Can it be proven in ZF that $\mathbb{N}$ is a universal size comparator?

Question

This is a follow-up to my previous question several years ago about whether every finite set is a universal size comparator. Now I am asking, can it be proven in ZF without the axiom of choice that the set $\mathbb{N}$ of natural numbers is a universal size comparator, meaning, it can be compared with every set in terms of cardinality?

Answer 1

No, it is consistent that there are infinite Dedekind finite sets.

Such set can easily shown to admin no injective to or from $ℕ$:

Let $D$ be that set, if $f:ℕ→D$ is injective, then $|D|=|D\setminus {f(0)}|$ by fixing a bijection between $f''ℕ\setminus\{0\}$ to $f''ℕ$, hence $ℕ\not≤D$.

If $g:D→ℕ$ is injective, then either $g''D$ is countable infinite or $g''D$ is finite, the first case implies it is Dedekind infinite, the second case implies it is finite, both cases lead to contradiction. ⇒ $D\not≤ℕ$

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That being said, we cannot prove that there exists a set $D$ that is not comparable with $ℕ$ from $ZF$(otherwise we could prove it exists in $ZFC$ as well, where we know all sets are comparable)