Cardinal equalities: $\aleph_0^\mathfrak c=2^\mathfrak c$

Question

$\aleph_0^\mathfrak c=2^\mathfrak c$

This is a question on my homework, and I am not sure how to show they equal each other, I tried Cantor-Schroeder-Bernstein but got stuck, any help would be great.

Answer 1

For all infinite cardinals $\kappa$, you have $2^{\kappa}=\lambda^{\kappa} \leq(\kappa^+)^{\kappa}$ for all cardinals $\lambda$, $2\leq \lambda\leq \kappa^+$, where $\kappa^+$ is the successor cardinal of $\kappa$.

Certainly, we know that since $2\leq \lambda\leq\kappa^+$ and $\kappa$ is infinite, then $2^{\kappa}\leq \lambda^{\kappa}\leq (\kappa^+)^{\kappa}$, since the set of all functions from $\kappa$ to $2$ naturally embeds into the set of all functions from $\kappa$ to $\lambda$, which naturally embeds into the set of all functions from $\kappa$ to $\kappa^+$.

Now, since $\kappa\lt 2^{\kappa}$ by Cantor's Theorem, we also know that $\kappa^+\leq 2^{\kappa}$, so $(\kappa^+)^{\kappa}\leq (2^{\kappa})^{\kappa} = 2^{\kappa\kappa} = 2^{\kappa}$, since $\kappa\kappa = \kappa$ (assuming the Axiom of Choice). So we have the chain of inequalities $$2^{\kappa}\leq \lambda^{\kappa} \leq (\kappa^+)^{\kappa} \leq 2^{\kappa},\qquad \text{if }2\leq \lambda\leq\kappa^+;$$ so all inequalities are equalities.

Since $\aleph_0\lt \mathfrak{c}$, it follows that $$\aleph_0^{\mathfrak{c}} = (\mathfrak{c}^+)^{\mathfrak{c}} = 2^{\mathfrak{c}}.$$

Note. The equality $\kappa\kappa=\kappa$ for arbitrary infinite cardinals/sets is equivalent to the Axiom of Choice; but one can prove that for alephs we have $\aleph_{\alpha}\aleph_{\alpha}=\aleph_{\alpha}$ without using AC, and from there we get that $\aleph_{\alpha}+\aleph_{\beta} = \aleph_{\alpha}\aleph_{\beta} = \max\{\aleph_{\alpha},\aleph_{\beta}\}$ without having to invoke AC. For the result you want, though, you don't need it because $\mathfrak{c}\mathfrak{c}= 2^{\aleph_0}2^{\aleph_0} = 2^{\aleph_0+\aleph_0} = 2^{\aleph_0} = \mathfrak{c}$ which does not require AC, as pointed out by Asaf in comments.

Answer 2

$2^{\mathfrak c} \le \aleph_0^{\mathfrak c}$ since $2\le \aleph_0$

If you know, that for any cardinal numbers $a^b \le 2^{ab}$ and that $\aleph_0\cdot\mathfrak c=\mathfrak c$ then you get

$$\aleph_0^{\mathfrak c} \le 2^{\aleph_0\cdot\mathfrak c} = 2^{\mathfrak c}$$

Now you can use Cantor-Bernstein (you have both inequalities).

EDIT:

The inequality $a^b \le 2^{ab}$ follows from the fact that $A^B \subseteq \mathcal P(B\times A)$. (Here $A^B$ denotes the set of all maps from $B$ to $A$; every such function is a subset of $B\times A$.)

EDIT2:

Arturo's post reminded me of another possibility of proving the inequality $a^b \le 2^{ab}$. Since we know $a<2^a$ from Cantor theorem, we get $a^b \le (2^a)^b = 2^{ab}$.

Answer 3

Here is an alternate "more elementary" proof that $\aleph_0^\mathfrak c \leq 2^\mathfrak c$:

We use that $c$ is the cardinality of both $\mathbb{R}$ and $(0,1)$.

Let $f: (0,1) \rightarrow \mathbb{N}$ be any function.

We define $F_f := \{ x+ f(x) | x \in (0,1) \}$.

Then it is easy to check that any function $f$ is uniquely identified by $F_f$ and thus

$f \to F_f$ is a 1-1 function from $\mathbb{N}^{(0,1)}$ to ${\mathcal P}(\mathbb{R})$.