$\mathcal P((-1,1))$ and the set of all functions $f:\Bbb R\to\Bbb R$ which attain every value in $\Bbb Z$ uncountably many times are equinumerous

elementary-set-theoryproof-writingsolution-verification

Prove that $\mathcal P((-1,1))$ and the set of all functions $f:\Bbb R\to\Bbb R$ which attain every value in $\Bbb Z$ uncountably many times are equinumerous.

My attempt:

My first claim:

Let $S=\{A\subseteq\Bbb R\mid A\text{ is uncountable}\}.\operatorname{card}(S)=2^{\mathfrak c}.$

$\boxed{\leq}:$ $$S\subseteq\mathcal P(\Bbb R)\implies\operatorname{card}(S)\le\operatorname{card}(\mathcal P(\Bbb R))=2^{\mathfrak c}.$$

$\boxed{\geq}:$ Let $\Phi:\mathcal P((-1,1))\to S,$ $$\Phi: A\mapsto A\cup B, B=\Bbb R\setminus((-1,1)\cup\Bbb Q).$$

We see $A\cap B=\emptyset,\forall A\in\mathcal P((-1,1)),$ therefore $$\Phi(A_1)=\Phi(A_2)\implies A_1\cup B=A_2\cup B\implies A_1=A_2,$$
so $\Phi$ is injective. $\implies 2^{\mathfrak c}=\operatorname{card}(\mathcal P((-1,1))\le\operatorname{card}(S).$

Therefore, $\operatorname{card}(S)=2^{\mathfrak c}.$

Let $T:=\{f:\Bbb R\to\Bbb R\mid f^{-1}(n)\text{ is uncountable },\forall n\in\Bbb Z\}.$

$\boxed{\leq }:$ $$T\subseteq \{f:\Bbb R\to\Bbb R\}\implies \operatorname{card}(T)\le 2^{\mathfrak c}.$$
$\boxed{\geq }:$ Let $S_{\Bbb Z}=\{A\in S\mid\ A\cap (n,n+1)\text{ is uncountable },\forall n\in\Bbb Z\}.$

Let $\Psi: S_{\Bbb Z}\to T, \Psi:A\mapsto \Psi_A,$ $$\Psi_A(x)=\begin{cases}n, &x\in A\cap (n,n+1)\\ 0,& \text{otherwise}\end{cases}$$

I think $\Psi$ is well-defined because $\operatorname{card}(\Psi^{-1}_A(n))=\operatorname{card}(A\cap (n,n+1))=\mathfrak c,\forall n\in\Bbb Z,\forall A\in S_{\Bbb Z}.$

We can write each $A\in S_{\Bbb Z}$ in a unique way as $$A=\bigcup_{n\in\Bbb Z}\underbrace{A\cap (n,n+1)}_{:=A^{(n)}}.$$

For $A_1,A_2\in S_{\Bbb Z}, A_1\ne A_2,\Psi_{A_1}$ and $\Psi_{A_2}$ must differ for some $x$ because $\exists n\in\Bbb Z, A_1^{(n)}\ne A_2^{(n)}.$ So, I believe, $\Psi$ is injective and hence $$\operatorname{card}\left(S_{\Bbb Z}\right)\le\operatorname{card}(T).$$

I think $\operatorname{card}\left(S_{\Bbb Z}\right)=2^{\mathfrak c}$ because we might similiarly prove that the set $S_n$ of all uncountable subsets of $(n,n+1)$ is also of the cardinality $2^{\mathfrak c},$ each $A\in S_{\Bbb Z}$ is of the form $$A=\bigcup_{n\in\Bbb Z}A^{(n)}, A^{(n)}\in S_n,n\in\Bbb Z$$

and $$\operatorname{card}\left(S_{\Bbb Z}\right)=\operatorname{card}\left(\displaystyle\prod_{n\in\Bbb Z} S_n\right)=\left(2^{\mathfrak c}\right)^{\aleph_0}=2^{\mathfrak c\cdot\aleph_0}=2^{\mathfrak c}$$ because $$\mathfrak c\cdot 1\le\mathfrak c\cdot\aleph_0\le\mathfrak c\cdot\mathfrak c=\mathfrak c.$$

So, $2^{\mathfrak c}\le\operatorname{card}(T).$

And finally, $\operatorname{card}(T)=2^{\mathfrak c}.$

On the other hand, $$\operatorname{card}(\mathcal P((-1,1)))=2^{\operatorname{card}((-1,1))}=2^{\mathfrak c}.$$

Are my conclusion and final answers valid? Also, is there any other way of proving this?

Best Answer

It looks that your answer is correct.

------ $\newcommand{\c}{\operatorname{card}}$

Here is a simpler way.

Let $T$ be the set of all functions $f:\Bbb R\to\Bbb R$ which attain every value in $\Bbb Z$ uncountably many times.

  • Since $T\subseteq \{f:\Bbb R\to\Bbb R\}$, $\c(T)\le \mathfrak c^\mathfrak c= 2^{\mathfrak c}=\c(\mathcal P((-1,1)).$
  • For all $s\in \mathcal P((-1,1))$, define $$\phi(s)(x)=\begin{cases}\lceil x\rceil &\text{if } x\le -1\\\lfloor x-1\rfloor &\text{if } x\ge1\\-1/2&\text{if }x\in(-1,1), x\not\in s\\1/2&\text{if } x\in s.\end{cases}$$ It is straightforward to verify that $\phi$ is an injective map from $\mathcal P((-1,1))$ to $T$. Hence, $\c(\mathcal P((-1,1)))\le \c(T)$.
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