Prove that if $B$ is a set and $\mathcal F$ is family of sets and $∪ \mathcal F ⊆ B$ then $\mathcal F ⊆ \mathscr P(B).$

elementary-set-theoryproof-verificationproof-writing

Suppose $B$ is a set and $\mathcal F$ is a family of sets. Prove that
if $∪ \mathcal F ⊆ B$ then $\mathcal F ⊆ \mathscr P(B).$

Note: $\mathscr P(B)$ stands for power set of $B$.

Suppose $\bigcup \mathcal F ⊆ B$.

$\bigcup F$ is the set that contains the elements of all subsets in $\mathcal F$. In other words, if arbitrary set, call it $A$, is a subset of $\mathcal F$, then all its elements will be in $\bigcup F$, or more formally: $\forall A(A \in \mathcal F \implies A \subseteq \bigcup F)$

$\bigcup \mathcal F ⊆ B$ means that all elements in $\bigcup \mathcal F$ are also in $B$. It follows that if arbitrary set, call it $A$, is the subset of $\bigcup \mathcal F$, then it also will be the subset of $B$, or more formally: $\forall A(A \subseteq \bigcup F \implies A \subseteq B)$

By definition, $\mathscr P(B)$ is the set containing all the subsets of $B$. In other words, $\forall A (A \subseteq B \implies A \in \mathscr P(B))$

To sum it all up, we have:

  1. $\forall A(A \in \mathcal F \implies A \subseteq \bigcup \mathcal F)$
  2. $\forall A(A \subseteq \bigcup F \implies A \subseteq B)$
  3. $\forall A (A \subseteq B \implies A \in \mathscr P(B))$

From this we can conclude that $\forall A(A \in F \implies A \subseteq \bigcup \mathcal F \implies A \subseteq B \implies A \in \mathscr P(B))$ and thus $\forall A(A \in \mathcal F \implies A \in \mathscr P(B))$.

Therefore, $\mathcal F \subseteq \mathscr P(B)$

Is it correct?

P.S Attempt to write more concise proof:

Suppose $\bigcup F \subseteq B$. Let $x$ be arbitrary set where $x \in \mathcal F$. If $x \in \mathcal F$ then all of its elements will be in $\bigcup F$, and since $\bigcup F \subseteq B$, then $x \subseteq B$. We know that $\mathscr P(B)$ is a power set, in other words , given arbitrary set A, $A \subseteq B \implies A \in \mathscr P(B)$. Thus $ x \in \mathscr P(B)$. Because $x$ was arbitrary, we can conclude that $\forall x(x \in \mathcal F \implies x \in \mathscr P(B))$. Therefore, $\mathcal F ⊆ \mathscr P(B)$

Best Answer

Yes, it is correct.

A bit more concise:

Let $F\in\mathcal F$.

Then consequently $F\subseteq\bigcup\mathcal F$.

Since $\bigcup\mathcal F\subseteq B$ we are allowed to conclude that $F\subseteq B$ or equivalently that $F\in\wp(B)$.

Proved is now that: $$\forall F\;[F\in\mathcal F\implies F\in\wp(B)]$$or equivalently that:$$\mathcal F\subseteq\wp(B)$$

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