# Find cardinality of $X = \left\{ A : A \subset \mathbb R \wedge \text{c}(A) \right\}$

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

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 $$|X|≤\frak c$$.

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).

• Why $|X|\ge 2^{\mathfrak c}$ ? – VirtualUser Jan 18 at 22:02

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 $$|X|=2^c.$$