Is $\mathcal P(\Bbb R)\sim$ the set of all real functions defined on $\Bbb R$ with values in $\{0,1\}$?

Question

Is $\mathcal P(\Bbb R)\sim$ the set of all real functions defined on $\Bbb R$ with values in $\{0,1\}$ ?

I'm pretty sure this is true because I believe that for function $\mathfrak F$,

card ($\mathcal P(\Bbb R))=$ card$(\mathfrak F(\Bbb R))=\{\mathfrak f:\Bbb R\mapsto \{0,1\}\}=\{0,1\}^\Bbb R $. This is correct right?

I've been looking up online information on proving the cardinality equivalence of infinite sets, and to prove this, I was thinking of using an indicator/characteristic function. I'm not sure if I am writing out the proof right, but here is my attempt:

$\mathcal P(\Bbb R)=\{A: A \subseteq \Bbb R \}$

Now, create a character function of A. Let $\mathfrak f(A)= \chi _A(x)=\begin{cases} 1, & \text{if $x\in A$} \\ 0, & \text{if $x\in \Bbb R\setminus A$} \end{cases}$

Let $B$ be another set such that $B \subseteq P(\Bbb R) $ , $ \mathfrak f(B)= \chi _B(x)$, and $\chi _B(x)=\chi _A (x)$

This implies $\chi _B(x)= \begin{cases} 1, & \text{if $x\in B$} \\ 0, & \text{if $x\in \Bbb R\setminus B$} \end{cases}$

So, for $x\in A$ and $x\in B$, $\chi _B(x) = \chi _A(x)=1$ and

for $x\in \Bbb R\setminus A$ and $x\in \Bbb R\setminus B$, $\chi _B(x) = \chi _A(x)=0$

Thus, $ \mathfrak f(A)= \mathfrak f(B) \implies A=B$ so is one-to-one.

Let $\mathfrak g \in \{0,1\}^\Bbb R = \{\mathfrak g:\Bbb R\mapsto \{0,1\}\} $.

We know $A \subseteq \Bbb R \implies \mathfrak g(x) = \chi _A(x)$ for all $x \in \Bbb R $ so is onto.

Therefore, by definition, $\mathcal P(\Bbb R)\sim \{\mathfrak f:\Bbb R\mapsto \{0,1\}\} $.

Is this correct?

Thanks a lot.

Answer 1

This statement is correct, though your proof is not entirely correct. In proving your map is one-to-one, it is not very clear how you are concluding that $A=B$. You need to state more explicitly that for all $x$, $x\in A$ if and only if $x\in B$ (and why this is true).

In the proof that your map is onto, you need to explain how you are finding a set $A$ such that $\mathfrak{g}=\chi_A$. This certainly isn't true for an arbitrary set $A\subseteq\mathbb{R}$. So you need to explicitly define what this set $A$ is in terms of $\mathfrak{g}$.