Is my proof to show that $\mathcal{P}(A) \subseteq\mathcal{P}(B) \implies A \subseteq B$ correct? $\mathcal{P}$ refers to the power set.

Suppose $A$ and $B$ are sets, and that $x$ is an arbitrary element of $A$.

The definition of the given $\mathcal{P}(A) \subseteq \mathcal{P}(B)$ means $$\forall y[(y \in \mathcal{P}(A) \rightarrow y \in \mathcal{P}(B)]$$ where $S$ denotes a subset and $y$ an element.

The antecedent means: $$\forall z(z \in y \rightarrow z \in A )$$

Likewise the consequent means: $$\forall z(z \in y \rightarrow z \in B )$$

Since $x\in A$ by the definition of the antecedent, it is trivially true So the consequent is also true.

So: $$A \subseteq B\,.$$

Edited: fixed my definitions, my initial definitions were wrong.

• I don't quite understand what $\forall y[(y \in S \rightarrow y \in A) \rightarrow (y \in S \rightarrow y \in B)]$ means - I've never seen this kind of notation. But doesn't $A \in \mathcal{P}(A) \implies A \in \mathcal{P}(B) \implies A \subseteq B$? – Fang Jing Jun 18 '16 at 12:35
• It's much easier to note that $A\in \mathcal P(A)$ so if $\mathcal P(A)\subset \mathcal P(B)$ then $A\in \mathcal P(B)$, so... – Thomas Andrews Jun 18 '16 at 12:36

Your definition of $\mathcal P(A) \subseteq \mathcal P(B)$ is wrong because it never quantifies $S$. It should be: $$(\forall y)(y \in \mathcal P(A) \implies y \in \mathcal P(B))$$
Now, what does $y \in \mathcal P(A)$ mean? It means $y \subseteq A$: $$(\forall y)(y \subseteq A \implies y \subseteq B)$$
Now, we have $A \subseteq A$. Therefore, we can conclude $A \subseteq B$.
• You still somehow conclude $x \in A$ from the antecedent. This is wrong. It does not follow from $A \subseteq B$ that there exists an $x$ such that $x \in A$. What if $A=\emptyset$? Instead, you have to start with $x \in A$ and then prove $x \in B$. This way, if $A=\emptyset$, then your proof is vacuously true. – Noble Mushtak Jun 18 '16 at 13:39