Prove that if $F$ is a family of sets and $A\in F$, then $\bigcap F\subseteq A$

Question

Question is from "How to Prove it" by Vellenman. I am struggling to even understand how this theorem could ever be true?

I get that the first step is to assume x is an arbitrary element of $\bigcap F$, as the definition of $\bigcap F\subseteq A$ breaks down to $\forall x(x\in\bigcap F \rightarrow x\in A)$. Suppose $x\in \bigcap F$.

From here I am now confused about the definition of $\bigcap F$ and how $\bigcap F$ could even be a proper subset of $A$?

Answer 1

A proof in Velleman's style would possibly look like this:

Let $x$ be an arbitrary element of $\bigcap\mathcal F$. From $x\in\bigcap\mathcal F$ and $A\in\mathcal F$ we obtain $x\in A$ $($by universal instantiation$)$. Since $x$ is arbitrary, $\forall x(x\in\bigcap\mathcal F\rightarrow x\in A)$ and so $\bigcap\mathcal F\subseteq A$. $Q.E.D.$

Universal instantiation is well explained in the book.

Answer 2

$\bigcap F$ is the set of all things that are members of every member of the family $F$: $$\forall x\left(x\in\bigcap F\leftrightarrow\forall S\in F(x\in S)\right)\;.$$

Suppose that $x\in\bigcap F$; then by definition $x\in S$ for every $S\in F$. In particular, $x\in A$, since $A\in F$. Thus, $\bigcap F\subseteq A$.

Answer 3

I assume $\bigcap F $ means $\bigcap_{X\in F}X=:B$ and $A$ is an element in $F$.

Now take any $x\in B$, then it is in all $X$ from $F$ in paritculary in $A$.

Since $x$ was arbitrary we have $B\subseteq A$ and we are done.