Find cardinality of all subsets of $\Bbb{R}, \Bbb{Q}$ that have both the biggest and the smallest element Let $X$ be the set of all subsets of $\Bbb{R}$ that have both the biggest and the smallest element.
Let $Y$ be the set of all subsets of $\Bbb{Q}$ that have both the biggest and the smallest element.
Find cardinalities of $X,Y$.
For the first one I tried to narrow down the situation to all the subsets that have $0$ and $1$ as their min and max elements. Let $A=\{\{0,1\}\cup p \ | \ p \in P((0,1))\}$, then $|A|=2^{|(0,1)|}=2^c$, hence $c < |A| \leq |X|$. Because there is $c$ such intervals then $|X|=2^ c \cdot c$.
I don't even know how to approach the $Y$.
 A: Let $E=\Bbb R$ or $\Bbb Q.$ Choosing a subset of $E$ that has a max and a min amounts to choosing $a\le b$ in $E$ and then some subset of the open interval $(a,b)\subset E.$
If $a=b,$ $(a,b)=\varnothing.$ If $a<b,$ $|(a,b)|=|E|.$
The number of singletons in $E$ is $|E|.$
The number of triples $(a,b,F)$ with $a,b\in E,a<b,F\subset(a,b)$ is the number $|E|$ of elements $a$ times the number $|E|$ of elements $b>a$ times the number $2^{|(a,b)|}=2^{|E|}$ of subsets of $(a,b).$
So the number of subsets of $E$ that have a max and a min is
$$|E|+|E|^2\cdot2^{|E|}=2^{|E|}.$$
A: Anne gives an answer to one possible interpretation of the question but I read it differently.  I took it as asking what cardinalities such a subset could have.
First, let's look at $\mathbb{Q}$.  $\{1\}$ qualifies so a cardinality of $1$ is possible.  Similarly, $\{1, 2, 3, ..., n\}$ qualifies so all finite cardinalities are possible.  The interval $[0, 1]$ qualifies so countable infinity is possible.  That's the cardinality of the whole of $\mathbb{Q}$ so we can't get any bigger.
Now, let's look at $\mathbb{R}$.  All of the previous subsets are also subsets of $\mathbb{R}$ so all of the previous answers are possible.  $[0, 1]$ as a subset of $\mathbb{R}$ also unqualified and it has the same cardinality as $\mathbb{R}$ so that is a new possibility.  What about between the cardinalities of $\mathbb{Q}$ and $\mathbb{R}$?  Well that is the Continuum Hypothesis.
