Hello friends. I was wondering if you could critique the following proof. Thank you very much.
means that have the same cardinal number.
Problem: Find the cardinality of the set of all functions such that .
Claim: The cardinality of this set is where is the cardinality of the continuum.
Proof: Since we know that there exists some such that is a bijection. Clearly then, the function given by is a bijection. The conclusion follows. .
Lemma: Let and . then .
Proof: Since we know there exists some such that is a bijection. Then define by . To see that this is a bijection suppose that
and since is an injection it follows that . And since the conclusion follows.
Now, define a mapping by where . To see that this is an injection assume that and since if and only if
we see that where it follows, by an analogous argument, that which tells us precisely that
from where it follows that is an injection.
Now, define a mapping by where, keeping with common notation, we have that is the graph of . It should be clear that this is an
injection, but if not assume that and let . We then see that but since we see that . But, since every element of is
of the form we see that and since was arbitrary it follows that . Thus is an injection.
Lastly, noting that (by previous problem) it follows by our lemma that which means that there exist some which is bijective. Clearly
then, is an injection. Thus, by virtue of the Schroder-Bernstein theorem we may conclude that . But, by our lemmas we see that from
which we may conclude, by the transitivity of the relation, that and since our proof is done.
Remark: I believe that this can be extended to show that if is uncountable and that . This can be concluded since, it is easy to prove that when and
that and a little finagling with the above proof shows that from where the conclusion follows since