Counterintuitive consequences of $\lnot AC$? I know that there are many counterintuitive situations that can't be disproved in $ZF$, some enumerated in this answer (I would add the existence of a finite undetermined game, from this answer  [by the same author]). However, some of these can be solved with a weaker choice principle than the full Axiom of Choice. I'm looking for ones that can't. What are some counterintuitive statements one can deduce from $ZF + \lnot AC$?
Examples of other statements that can be proven in $ZFC$ but not in $ZF+DC$ (dependent choice) would also be interesting.
 A: The thing is that $\lnot\sf AC$ is just as non-constructive as $\sf AC$. Just like $\sf AC$ tells you that every set can be well-ordered, but it doesn't specify the well-ordering; $\lnot\sf AC$ simply tells you that some sets cannot be well-ordered, but it does not specify which sets.
So in general, all you can prove is that statements that are equivalent to $\sf AC$ fail. This may end up as counterintuitive in some cases, but that really depends on your intuition. Some examples:

*

*Some family of non-empty sets does not have any choice functions.

*Some two sets cannot be compared in cardinality.

*Some vector space does not have a basis.

*Some tree (a partial order where every downwards cone is well-ordered) is closed, in the sense that every increasing sequence has an upper bound, but there are no chain which meets every level of the tree.

*Some forcing notion $\Bbb P$ and some formula $\varphi(x)$ satisfy that $1\Vdash\exists x\varphi(x)$, but there is no $\Bbb P$-name $\dot x$ such that $1\Vdash\varphi(\dot x)$.

*Some surjection does not admit an inverse.

*Some set does not admit a group structure.

*Some commutative ring with a unit does not have any maximal ideals.

Again, we can just enumerate all the equivalences of the axiom of choice and consider their negation. That's pretty much all we can do here.
We cannot prove that $\sf BPI$ holds, but we cannot prove that it fails. We cannot prove that $\sf DC$ holds, but we cannot prove that it fails either. We cannot prove that $\mathcal{PPPPPPPPPPPPPP}\Bbb R$ can be well-ordered, but we cannot prove that it cannot be well-ordered either. And so on.
But, here are a few things that are at the very least consistent with $\sf ZF+DC+\lnot AC$.

*

*$\Bbb R$ can be partitioned into strictly more parts than elements. Or, just as well, it can be partitioned so that the partition is incomparable with $\Bbb R$ itself.

*Some infinite set $A$ satisfies that $|A|<|A\times 2|$.

*$\mathcal P(\Bbb R)$ cannot be linearly ordered.

*$\Bbb R$ does not have a Hamel basis over $\Bbb Q$.

*Some vector space over some field satisfies that every proper subspace has a countable basis, but the space itself does not.

*For every set $x$, there is some $A_x$ such that $A_x$ is the union of $\aleph_1$ sets of size $\aleph_1$, and $\mathcal P(A_x)$ can be mapped onto $x$.

And, again, there are a lot more.
