Let there be 2 players, p$\mathbb{N}$ and p$\mathbb{R}$. They are playing the Set Cardinality Game where p$\mathbb{N}$ has to present some number n that has not been used in the game so far by either of the players and it is in the set of natural numbers $\mathbb{N}$. Afterwards the player p$\mathbb{R}$ has to take some real number from the set $\mathbb{R}$ that has not been used before too. E.g If p$\mathbb{R}$ chooses 4, then both p$\mathbb{R}$ and p$\mathbb{N}$ cannot use it any more. By taking one number at time from their sets they are passing the turns.



If the player p$\mathbb{N}$ cannot find such a number n anymore, he loses and p$\mathbb{R}$ wins and vice versa. The game finishes only if one of the players wins.




1. Given that any of them starts first, is there any strategy for p$\mathbb{R}$ to defeat p$\mathbb{N}$ in an infinite amount of time?



2. If not, how should the game be modified so that p$\mathbb{R}$ could win? The game rules cannot rely on the properties of the numbers in the sets, but only on their amount.



3. Can other players like p$\aleph\_N$ defeat p$\mathbb{N}$ where p$\aleph\_N$ has the set which is a superset of $\mathbb{N}$ and its cardinality is $\aleph\_N$? Players with other cardinalities are also welcome.
What is the precise amount of the steps p$\aleph\_N$ would need to win?



I will appreciate solutions, explanations and also references what I should study to know more on the subject. I am still an undergraduate student.



Set Cardinality Game - Can a player with numbers in R win over a player with numbers in N as each of them in one turn has to present a new number?
1. Yes.
2. Any proper superset of $N$ will do. Just exhaust $N$ first and only play a non-member of $N$ when $N$ is exhausted, winning in the next step.


