Is my proof correct ? $ \bigcup \{A - B\mid A \in F\} \subseteq \bigcup(F - P(B)) $

Question

Suppose $B$ is a set and $F$ is family os sets. Prove the following

$$ \bigcup \{A - B \mid A \in F\} \subseteq \bigcup(F - \mathcal P(B)) $$

where $\mathcal P(B) = \text{Power set of } B$

Proof :

Let $ x \in \bigcup \{A - B \mid A \in F\}$. So, $\:\exists A_1 \in F $ such that $ x \in A_1 - B $. So $x \in A_1$ and $x \notin B. $ which means $ x \notin\mathcal P(B)$

Claim : $A_1 \notin \mathcal P(B)$. Assume not true that is $A_1 \in \mathcal P(B)$. Since $x \in A_1$, so $ x \in \cup P(B)$. So $x \in B$. which is contradiction.

So for set $ A_1 $, we have $A_1 \in F$ and $A_1 \notin \mathcal P(B)$: we have shown that for arbitrary $x \in A_1 $ and $x \notin\mathcal P(B) $

Answer 1

In order to be unambiguous in notation, let us use $\mathscr{F}$ to denote our family of sets, and let us use $\mathscr{P}(B)$ to denote the power set of $B$.

On the left-hand-side, you have $$ \bigcup_{A \in \mathscr{F}} (A \setminus B), $$ whereas on the right-hand-side, you have $$ \bigcup_{S \in \mathscr{F} \setminus \mathscr{P}(B) } S. $$

Now let us suppose that $$ x \in \bigcup_{A \in \mathscr{F}} (B\setminus A). $$ Then there exists at least one set $A_x \in \mathscr{F}$ such that $x \in A_x \setminus B$, which implies that $x \in A_x$ and $x \not\in B$, and so the set $A_x \not\subset B$, which is the same thing as saying that $A_x \not\in \mathscr{P}(B)$, and thus we have $x \in A_x$ and $A_x \in \mathscr{F} \setminus \mathscr{P}(B)$, which implies that $$x \in \bigcup_{S \in \mathscr{F} \setminus \mathscr{P}(B) } S. $$

Therefore we obtain $$ \bigcup_{A \in \mathscr{F}} (A \setminus B) \subset \bigcup_{S \in \mathscr{F} \setminus \mathscr{P}(B) } S. $$

Conversely, suppose that $$ x \in \bigcup_{S \in \mathscr{F} \setminus \mathscr{P}(B) } S. $$ Then there exists at least one set $S_x \in \mathscr{F} \setminus \mathscr{P}(B)$ such that $x \in S_x$, and that particular set $S_x \in \mathscr{F}$ but $S_x \not\in \mathscr{P}(B)$, that is, $S_x \in \mathscr{F}$ but $S_x \not\subset B$. However, from this we cannot conclude that $x \not\in B$.

For example, let $$ \mathscr{F} \colon= \big\{ \ \{ 2 \}, \{1, 3, 5 \}, \{ 7 \} \ \big\}, $$ and let $$ B \colon= \{ 1, 3 \}. $$ Then $$ \mathscr{P}(B) = \big\{ \ \emptyset, \{ 1 \}, \{ 3 \}, B \ \big\}, $$ and so $$ \mathscr{F} \setminus \mathscr{P}(B) = \big\{ \ \{ 2 \}, \{1, 3, 5 \}, \{ 7 \} \ \big\}, $$ which is a family of sets with three set in it. Therefore we have $$ \bigcup_{ S \in \mathscr{F} \setminus \mathscr{P}(B) } S = \{ 2 \} \cup \{ 1, 3, 5 \} \cup \{ 7 \} = \{ 1, 2, 3, 5, 7 \}. $$ On the other hand, $$ \big\{ \ A-B \, \colon \, A \in \mathscr{F} \ \big\} = \big\{ \ \{ 2 \}, \{ 5 \}, \{ 7 \} \ \big\}, $$ which yields $$ \bigcup_{A \in \mathscr{F} } (A-B) = \{ 2 \} \cup \{ 5 \} \cup \{ 7 \} = \{ 2, 5, 7 \}. $$ Thus we see that $$ \bigcup_{A \in \mathscr{F} } (A-B) \subsetneqq \bigcup_{ S \in \mathscr{F} \setminus \mathscr{P}(B) } S. $$

Hope this helps.

Answer 2

In your proof (I would classify it as "okay, but not spotless") you state that:

"So $x \in A_1$ and $x \notin B $ which means $ x \notin\mathcal P(B)$..."

But it is not relevant (and wrong) to conclude that $x\notin\mathcal P(B)$.

It is enough to conclude on base of this that $A_1\notin\mathcal P(B)$ (as you do later).

Something else: the notation $ \bigcup (A - B \mid A \in F)$ is IMV not okay.

The sign $\bigcup$ must be followed by a set/collection and in $(A - B \mid A \in F)$ we cannot recognize a set/collection.

In the sequel I preassume that you meant to write: $\bigcup\left\{ A-B\mid A\in\mathcal{F}\right\} $.

---

My solution:

The following statements are equivalent:

• $x\in\bigcup\left\{ A-B\mid A\in\mathcal{F}\right\} $
• $x\in A-B$ for some $A\in\mathcal{F}$
• $x\notin B$ and $x\in A$ for some $A\in\mathcal{F}$

From the last statement we conclude that $A$ is not a subset of $B$ or equivalently $A\notin\mathcal{P}\left(B\right)$ so we can state that $x\in A$ for some $A\in\mathcal{F}-\mathcal{P}\left(B\right)$ or equivalently: $$x\in\bigcup\left(\mathcal{F}-\mathcal{P}\left(B\right)\right)$$