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

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 :(

• You have infinitely many (uncountable many) $\mathfrak c$ factors, whereas with your argument you can only remove finitely many. – Dog_69 Jan 20 at 16:09
• so should I write that as $\mathfrak{c}^{\mathfrak{c}}$? – VirtualUser Jan 20 at 16:29

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 $$(\aleph_0)^{\aleph_0} =\mathfrak c > \aleph_0$$.)