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.