinality, which are not isomorphic.

Question. Suppose that whenever $W,W'$ are uncountable $F$-vector spaces and $|W|=|W'|>|F|$, we have that $W\cong W'$. Does the axiom of choice holds?

One of the classical uses of the existence of bases of vector spaces (which is equivalent to the axiom of choice) is the following theorem:

If $V$ is an infinite vector space over a field $F$, and $|V|>|F|$, then $|V|=\dim V$.

As a corollary, we immediately get that any two uncountable fields of the same characteristics are isomorphic as additive groups if and only if they have the same cardinality. Proof? They have the same dimension over their prime field. So for example $(\Bbb R,+)\cong(\Bbb C,+)$.

Without the axiom of choice, things are not that nice. If every set is Lebesgue measure, or every set has the Baire property, then $(\Bbb R,+)$ and $(\Bbb C,+)$ are not isomorphic. Therefore there are two $\Bbb Q$-vector spaces of the same uncountable cardinality, which are not isomorphic.

Question. Suppose that whenever $W,W'$ are uncountable $F$-vector spaces and $|W|=|W'|>|F|$, we have that $W\cong W'$. Does the axiom of choice holds?

One of the classical uses of the axiom of choice is the following theorem:

If $V$ is an infinite vector space over a field $F$, and $|V|>|F|$, then $|V|=\dim V$.

As a corollary, we immediately get that any two uncountable fields are isomorphic as additive groups if and only if they have the same cardinality. Proof? They have the same dimension over their prime field. So for example $(\Bbb R,+)\cong(\Bbb C,+)$.

Without the axiom of choice, things are not that nice. If every set is Lebesgue measure, or every set has the Baire property, then $(\Bbb R,+)$ and $(\Bbb C,+)$ are not isomorphic. Therefore there are two $\Bbb Q$-vector spaces of the same uncountable cardinality, which are not isomorphic.

Question. Suppose that whenever $W,W'$ are uncountable $F$-vector spaces and $|W|=|W'|>|F|$, we have that $W\cong W'$. Does the axiom of choice holds?

One of the classical uses of the existence of bases of vector spaces (which is equivalent to the axiom of choice) is the following theorem:

If $V$ is an infinite vector space over a field $F$, and $|V|>|F|$, then $|V|=\dim V$.

As a corollary, we immediately get that any two uncountable fields are isomorphic as additive groups if and only if they have the same cardinality. Proof? They have the same dimension over their prime field. So for example $(\Bbb R,+)\cong(\Bbb C,+)$.

Without the axiom of choice, things are not that nice. If every set is Lebesgue measure, or every set has the Baire property, then $(\Bbb R,+)$ and $(\Bbb C,+)$ are not isomorphic. Therefore there are two $\Bbb Q$-vector spaces of the same uncountable cardinality, which are not isomorphic.

Question. Suppose that whenever $W,W'$ are uncountable $F$-vector spaces and $|W|=|W'|>|F|$, we have that $W\cong W'$. Does the axiom of choice holds?