Does $\mathsf{P} \ne \mathsf{NP}$ imply that $|\mathsf{NP}| > |\mathsf{P}|$?

Question

Is it possible that $\mathsf{P} \not = \mathsf{NP}$ and the cardinality of $\mathsf{P}$ is the same as the cardinality of $\mathsf{NP}$? Or does $\mathsf{P} \not = \mathsf{NP}$ mean that $\mathsf{P}$ and $\mathsf{NP}$ must have different cardinalities?

Answer 1

It is known that P$\subseteq$NP$\subset$R, where R is the set of recursive languages. Since R is countable and P is infinite (e.g. the languages $\{n\}$ for $n \in \mathbb{N}$ are in P), we get that P and NP are both countable.

Answer 2

If you are concerned about the size of two sets P and NP, the size of both these sets is infinite and equal.

If these two sets are equal, then their size is equal as well. If they are not equal, since they are countable then their cardinality is equal to the cardinality of natural numbers and equal.

So, in either case, their cardinality is equal.

Answer 3

I work in mathematics mainly and have only a bit of familiarity with this type of problem. However, set theory is one of my favorite areas of study, and this seems to be a set theory question.

So, to begin with, both P and NP are countably infinite as others have pointed out before. So, it does not make sense to discuss the cardinality of P and NP any further.

However, in general:

Set inequality does not inform one about the size of a set. Take for instance, $A=\{1,2,3\}$ and $B=\{4,5,6\}$. $A\neq B$, but $|A|=|B|$. Consider also, $C=\{1,2,3\}$ and $D=\{4,5\}$. $C\neq D$, and $|C|\neq|D|$.

However, by definition, set equality does inform us about cardinality. If $A=B$, then $|A|=|B|$. Consider the case of $A=\{1,2,3\}$ and $B=\{1,2,3\}$. $A=B$, and $|A|=|B|$.

If two sets are countably infinite, then they share the same cardinality. P and NP are both countably infinite, so that pretty much sums that up.