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Well, this depends on what do you mean by "a form of choice". If you mean a statement of the form "Every family with property ... admits a choice function", then the answer, to my best of knowledge, is unknown to this date.

It is very important to point out that any such implication will be very local. That is, even if such a choice principle is provable it is likely one which only applies to particular sets of real numbers. It is easily arrangeable for any global choice principle (i.e. one which does not require all the sets we choose from to be subsets of a particular set) to fail, but the real numbers to be well-orderedable, and so all the $\sf ZFC$ theorems hold for it.

In the diagrams appearing in both Herrlich's The Axiom of Choice and Moore's Zermelo's Axiom of Choice there is really just the one obvious conclusion:

If there exists such endomorphism, then there exists a non-measurable set. I believe that we can also conclude the existence of a set of real without the Baire property, by the same argument.

To my knowledge, we don't even know if this assumption implies the existence of a Hamel basis (although my guess would be that it doesn't).

For sake of completeness, I should also point out that the statement itself is unprovable from $\sf ZF$, and in fact even when augmented by $\sf DC$, we still can't prove that there is a discontinuous endomorphism. The reason is that it is consistent (with $\sf ZF+DC$) that every set of reals has the Baire property, and from this follows that every homomorphism between Polish groups is continuous.

Well, this depends on what do you mean by "a form of choice". If you mean a statement of the form "Every family with property ... admits a choice function", then the answer, to my best of knowledge, is unknown to this date.

It is very important to point out that any such implication will be very local. That is, even if such a choice principle is provable it is likely one which only applies to particular sets of real numbers. It is easily arrangeable for any global choice principle (i.e. one which does not require all the sets we choose from to be subsets of a particular set) to fail, but the real numbers to be well-orderedable, and so all the $\sf ZFC$ theorems hold for it.

In the diagrams appearing in both Herrlich's The Axiom of Choice and Moore's Zermelo's Axiom of Choice there is really just the one obvious conclusion:

If there exists such endomorphism, then there exists a non-measurable set. I believe that we can also conclude the existence of a set of real without the Baire property, by the same argument.

To my knowledge, we don't even know if this assumption implies the existence of a Hamel basis (although my guess would be that it doesn't).

For sake of completeness, I should also point out that the statement itself is unprovable from $\sf ZF$, and in fact even when augmented by $\sf DC$, we still can't prove that there is a discontinuous endomorphism. The reason is that it is consistent (with $\sf ZF+DC$) that every set of reals has the Baire property, and from this follows that every homomorphism between Polish groups is continuous.

Well, this depends on what do you mean by "a form of choice". If you mean a statement of the form "Every family with property ... admits a choice function", then the answer, to my best of knowledge, is unknown to this date.

In the diagrams appearing in both Herrlich's The Axiom of Choice and Moore's Zermelo's Axiom of Choice there is really just the one obvious conclusion:

If there exists such endomorphism, then there exists a non-measurable set. I believe that we can also conclude the existence of a set of real without the Baire property, by the same argument.

To my knowledge, we don't even know if this assumption implies the existence of a Hamel basis (although my guess would be that it doesn't).

For sake of completeness, I should also point out that the statement itself is unprovable from $\sf ZF$, and in fact even when augmented by $\sf DC$, we still can't prove that there is a discontinuous endomorphism. The reason is that it is consistent (with $\sf ZF+DC$) that every set of reals has the Baire property, and from this follows that every homomorphism between Polish groups is continuous.

Well, this depends on what do you mean by "a form of choice". If you mean a statement of the form "Every family with property ... admits a choice function", then the answer, to my best of knowledge, is unknown to this date.

It is very important to point out that any such implication will be very local. That is, even if such a choice principle is provable it is likely one which only applies to particular sets of real numbers. It is easily arrangeable for any global choice principle (i.e. one which does not require all the sets we choose from to be subsets of a particular set) to fail, but the real numbers to be well-orderedable, and so all the $\sf ZFC$ theorems hold for it.

In the diagrams appearing in both Herrlich's The Axiom of Choice and Moore's Zermelo's Axiom of Choice there is really just the one obvious conclusion:

If there exists such endomorphism, then there exists a non-measurable set. I believe that we can also conclude the existence of a set of real without the Baire property, by the same argument.

To my knowledge, we don't even know if this assumption implies the existence of a Hamel basis (although my guess would be that it doesn't).

For sake of completeness, I should also point out that the statement itself is unprovable from $\sf ZF$, and in fact even when augmented by $\sf DC$, we still can't prove that there is a discontinuous endomorphism. The reason is that it is consistent (with $\sf ZF+DC$) that every set of reals has the Baire property, and from this follows that every homomorphism between Polish groups is continuous.

No, one cannot prove that from $\sf ZF$, or even $\sf ZF+DC$, where $\sf DC$ is the principle of dependent choice, a strengthening of the axiom of countable choice.

This is because it is consistent with $\sf ZF+DC$ that every set of real numbers have the Baire property, and therefore every endomorphism is Baire measurable, and therefore automatically continuous.

The consistency proof is due to Solovay and Shelah. Solovay constructed a model where every set of real numbers has the Baire property, and is Lebesgue measurable. This construction used an inaccessible cardinal (which is a greater consistency requirement than just $\sf ZFC$), and later Shelah proved that for measurability one has to have an inaccessible cardinal, but not for the Baire property.

Well, this depends on what do you mean by "a form of choice". If you mean a statement of the form "Every family with property ... admits a choice function", then the answer, to my best of knowledge, is unknown to this date.

In the diagrams appearing in both Herrlich's The Axiom of Choice and Moore's Zermelo's Axiom of Choice there is really just the one obvious conclusion:

If there exists such endomorphism, then there exists a non-measurable set. I believe that we can also conclude the existence of a set of real without the Baire property, by the same argument.

To my knowledge, we don't even know if this assumption implies the existence of a Hamel basis (although my guess would be that it doesn't).

For sake of completeness, I should also point out that the statement itself is unprovable from $\sf ZF$, and in fact even when augmented by $\sf DC$, we still can't prove that there is a discontinuous endomorphism. The reason is that it is consistent (with $\sf ZF+DC$) that every set of reals has the Baire property, and from this follows that every homomorphism between Polish groups is continuous.