Understanding proof of $\mathcal P(A) \subseteq \mathcal P(B) \to A \subseteq B$.

Question

Working on Richard Hammock. "Book of Proof" (p. 162), appears the following proof:

Example 8.9 Suppose A and B are sets. If $\mathcal P(A) \subseteq \mathcal P(B) \to A \subseteq B$.

Proof. We use direct proof. Assume $P(A) \subseteq \mathcal P(B)$. Based on this assumption, we must now show that $A \subseteq B$. To show $A \subseteq B$, suppose that $a \in A$. Then the one-element set $\{a\}$ is a subset of $A$, so $\{a\} \in \mathcal P(A)$. But then, since $P(A) \subseteq \mathcal P(B)$, it follows that $\{a\} \in \mathcal P(B)$. This means that $\{a\} \subseteq B$, hence $a \in B$. We’ve shown that $a \in A \to a \in B$, so therefore $A \subseteq B$.

As I understand, the definition of inclusion says that whenever an arbitrary element belongs to $\mathcal P(A)$, then that same element belongs to $\mathcal P(B)$. But the author talks about a fixed element {a}. Does this invalidate the proof ? Now, that element is not arbitrary — it is going to be a singleton set.

Answer 1

The hypothesis that $\wp(A)\subseteq\wp(B)$ says that every subset of $A$ is also a subset of $B$. In particular, every singleton subset of $A$ is a singleton subset of $B$. And as it happens, that part of the hypothesis is all that we need in order to conclude that every element of $A$ is also an element of $B$, i.e., that $A\subseteq B$.

The fact that we use only part of the hypothesis does not invalidate the argument. If it’s true that every subset of $A$ is a subset of $B$, then it’s certainly true that every singleton subset of $A$ is a subset of $B$; if the desired conclusion follows from that part of the hypothesis, it certainly follows from the whole hypothesis. More generally, if $P$ implies $Q$, and $Q$ implies $R$, then $P$ implies $R$. Here $P$ is the hypothesis that $\wp(A)\subseteq\wp(B)$, $Q$ is the assertion that every singleton subset of $A$ is a subset of $B$, and $R$ is the assertion that $A\subseteq B$.

In fact in this case the hypothesis that $\wp(A)\subseteq\wp(B)$ is actually equivalent to the statement that every singleton subset of $A$ is a subset of $B$. It would be a good exercise for you to try to prove that if every singleton subset of $A$ is a subset of $B$, then in fact $\wp(A)\subseteq\wp(B)$.

Answer 2

Here's another approach that might provide some clarity.

Recall that $E\in\mathcal{P}(S)$ means $E\subseteq S$.

Now, note that $A\in\mathcal{P}(A)$. Since $\mathcal{P}(A)\subseteq\mathcal{P}(B)$, it follows that $A\in\mathcal{P}(B)$. This means that $A\subseteq B$.

Answer 3

• You seem to assume that $a$ is some fixed element, while it is, in fact, an " arbitrary " element.

• The author makes no assumption as to this object $a$ that would individuate it, or make it " special" in whatever way. He only assumes what is necessary for this object $a$ to do its job, namely, that $a$ is an element of $A$ ( implying that $\{a\}$ is an element of $P(A)$).

• This is a general technique in mathematics: considering, so to say, generic individuals, that is, individuals capable of representing a whole class : what the author will prove about this arbitrary object $a$ will therefore also hold " for all $x$ belonging to $A$" .

• The aim of this procedure is to use " universal generalization" at the end of the proof. As you can see, the goal here is to end with :

" for all x, if $x$ is an element of $A$, then $x$ is an element of $B$"