what is the cardinality of the injective functuons from R to R?

Question

what is the cardinality of the injective functuons from R to R? lets say A={he injective functuons from R to R} obviously, A<= $2^א$ I have no Idea from which group I have to find an injective function to A to show (The Cantor-Schroeder-Bernstein theorem) that A=> $2^א$

Answer 1

We need Beth numbers for this. Knowing such a function's images at all reals $\lt a$, there are $\beth_1$ values left to choose for the image of $a$. Therefore, there are $\beth_1^{\beth_1}=\beth_2$ such functions. Note that if the functions are also required to be continuous the answer falls to $\beth_1^{\beth_0}=\beth_1$, since we determine the function with its image of $\Bbb Q$. (In particular, the functions of the form $kx,\,k\in\Bbb R\setminus\{0\}$ are a size-$\beth_1$ subset of such functions.)

Answer 2

Let $\kappa$ be any infinite cardinal. Clearly there are less than $\kappa^\kappa = 2^\kappa$ injective functions $\kappa\to \kappa$, so let's show that there are at least $2^\kappa$ as well, so we may conclude by Cantor-Bernstein.

Let $F\subset \kappa$ be any subset of $\kappa$ that isn't the complement of a singleton. Then I claim there is a bijection $\kappa \to \kappa$ whose fixed point set is precisely $F$.

For this, it suffices to show that $\kappa \setminus F$ has a self-bijection with no fixed points. Let $A=\kappa \setminus F$; by choice of $F$, $A$ is not a singleton. If $A$ is finite, it is easy to find such a permutation (for instance a cyclic permutation).

If $A$ is infinite, then there is a bijection $A\sim A\times \{0,1\}$ and then switching $0$ and $1$ on the RHS gives a bijection with no fixed point, so by transfer there must be one on $A$ as well.

It follows that $\{$ bijections $\kappa\to \kappa\}\to 2^\kappa, f\mapsto \{$ fixed points of $f\}$ is surjective onto the set of subsets that aren't complements of singletons. But now there are only $\kappa$ complements of singletons, so the set of subsets that aren't complements of singletons has size $2^\kappa$, so there are at least $2^\kappa$ bijections, and so at least $2^\kappa$ injections .

Answer 3

There are $\beth_2 = \mathfrak c ^{\mathfrak c} = 2^{\mathfrak c}$ functions (injective or not) from $\mathbb R$ to $\mathbb R$. For each such function $\phi$, there is an injective function $\hat\phi : \mathbb R \to \mathbb R^2$ given by $\hat\phi(x) = (x,\phi(x))$. If $\phi_1 \ne \phi_2$, then $\hat\phi_1 \ne \hat\phi_2$. So there are at least $\beth_2$ injective maps from $\mathbb R$ to $\mathbb R^2$. Finally since $\mathbb R$ and $\mathbb R^2$ have the same cardinality, there are at least $\beth_2$ injective maps from $\mathbb R$ to $\mathbb R$.

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Explanation of $\mathfrak c ^ \mathfrak c = 2^{\mathfrak c}$. $$ \mathfrak c ^ \mathfrak c = \big(2^{\aleph_0}\big)^{\mathfrak c} = 2^{\aleph_0\cdot\,\mathfrak c} = 2^{\mathfrak c} $$

Answer 4

Clearly there are at most $2^{\mathfrak{c}}$ injections $\mathbb{R} \to \mathbb{R}$.

On the other hand, for every $S \subseteq \langle 0,1\rangle$ define $f_S : \mathbb{R} \to \mathbb{R}$ with $$f_S(x) = \begin{cases} -x, &\text{ if $x \in S$ or $-x \in S$}\\x, &\text{otherwise}\end{cases}$$

Then $f_S$ is injective and $S \mapsto f_S$ is an injection so there are at least $2^\mathfrak{c}$ injections $\mathbb{R} \to \mathbb{R}$.