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I have some doubts with this task:

Find cardinality of a) $X = \left\{ A : A \subset \mathbb R \wedge \text{c}(A) \right\} $
b) $X = \left\{ A : A \subset \mathbb Q \wedge \text{c}(A) \right\} $
where $\text{c}(A)$ means that set contains maximum and minimum element

I think that the result is $ \mathfrak{c} $, so I have decided to show two injectives:
$$f:\mathbb R \rightarrow X $$ and $$g:X \rightarrow \mathbb R $$ If it comes to $f$ it may be $$ f = \lambda x.\left\{x \right\} $$ and that set contains maximum and minimum element so I think that it is good example (both in a) and b) ).

But I am trying to show example of function $g$ and I have stuck there. One idea was to take $$ g = \lambda A. \frac{1}{2}(\min+\max) $$ but it is not injective :( thanks for your time

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For (a) let $H:\mathbb R\to(0,1)$ be bijective.

Then define $f:2^{\mathbb R}→ X$: $f(A)=H[A]∪\{0,1\}$, because $H$ is bijective $f$ is injective so $2^{\frak{c}}≤|X|$, and $X⊆ 2^{\Bbb R}$ so $|X|≤2^\frak c$.

For (b) let $H:\mathbb Q\to \mathbb Q\cap (0,1)$ be bijective.

Then define $f:2^{\mathbb Q}→ X$: $f(A)=H[A]∪\{0,1\}$, because $H$ is bijective $f$ is injective so ${\frak c}≤|X|$, and $X⊆ 2^{\Bbb Q}$ so $|X|≤\frak c$.

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For a), here's an injective map $\mathfrak P(\Bbb R)\to X$: $$ S\mapsto \{\,\tfrac x{1+|x|}\mid x\in S\}\cup \{-2,2\}$$ showing that $|X|\ge 2^{\mathfrak c}$ (and of course this means equality).

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  • $\begingroup$ Why $ |X|\ge 2^{\mathfrak c} $ ? $\endgroup$ – VirtualUser Jan 18 at 22:02
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Let $J= (-\pi/2,\pi/2)\}$ and let $Y$ be the set of all subsets of $J.$

Now $|J|=|\Bbb R|=c.$ E.g. $f(x)=\tan x$ is a bijection from $J$ to $\Bbb R.$ So $|Y|=\{t:t\subset \Bbb R\}|=2^c>c.$

For $s\in Y$ let $G(s)=s\cup\{-\pi/2,\pi/2\}.$ Since $G$ is one-to-one and since $\{G(s):s\in Y\}\subset X,$ we have $$2^c=|Y|=|\{G(s):s\in Y\}|\le |X|\le 2^c.$$ So $|X|=2^c.$

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