This one is purely for fun. Tell me what you think of this.
Problem: Call a mapping eventually constant if there exists some such that if then for some fixed . Let be the set of all eventually constant function.
Find the cardinality of
Claim: ( is used here to denote equipotence)
Proof: Let , and construct a mapping by . To see that this mapping is injective assume that
where it clearly follows that where it follows that . To see that it's surjective one merely needs to
know that for any the function is in and . The conclusion follows, so where the last equipotence is
obvious. Thus, for a fixed all the possible functions that are eventually is . Now, since each is countable it follows (since the union of countably many countable sets is
countable) that for a fixed that is countable. Lastly, noticing then that and reapplying the theorem that the union of countably many countably sets is countable
finishes the argument.
What do you think? It's late, and maybe I'm paranoid but I feel I am overlooking something.
Thank you very much for taking the time to read it.