# Power sets : Do the relations between P(A) and P(B) always mirror the relations between the sets A and B?

If I am correct it is true that:

(1) "$$P(A)$$ is included in $$P(B)$$" implies "$$A$$ is included in $$B$$".

(2) "$$P(A) = P(B)$$" implies "$$A = B$$".

Might I conclude from this that the power sets of two sets always have the same relations as these two sets have with one another?

Are there classical counterexamples to this (hasty) generalization?

I can think of this as a counterexample :

The fact that $$A$$ and $$B$$ are disjoint does NOT imply that $$P(A)$$ and $$P(B)$$ are disjoint.

Attenpt to prove (1) using the theorem : "$$\{ x \}$$ belongs to $$P(S)$$" $$\Longleftrightarrow$$ "$$x$$ belongs to $$S$$.

Let's admit that : $$P(A)$$ is included in $$P(B)$$.

Now, suppose (in view of refutation) that $$A$$ is not included in $$B$$.

It means that there exists an x such that x belongs to $$A$$ but not to $$B$$. And consequently that there is an $$x$$ such that $$\{ x \}$$ belongs to $$P(A)$$ but not to $$P(B)$$. If this were true, there would be a set $$S$$ such that $$S$$ belongs to $$P(A)$$ but not to $$P(B)$$. This contradicts our hypothesis according to which $$P(A)$$ is included in $$P(B)$$.

Conclusion: "$$P(A)$$ is included in $$P(B)$$" implies "$$A$$ is included in $$B$$".

• Indeed $P(A)\subseteq P(B)\iff A\subseteq B$: If $A\subseteq B$, then every subset of $A$ is also a subset of $B$, and if $A\not\subseteq B$, then $A$ is a subset of $A$ that is not a subset of $B$. – Hagen von Eitzen Apr 18 '19 at 10:00
• @HagenvonEitzen. Thanks. Do you think there are counter-examples to the generalization : power sets of A and of B always have the same relations as the original sets A and B? – user654868 Apr 18 '19 at 10:04

## 2 Answers

A very natural relation to consider would be that of the cardinality of the sets. We certainly have that $$|A| = |B|$$ implies that $$|P(A)| = |P(B)|$$. You may ask whether or not the converse is true, so does $$|P(A)| = |P(B)|$$ imply $$|A| = |B|$$?

This turns out to be not necessarily true, it is in fact independent from ZFC. That means that there can be set-theoretic universes where $$|P(A)| = |P(B)|$$ implies $$|A| = |B|$$ (e.g. when GCH holds), but there can also be set-theoretic universes where this fails. So then there are $$A$$ and $$B$$ such that $$|P(A)| = |P(B)|$$ while $$|A| \neq |B|$$. See also this answer: https://math.stackexchange.com/a/244873/661457.

Let's take these sets as an example: \begin{align} A &= \{1\} \\ B &= \{2\} \\ P(A) &= \{\emptyset, \{1\}\} \\ P(B) &= \{\emptyset, \{2\}\} \end{align} A relation is in this context a function that takes two sets and gives true or false.

A simple counterexample relation is: $$f(X, Y) = X \text{ contains a set}$$ Then we have: \begin{align} f(A, B) &= \text{false} \\ f(P(A), P(B)) &= \text{true} \\ \end{align}