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.
I think you are quite right.
If I am supposed to prove this, I will bigin with proving that the collection of all finite subset of N is countable, and then the collection of eventually constant mapping is embeded in the cartesian product of the collection of all finite subset of N, and N.
Your proof looks fine. It is, essentially, what I would do. In fact, I spend a wee while writing up here what I would do, posted it, then re-read your post to discover what I had written was, essentially, the same as what you had written. Essentially.
Originally Posted by Drexel28
That said, what I did was instead of defining the function I wasn't explicit with the . When you are defining your bijection near the top of your proof you can just forget about the 's as taking a cross product with the natural numbers will preserve your countability. This just makes defining everything neater and should, I think, remove a few lines afterwards.