MHB Proof of Sets X,Y: X⊆Y <=> P(X)⊆P(Y)

[cocoabeens]

If I have finite sets X,Y, and need to prove that X ⊆ Y <=> P(X) ⊆ P(Y), where P() denotes the power set of a set.

I started out saying that for infinite sets X,Y, x⊆X, and y⊆Y.
Given that X⊆Y, we want to show that P(B)⊆P(Y).
x⊆X, so through transitivity, x⊆Y (is this correct?). From here, I wasn't quite sure how to complete the rest.

And then I need to show the statement is true the other way, so
given P(X)⊆P(Y), show that X⊆Y.
X⊆P(X), and Y⊆P(Y), by definition of power set, so for some x⊆X, and y⊆Y, x⊆P(X), and y⊆P(Y). Am I on the right track here, or did I mess up some rules?

 

[Deveno]

Suppose we are given that $X \subseteq Y$. This means that if $x \in X$,then $x \in Y$.

Now we need to prove that $P(X) \subseteq P(Y)$. So let $A$ be any element of $P(X)$ so that: $A \subseteq X$.

This means for any $a \in A$, we have $a \in X$. Since $X \subseteq Y$, it follows then that $a \in Y$.

Since this is true for ANY $a \in A$, we conclude that $A \subseteq Y$, that is: $A \in P(Y)$. Since $A$ was arbitrary, this establishes that $P(X) \subseteq P(Y)$.

Note that finiteness did not play a role here.

To go the other way, suppose $P(X) \subseteq P(Y)$ and consider, for any $x \in X$, the element $\{x\} \in P(X)$.

 

[cocoabeens]

Okay, it seems I was confusing it up with properties of power sets.

For the second part, I just work backwards, correct?

I show that the element {x}∈P(Y) because of the given condition, and thus x∈Y. Because x∈X, therefore X⊆Y? Did I confuse up some symbols?

 

[Deveno]

That looks OK to me...in your conclusion, I would write:

"Because $x \in Y$ whenever $x \in X$, we have $X \subseteq Y$" instead of:

"Because $x \in X$, therefore $X \subseteq Y$".