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 $X \subseteq Y$".

 

[blue_raver22]

Your approach is on the right track. Here is a more detailed explanation of the proof:

Proof of X⊆Y => P(X)⊆P(Y):
Assume X⊆Y. We want to show that for any subset B of X, B is also a subset of Y. Let B be an arbitrary subset of X. Since X⊆Y, every element in B is also an element of Y. Therefore, B⊆Y. By definition of the power set, P(B) is the set of all subsets of B. Since B⊆Y, every subset of B is also a subset of Y. Therefore, P(B)⊆P(Y). Since B was chosen arbitrarily, this holds for any subset B of X. Hence, X⊆Y => P(X)⊆P(Y).

Proof of P(X)⊆P(Y) => X⊆Y:
Assume P(X)⊆P(Y). We want to show that for any element x in X, x is also an element of Y. Let x be an arbitrary element of X. Since x is an element of X, {x} is a subset of X. By assumption, P(X)⊆P(Y), so {x} is also a subset of Y. This means that x is an element of Y. Since x was chosen arbitrarily, this holds for any element in X. Hence, X⊆Y.

Therefore, we have proved that X⊆Y <=> P(X)⊆P(Y).