Problem: Prove that $\aleph_0^c=2^c$
My attempt: $\aleph_0^c=\aleph_0^{2^{\aleph_0}}=\aleph_0^{\aleph_1}=\aleph_2$
Then $2^{2^{\aleph_0}}=2^{\aleph_1}=\aleph_2$, so $\aleph_0^c=2^c$.
But my intuition is telling me I may have made a mistake or an assumption that I can't make. Did I mess up? Or is this a valid proof?
On
There is a theorem tells:
Assume $A, B$ are two groups, if there is an injective function from $A$ to $B$ and an injective function from $B$ to $A$, then the cardinality of $A$ and $B$ equals.
This theorem is not very easy to prove. If you know this theorem, by using it we can prove this way:
I will prove $\mathbb N ^c = 2^c$. the injection from $2^c$ to $\mathbb N ^c$ is easy to construct. I only give the construction of injection from $\mathbb N ^c$ to $2^c$.
Focus on this function $$f(x) = \tan\left(\pi x - \frac{\pi}2\right)$$
It is a bijective function from $(0,1)$ to $\mathbb R$. We define the map $\mathbb N^{\mathbb R} \rightarrow \mathbb N^{(0,1)}$ by mapping $g: \mathbb R \rightarrow \mathbb N$ to $g \circ f: (0,1) \rightarrow \mathbb N$. You can check that it is a bijection. Hence we now only need to construct an injective map from $N^{(0,1)}$ to $2^c$.
Now assume $h \in N^{(0,1)}$ is a function, we define a map $\mathcal F: N^{(0,1)} \rightarrow 2^c$ by mapping $h$ to $$\mathcal F(h): x \rightarrow \begin{cases} 0 &\text{if } x \in \mathbb Z\text{ or } x<0 \\ \lceil x \rceil \text{th digital of binary number of } h(x-\lfloor x \rfloor) & \text{if } x \in [0,\infty) - \mathbb N \end{cases} $$
You can check that $\mathcal F$ is an injection. Hence we can use the previous theorem to complete our proof.
Obviously $2^c \le \aleph_0^c$. For the other direction, it may be helpful to consider $2^c$ to be the (cardinality of the) set of functions $f\colon \mathbb{R} \rightarrow \{0,1\}$, and similarly $\aleph_0^c$ to be the cardinality of the set of functions $f \colon \mathbb{R} \rightarrow \mathbb{N}$. Try and contruct a surjection from the former into the latter (hint: use binary expansions).