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

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.

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    Please try to write your questions clearer and better titles, also try and be correct about the theorems you quote, especially their names. It should not be very hard to find the correct name on Wikipedia and copy-paste it here.2011-05-04

3 Answers 3

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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.

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    @Arturo, thanks, the title is enough. I found it as Theorem 3.5 on page 30 for anyone else interested.2011-05-09
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$2^{\mathfrak c} \le \aleph_0^{\mathfrak c}$ 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 $a^b \le (2^a)^b = 2^{ab}$.

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    About Edit 2: Yes, but the proof of the last equality then requires a little argument (any function $f:B\to(A^C)$ can be identified with a function $g_f:B\times C\to A$).2011-05-04
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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 ${\mathcal P}(\mathbb{R})$.

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    @user9176: The common shortcuts `\N` and `\R` for blackboard bold are not part of the standard LaTeX packages; you have to write `\mathbb{N}` and `\mathbb{R}`.2011-05-05