1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $A \in \mathcal{P}(A) \implies A \in \mathcal{P}(B) \implies A \subseteq B$? $\endgroup$ – Fang Jing Jun 18 '16 at 12:35
  • 2
    $\begingroup$ 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... $\endgroup$ – Thomas Andrews Jun 18 '16 at 12:36
2
$\begingroup$

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 $A \subseteq B$.

$\endgroup$
  • $\begingroup$ You're right my definition was wrong. However, I modified the question taking your comment into account. $\endgroup$ – Nate Jun 18 '16 at 13:37
  • $\begingroup$ 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. $\endgroup$ – Noble Mushtak Jun 18 '16 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.