Proof that the infinity of all continuous functions is greater than the infinity of reals

Question

I've been trying to prove the above statement and found the below link in the course of my research.

Proof that the set of all possible curves is of cardinality $\aleph_2$?

However, I do not understand part of the explanation provided by Eric Wofsey, and I attach a screenshot of this part below.

screenshot

If anybody could help me understand this, I would be grateful.

Tia,

Yang

Answer 1

$Y^X$ is the space of all functions from $X\to Y$.

Furthermore, $\lvert Y^X\rvert=\lvert Y\rvert ^{\lvert X\rvert}$.

Finally, $\aleph_0\cdot 2^{\aleph_0}=2^{\aleph_0}$, since $2^{\aleph_0}$ has higher cardinality than $\aleph_0$.

So we arrive at $2^{2^{\aleph_0}}$, which is in fact not equal to $\aleph_2$ without making some additional assumptions...

Answer 2

There are a few different points to make here.

First of all, the notation "$\aleph_2$" is being horribly misused here (it reflects an implicit assumption of the generalized continuum hypothesis). Leaving that aside, the claim you want to prove is actually false. That is:

• There are $2^{2^{\aleph_0}}$ (also denoted "$\beth_2$") many arbitrary functions from $\mathbb{R}$ to $\mathbb{R}$; by Cantor's theorem, $2^{2^{\aleph_0}}>2^{\aleph_0}$ and this latter (also called "$\beth_1$" or "$\mathfrak{c}$") is the cardinality of the continuum.

• However, there are only $2^{\aleph_0}$ many continuous functions from $\mathbb{R}$ to $\mathbb{R}$. This is a standard exercise; one way to see it is to note that any continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ is determined entirely by the set $$S_f=\{(a,b,c)\in\mathbb{Q}^3: \vert f(a)-b\vert<c\}$$ (that is, if $f$ and $g$ are continuous then $f=g$ iff $S_f=S_g$) and then note that $\mathbb{Q}^3$ is countable so there only $2^{\aleph_0}$ many sets of rational triples at all (strictly speaking this only shows that there are at most $2^{\aleph_0}$ continuous functions from $\mathbb{R}$ to $\mathbb{R}$, but it's easy to show that's also a lower bound).

Finally, as to the step you mention, the point is that the meaning of cardinal exponentiation is exactly the size of the corresponding set of functions: $\kappa^\lambda$ is defined to be the cardinality of the set of functions from a set of cardinality $\lambda$ to a set of cardinality $\kappa$. So the set of all functions from reals to reals has cardinality $$(2^{\aleph_0})^{(2^{\aleph_0})}$$ since $\mathbb{R}$ has cardinality $2^{\aleph_0}$. Note that here we're looking at the set of all functions, not just the continuous ones.

It's then a basic result of set theory that the usual rule $(a^b)^c=a^{bc}$ holds for cardinal exponentiation, and that the product of two infinite cardinals is simply their maximum. So we get $$(2^{\aleph_0})^{2^{\aleph_0}}=2^{\aleph_0\cdot 2^{\aleph_0}}=2^{2^{\aleph_0}}$$ as Eric wrote.