Understanding power of $\left\{ f | f: \mathbb R \rightarrow \mathbb R \right\}$

Question

I have problem with understanding power of $\left\{ f | f: \mathbb R \rightarrow \mathbb R \right\}$
Generally, from lecture I know that power of set $$\left\{ f | f: A \rightarrow B \right\}$$ is $ |B|^{|A|} $ - ok, so it seems that power of $$\left\{ f | f: \mathbb R \rightarrow \mathbb R \right\}$$ should be $ |\mathbb R|^{|\mathbb R|} = \mathfrak{c}^{\mathfrak{c}} = 2^{\mathfrak{c}} $
Ok, but from the other hand, I can interpret that function as infinity set of pairs $ (x,y) $
So for each $ x $ I choose $y$. I can do this on $\mathfrak{c}$ ways. After that I repeat this process $\mathfrak{c} $ times so I have $$\mathfrak{c} \cdot \mathfrak{c} \cdot \mathfrak{c} \cdot ... \ $$ But from lecture I know that $$\mathfrak{c} \cdot \mathfrak{c} = \mathfrak{c} $$ So I reduce it to $\mathfrak{c}$.
I know that somewhere I have failed, but I have some doubts about that :(

Answer 1

Multiply a finite number of $\mathfrak c$'s together and you get $\mathfrak c^n =\mathfrak c.$ Even for a countably infinite number of factors, $\mathfrak c^{\aleph_0}=\mathfrak c.$ And yet, when we multiply together $\mathfrak c$ factors of $\mathfrak c$, we get $\mathfrak c^{\mathfrak c} > \mathfrak c.$

There is nothing contradictory about this. You simply need to multiply together a lot of factors of $\mathfrak c$ to get something larger than $\mathfrak c.$ $\aleph_0$-many doesn't suffice.

All you can conclude from iterating the binary idempotence $\mathfrak c^2 = \mathfrak c$ is the finite case $\mathfrak c^n=\mathfrak c$ for arbitrary $n.$ You cannot extend this to infinite powers. Since $\mathfrak c-n=\mathfrak c,$ pulling out a finite number of factors does nothing: $$\mathfrak c^{\mathfrak c} = \mathfrak c^n \cdot \mathfrak c^{\mathfrak c} = \mathfrak c \cdot \mathfrak c^{\mathfrak c} = \mathfrak c^{\mathfrak c}$$ We're just going in circles.

(Note, again, $\mathfrak c ^{\aleph_0}=\mathfrak c$ happens to be true but requires a different argument. For instance, similarly, $(\aleph_0)^n= \aleph_0$ for any finite $n$, but $(\aleph_0)^{\aleph_0} =\mathfrak c > \aleph_0$.)