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Let $A,B \subseteq E $. Show that $ \mathcal{P}(A)=\{X \cup Y \subseteq E| X \in \mathcal{P}(A\cap B) \land Y \in \mathcal{P}(A\backslash B) \} $

Tranlating the set from set builder notation to quantified statements I get: $$\forall X \cup Y \subseteq E, X \cup Y \in \mathcal{P}(A) \iff X\in\mathcal{P}(A\cap B) \land Y \in \mathcal{P}(A\backslash B) \ $$

Using double implication and the fact that $ \forall $ distrubutes with respect to $ \land $ I get:

$$(\forall X \cup Y \subseteq E, X \cup Y \in \mathcal{P}(A) \implies X\in\mathcal{P}(A\cap B) \land Y \in \mathcal{P}(A\backslash B)) \ \land (\forall X \cup Y \subseteq E, X\in\mathcal{P}(A\cap B) \land Y \in \mathcal{P}(A\backslash B) \implies X \cup Y \in \mathcal{P}(A)) $$

I had no trouble to prove the second statement, but I'm not sure about the first. I tried to assume what X and Y are such that $ X \cup Y \in \mathcal{P}(A)$ but I think that's not possible. Another idea I tried is to prove the statement by cases, considering $ B= \emptyset, A \cap B = \emptyset $ and $ A\cap B \neq \emptyset $,

Does anyone have any recommendations on how to prove this?

Thanks for reading!

Community
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George25
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2 Answers2

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You would be better off if you got rid of some of the notation. There is a tendency when people are just starting to write mathematics to go overboard on notation, and for the most part it just makes the mathematics harder to read. I stand by what I wrote on the subject in this answer.

You want to show two things:

  • if $S\subseteq A$, then there are $X\subseteq A\cap B$ and $Y\subseteq A\setminus B$ such that $S=X\cup Y$, and
  • if $X\subseteq A\cap B$ and $Y\subseteq A\setminus B$, then $X\cup Y\subseteq A$.

The second is completely straightforward, but the first takes a little work: given $S\subseteq A$, we have to find suitable sets $X$ and $Y$. Fortunately, there are natural candidates for both. We want $X$ to be a subset of $A\cap B$, and it’s already a subset of $A$, so we’ll just let $X=S\cap B$: then clearly $X\subseteq A\cap B$, and we’ve made $X$ as large as possible within that constraint. And we want $Y$ to be disjoint from $B$, so again we do the simplest thing and let $Y=S\setminus B$: then $Y\subseteq S\subseteq A$, and clearly $Y\cap B=\varnothing$, so $Y\subseteq A\setminus B$. It only remains to check that $X\cup Y=S$, and I’ll leave that for you to check.

Brian M. Scott
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Tip: $\{X\cup Y\subseteq E\mid X\in\mathcal P(A\cap B)\wedge Y\in\mathcal P(A\smallsetminus B)\}$ can be rewritten as:

$$\{Z\subseteq E\mid\exists X\subseteq E~\exists Y\subseteq E~\big[Z=X\cup Y\wedge X\subseteq (A\cap B)\wedge Y\subseteq (A\smallsetminus B)\big]\}$$

And of course $\mathcal P(A)=\{Z\subseteq E\mid Z\subseteq A\}$

So what you seek to prove is that $$\forall Z\subseteq E:\Big[Z\subseteq A\leftrightarrow \exists X\subseteq E~\exists Y\subseteq E~\big[Z=X\cup Y\wedge X\subseteq(A\cap B)\wedge Y\subseteq (A\smallsetminus B)\big]\Big]$$

Therefore you need to show is that for any arbitrary $Z$ where $Z\subseteq E$ that

  • $\exists X\subseteq E~\exists Y\subseteq E~\big[Z=X\cup Y\wedge X\subseteq(A\cap B)\wedge Y\subseteq (A\smallsetminus B)\big]\to Z\subseteq A$

  • $Z\subseteq A\to \exists X\subseteq E~\exists Y\subseteq E~\big[Z=X\cup Y\wedge X\subseteq(A\cap B)\wedge Y\subseteq (A\smallsetminus B)\big]$

Assume $Z$ is a union of some subsets of $A\cap B$ and $A\smallsetminus B$. ... Then $Z$ will be a subset of $A$.

Assume $Z$ is a subset of $A$. ... Then it will be the union of some subsets of $A\cap B$ and $A\smallsetminus B$.

Graham Kemp
  • 129,094