Proper subset of power set

Question

I am working on basic set theory, more precisely, I am learning about the concepts of inclusion and power sets. I know that a power set $\mathcal P(A)$ of the set $A$ is the set of all subsets of $A$ and that a proper subset $X$ of a set $Y$ is such that $X \subset Y$ and $X \neq Y$.

If $X$ is a proper subset of the power set $\mathcal P(A)$ and $\mathcal P(A)$ is a proper subset of $Y$ such that $X$ contains exactly $2$ elements, how do I show the minimum number of elements that are in $Y$?

$$X\subset \mathcal P(A)\subset Y : |X|=2$$ $$X=\lbrace x_1, x_2 \rbrace, \; \text{find} \;|Y|$$ Basically what I am seeing is that if $X$ is a proper subset of $\mathcal P(A)$, then $\mathcal P(X)$ is also a proper subset of $\mathcal P(A)$ and $|\mathcal P(A)| \gt |\mathcal P(X)|\gt |X|$ so $|\mathcal P(A)|$ contains at least $|\mathcal P(X)|+1 = $ $2^{|X|}+1$ elements. $$X\subset \mathcal P(A) \implies \mathcal P(A) = \lbrace \,\ldots,\underbrace{ X, \lbrace x_1\rbrace, \lbrace x_2 \rbrace, \emptyset }_{\mathcal P(X)}\; \rbrace $$ Then, $$X\subset \mathcal P(A)\subset Y \implies X\subset Y$$ Since $X$ is already in $Y$ I do not count it twice and so we only know $ \lbrace x_1 \rbrace , \lbrace x_2 \rbrace$ and the empty set $\emptyset$ from $\mathcal P(A)$ that are included in $Y$, all the other elements of $\mathcal P(A)$ but $X$ are also included in $Y$ but they are unknown hence, $$Y=\lbrace\ldots, X, \lbrace x_1 \rbrace, \lbrace x_2 \rbrace, \emptyset \, \rbrace $$ Therefore I would claim that $$|Y| \ge 2^{|X|}+1$$ Is this a correct reasoning and does it apply to any $|X|$ in such conditions?

Answer 1

It is not true that if $X\subset\mathcal{P}(A)$ then $\mathcal{P}(X)\subset\mathcal{P}(A)$. For example, consider $A=\{0,1\}$ and $X=\{\{0\},\{1\}\}$. Then, for instance, $\{\{1\}\}$ is an element of $\mathcal{P}(X)$ but is not an element of $\mathcal{P}(A)$. Really, there is no reason at all for $\mathcal{P}(X)$ to be a subset of $\mathcal{P}(A)$ in general: an element of $\mathcal{P}(X)$ is a set of subsets of $A$, rather than a set of elements of $A$.

Instead, all you know from $X\subset\mathcal{P}(A)$ is that $2<|\mathcal{P}(A)|$, since $\mathcal{P}(A)$ contains the two elements of $X$ and also at least one more element (or in general, all you know is that $|X|+1\leq|\mathcal{P}(A)|$). Given this, what is the smallest cardinality that $\mathcal{P}(A)$ could have (keeping in mind that the cardinality of a power set can't be just any number)? And then given that $\mathcal{P}(A)\subset Y$, what is the smallest cardinality $Y$ could have?