Set theory with infinite subsets of N but without an uncountable power set

Question

If we allow infinite subsets of $N$, do we necessarily have an uncountable power set$?$ Do we have any axiomatisation which allows for infinite subsets of $N$ $($ or any countably infinite set$)$, but where the set of all permissible subsets cannot be proved to be uncountable?

Answer 1

If the full powerset of $\mathbb{N}$ exists in any (sufficiently usual) set theory, we are able to apply Cantor's argument to show that this set is uncountable.

But if we say "the full powerset of $\mathbb{N}$ does not exist in the set theory" then we are fine.

To do this, take ZF or ZFC but remove the axiom of the powerset. Then we still have all natural numbers and all their subsets (using the axiom of separation), but the powerset of the naturals will be a proper class and not a set. This will make real analysis inside the theory more difficult.

Another "solution" is this: (edited, because I made false claims) there are the realizability topoi (a kind of universe of sets) in which all functions and definable predicates are computable wrt. some notion of computability. The logic of these topoi is intuitionistic, so a lot of unusual behaviour can occur. Andrej Bauer constructed an example where there is an injection $\mathbb{N}^\mathbb{N} \hookrightarrow \mathbb{N}$.

The difference between the powerset $\mathcal{P}(\mathbb{N})$ and the set $\{0, 1\}^\mathbb{N}$ appears in intuitionistic logic, because the powerset is defined as $\mathcal{P}(\mathbb{N})=\Omega^\mathbb{N}$ where $\Omega$ is the set of truth values. And a topos is intuitionistic (and not-classical) if $\Omega$ is not simply (in bijection to) $\{0,1\}$.

Here a list of relevant StackExchange questions:

• Is there a constructive proof that $2^$ can't inject into ?
• Are real numbers countable in constructive mathematics?