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 :(

  • $\begingroup$ You have infinitely many (uncountable many) $\mathfrak c$ factors, whereas with your argument you can only remove finitely many. $\endgroup$ – Dog_69 Jan 20 at 16:09
  • $\begingroup$ so should I write that as $\mathfrak{c}^{\mathfrak{c}} $? $\endgroup$ – 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 $(\aleph_0)^{\aleph_0} =\mathfrak c > \aleph_0$.)

  • $\begingroup$ Ok,thanks, now I understand my fail $\endgroup$ – VirtualUser Jan 20 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.