Is it possible to find Bernstein set without Axiom of Choice?

Question

Statement(True/False) : Without Axiom of of choice there doesn't exists any set $B$ such that both $B$ and $B^c$ intersect every closed uncountable subsets Or every perfect sets in $\Bbb{R}$.

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Any set $B\subset \Bbb{R}$ which intersect every closed uncountable sets is known as Bernstein set.

If $B$ is Bernstein set then $B$ complement is also Bernstein . Also any subset of Bernstein that that are measurable must be meager, hence Bernstein set is not measurable infact has full outer measure.

Since without Axiom of choice it is impossible to construct a non measurable set (here) and Bernstein set are non measurable , without Axiom of Choice existence of Bernstein set is impossible.

My Question: Is my argument correct?

Is there any alternative way to prove non existence of Bernstein set without Axiom of Choice?

Answer 1

There is a subtle distinction. If I say that you don't have a cat, does that mean that you have a dog?

No. It doesn't. Without the axiom of choice means just not assuming it, but it means nothing about whether or not we assume its negation. It also doesn't tell you how it fails.

Bernstein sets can exist with the axiom of choice being false, simply because choice might fail "elsewhere", while the real numbers can still be well ordered in the universe.

Indeed, Bernstein sets can exist even if the real numbers cannot be well ordered! That much is consistent.

What you really want to say, though, is that you cannot prove the existence of Bernstein sets in ZF alone. And indeed, your argument is fine. Bernstein sets are non measurable, they are uncountable and without a perfect subset, etc. And indeed it is consistent that all sets are Lebesgue measurable, or have perfect subsets, etc.