Hello friends. I was wondering if you could critique the following proof. Thank you very much.

__Notation:__ 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:__ **Lemma:** **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