Halmos's minimality condition I'm working on the last bit of Halmos' Families exercises, page 37 in Naive Set Theory. He states that we have $J \subset I$ and a nonempty family $\{X_i\}_{i \in I}$, which gives rise to $\bigcap_{i} X_i \subset X_j \subset \bigcup_{i} X_i$ (with $i \in I, j \in J$). 
It then goes on to say that if $X_j \subset Y$ for all $j$ then $\bigcup_{i} X_i \subset Y$. 
$Y$ hasn't been otherwise described before that I see. So I'm not sure how we arrive at this implication. By hypothesis $Y$ can have members that are not in any $X_j$, but that doesn't imply that these members must be in some $X_i$. 
Looking at other threads on this exercise on the internet, it looks like some people assume $\bigcup_i X_i \subset Y$. But I don't see in the book that this subset relationship is given. 
 A: According to Halmos's book (p. 37), there are three things to prove (assume $I \neq \emptyset$; note that your reading of the exercise is wrong, there is only one index set $I$):

*

*$\bigcap_{i \in I} X_i \subseteq X_j \subseteq \bigcup_{i \in I} X_i$ for all $j \in I$;

*minimality of $\bigcup_{i \in I} X_i$: if $X_j \subseteq Y$ for all $j \in I$ then $\bigcup_{i \in I} X_i \subseteq Y$;

*maximality of $\bigcap_{i \in I} X_i$: if $Y \subseteq X_j$ for all $j \in I$ then $Y \subseteq \bigcap_{i \in I} X_i$.

Proof.

*

*Let $x \in \bigcap_{i \in I} X_i$: then $x \in X_j$ for all $j \in I$; therefore, $\bigcap_{i \in I} X_i \subseteq X_j$ for all $j \in I$. 
Let $j \in I$ and $x \in X_j$: then $x \in \bigcup_{i \in I} X_i$; therefore $X_j \subseteq \bigcup_{i \in I} X_i$ for all $j \in I$.


*Assume $Y$ is such that $X_j \subseteq Y$ for all $j \in I$. Let $x \in \bigcup_{i \in I} X_i$. Then $x \in X_{j_0}$ for some $j_0 \in I$, and hence $x \in Y$ (since $X_{j_0} \subseteq Y$). Therefore, $\bigcup_{i \in I} X_i \subseteq Y$.


*Assume $Y$ is such that $Y \subseteq X_j$ for all $j \in I$. Let $x \in Y$. Then $x \in X_{j}$ for all $j \in I$, and hence $x \in \bigcap_{i \in I} X_i$. Therefore, $Y \subseteq \bigcap_{i \in I} X_i$.
A: Without the assumption $X_i\subseteq Y$ for all $i\in I$, this is false: take $X_0=\{0\}$, $X_1=\{1\}$, and $Y=X_0$ with $J=\{0\}\subseteq \{0,1\}=I$.  Clearly $X_j\subseteq Y$ for all $j\in J$, but $$\bigcup_{i\in I}X_i=X_0\cup X_1=\{0,1\}\not\subseteq \{0\}=Y\text{.}$$
But assuming $X_i\subseteq Y$ for each $i\in I$ (which is probably how the question should be read), we can show $\bigcup_i X_i\subseteq Y$ pretty directly.  Suppose $x\in \bigcup_i X_i$.  Thus there is some $i'\in I$ with $x\in X_{i'}$.  Since $X_{i'}\subseteq Y$, it follows that $x\in Y$.  This means that $x\in\bigcup_{i}X_i$ implies $x\in Y$, and thus $\bigcup_{i}X_i\subseteq Y$.
