Hello friends. I was wondering if you could critique the following proof. Thank you very much.

Notation:

$\displaystyle X\cong Y$ means that $\displaystyle X,Y$ have the same cardinal number.

: Find the cardinality of the set of all functions $\displaystyle f$ such that $\displaystyle f:[0,1]\mapsto\mathbb{R}$.Problem

: The cardinality of this set is $\displaystyle 2^{\mathfrak{c}}$ where $\displaystyle \mathfrak{c}$ is the cardinality of the continuum.Claim

Proof:

Lemma:$\displaystyle X\cong Y\implies \wp\left(X\right)\cong\wp\left(Y\right)$

Proof:Since $\displaystyle X\cong Y$ we know that there exists some $\displaystyle f:X\mapsto Y$ such that $\displaystyle f$ is a bijection. Clearly then, the function $\displaystyle \tilde{f}:\wp\left(X\right)\mapsto\wp\left(Y\right )$ given by $\displaystyle A\mapsto f\left(A\right)$ is a bijection. The conclusion follows. $\displaystyle \blacksquare$.

Lemma:Let $\displaystyle A=\left\{f:\text{ }f:[0,1]\mapsto\mathbb{R}\right\}$ and $\displaystyle B=\left\{f:\text{ }f:[0,1]\mapsto[0,1]\right\}$. then $\displaystyle A\cong B$.

Proof:Since $\displaystyle \mathbb{R}\cong[0,1]$ we know there exists some $\displaystyle g:\mathbb{R}\mapsto[0,1]$ such that $\displaystyle g$ is a bijection. Then define $\displaystyle \varphi:A\mapsto B$ by $\displaystyle f\mapsto gf$. To see that this is a bijection suppose that

$\displaystyle \varphi\left(f\right)=\varphi\left(\tilde{f}\right )\implies gf=g\tilde{f}$ and since $\displaystyle g$ is an injection it follows that $\displaystyle f=\tilde{f}$. And since $\displaystyle B\subseteq A$ the conclusion follows. $\displaystyle \blacksquare$

Now, define a mapping $\displaystyle \varphi:\wp\left([0,1]\right)\mapsto B$ by $\displaystyle A\mapsto f_A$ where $\displaystyle f_A=\begin{cases} 1 & \mbox{if}\quad x\in A \\ 0 & \mbox{if} \quad x\notin A\end{cases}$. To see that this is an injection assume that $\displaystyle \varphi\left(A\right)=\varphi\left(B\right)\implie s f_A=f_B$ and since $\displaystyle f_A=f_B$ if and only if

$\displaystyle f_A(x)=f_B(x)\quad\forall x\in [0,1]$ we see that $\displaystyle x\in A\implies f_A(x)=1=f_B(x)\implies x\in B$ where it follows, by an analogous argument, that $\displaystyle x\in B\implies x\in A$ which tells us precisely that $\displaystyle A=B$

from where it follows that $\displaystyle \varphi:\wp\left([0,1]\right)\mapsto B$ is an injection.

Now, define a mapping $\displaystyle \Delta:B\mapsto\wp\left([0,1]^2\right)$ by $\displaystyle f\mapsto\Gamma_f$ where, keeping with common notation, we have that $\displaystyle \Gamma_f=\left\{(x,y)\in[0,1]^2:y=f(x)\right\}$ is thegraphof $\displaystyle f$. It should be clear that this is an

injection, but if not assume that $\displaystyle \Delta\left(f\right)=\Delta\left(\bar{f}\right)\im plies \Gamma_f=\Gamma_{\bar{f}}$ and let $\displaystyle x\in[0,1]$. We then see that $\displaystyle \left(x,f(x)\right)\in\Gamma_f$ but since $\displaystyle \Gamma_f=\Gamma_{\bar{f}}$ we see that $\displaystyle \left(x,f(x)\right)\in\Gamma_{\bar{f}}$. But, since every element of $\displaystyle \Gamma_{\bar{f}}$ is

of the form $\displaystyle \left(x,\bar{f}(x)\right)$ we see that $\displaystyle \left(x,f(x)\right)=\left(x,\bar{f}(x)\right)\impl ies f(x)=\bar{f}(x)$ and since $\displaystyle x$ was arbitrary it follows that $\displaystyle f(x)=\bar{f}(x)\quad\forall x\in[0,1]\implies f=\bar{f}$. Thus $\displaystyle \Delta:B\mapsto\wp\left([0,1]^2\right)$ is an injection.

Lastly, noting that $\displaystyle [0,1]\cong [0,1]^2$ (by previous problem) it follows by our lemma that $\displaystyle \wp\left((0,1)\right)\cong\wp\left([0,1]^2\right)$ which means that there exist some $\displaystyle g:\wp\left([0,1]^2\right)\mapsto\wp\left([0,1]\right)$ which is bijective. Clearly

then, $\displaystyle g\Delta:B\mapsto\wp\left([0,1]\right)$ is an injection. Thus, by virtue of the Schroder-Bernstein theorem we may conclude that $\displaystyle B\cong\wp\left([0,1]\right)$. But, by our lemmas we see that $\displaystyle A\cong B,\wp\left([0,1]\right)\cong\wp\left(\mathbb{R}\right)$ from

which we may conclude, by the transitivity of the $\displaystyle \cong$ relation, that $\displaystyle A\cong\wp\left(\mathbb{R}\right)$ and since $\displaystyle \left|\wp\left(\mathbb{R}\right)\right|=2^{\mathfr ak{c}}$ our proof is done.

Remark:I believe that this can be extended to show that if $\displaystyle A$ is uncountable and $\displaystyle \left|B\right|\geqslant2$ that $\displaystyle \left|\left\{f:\text{ }f:A\mapsto B\right\}\right|=2^{\mathfrak{c}}$. This can be concluded since, it is easy to prove that when $\displaystyle A=\mathbb{R}$ and

$\displaystyle B=\{0,1\}$ that $\displaystyle \left\{f:\text{ }f:A\mapsto B\right\}\cong\wp\left(\mathbb{R}\right)$ and a little finagling with the above proof shows that $\displaystyle \left|\left\{f:\text{ }f:\mathbb{R}\mapsto\mathbb{R}\right\}\right|=2^{\ mathfrak{c}}$ from where the conclusion follows since

$\displaystyle 2^{\mathfrak{c}}=\left|\left\{f:\text{ }\mathbb{R}\mapsto\{0,1\}\right\}\right|\leqslant\ left|\left\{f:\text{ }f:A\mapsto B\right\}\right|\leqslant\left|\left\{f:\text{ }f:\mathbb{R}\mapsto\mathbb{R}\right\}\right|=2^{\ mathfrak{c}}$