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Find the powers of the sets:

(a) = {$ : ⊆ \mathbb R$ and $$ has a smallest and largest element};

(b) = {$ : ⊆ \mathbb Z$ and $$ has a smallest and largest element};

(c) = {$ : ⊆ \mathbb Q$ and $$ has a smallest and largest element}.

I understand that for (a) we can use $\{−2\} ∪ (whatever ∩ (−1,1)) ∪ \{2\} ⊆ X ⊆ P(\mathbb R)$. But i don't know how to use it and also what to do with (b) and (c).

Any help would be appreciated.

P.S To clarify, in this question power of set means the collection of all subsets of set.

1 Answers1

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I assume that by power of a set you mean its cardinality.

(a) Using your construction we obtain that for any set $A\subseteq (-1,1)$ the set $A' := A\cup \{-2, 2\}$ lies in $X$. Moreover, the constructed sets $A'$ are pairwise distinct. Hence the set $X' = \{A\cup \{-2, 2\}\mid A\subseteq(-1,1)\}$ of all such sets is a subset of $X$, so we have $X'\subset X\subset 2^{\mathbb{R}}$. Now, using the fact that $|X'| = |2^{(-1,1)}| = |2^{\mathbb{R}}|$, we may use Schröder–Bernstein Theorem and conclude that $|X| = |2^{\mathbb{R}}|$.

(b) $Y$ is clearly infinite. However, for any two integers $a \leq b \in \mathbb{Z}$ the amount of sets $A\subset\mathbb{Z}$ such that $a = \min A$ and $b = \max A$ is finite. Thus $Y$ is a union of a countable amount of finite sets and hence is countable: $|Y| = |\mathbb{N}|$.

(c) $|Z| = |2^{(-1,1)\cap\mathbb{Q}}| = |2^{\mathbb{Q}}| = |\mathbb{R}|$ by the same argument as in (a).

Eugene Kogan
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    Thank you very much. Also I have a question: (a) is similiar to (c)? –  Nov 25 '23 at 19:36
  • @PoetryLover The argument is the same, we only use that the cardinality of $(0,1)\cap S$ the same as the cardinality of $S$.

    Another thing. I have just noticed that by the power of a set you probably mean the cardinality of the set (https://en.wikipedia.org/wiki/Cardinality). I will now correct my answer. Whoops :)

    UPD. Turns out I need not correct the answer.

    – Eugene Kogan Nov 25 '23 at 19:47
  • @PoetryLover Sorry, my bad. I added the explanation for (a). – Eugene Kogan Nov 25 '23 at 20:13
  • Thank you very much! Now I understand everything. Unfortunately, I can't vote your answer up, but I will do so in future. –  Nov 25 '23 at 20:21