# Exercise in Halmos set theory book.

I have other question about families. At the end of the chapter of families, there is a paragraph in the Halmos book of sets where the author says:

Prove also with appropriate provisos about empty families that $\bigcap_i X_i \subset X_j \subset \bigcup_i X_i$ for each index $j$ and the intersection and the union can in fact characterized as the extreme solution of these inclusions. This means that if $X_j \subset Y$ for each index $j$, then $\bigcup_i X_i \subset Y$ and $\bigcup_i X_i$ is the only set satisfying this minimality condition; the formulation for intersection is similar.

The first part is pretty easy to prove.

Suppose that $z\in \bigcap_i X_i.$ Then for each $i\in I$, we have that $z\in X_i$. So, clearly if $j\in I$ that means $z\in X_j$; which prove the first inclusion, $\bigcap_i X_i \subset X_j.$

Now suppose that $z\in X_j$. Then, $z\in X_i$ for at least one $i\in I$, i.e., $z\in \bigcup_i X_i$; which prove the second inclusion $X_j \subset \bigcup_i X_i$

So my question in specific is about the last part, where Halmos says: "if $X_j \subset Y$ for each index $j$, then $\bigcup_i X_i \subset Y$ and $\bigcup_i X_i$ is the only set satisfying this minimality condition".

How do I prove the uniqueness part?

Proposition: If $X_j \subset Y$ for each index $j$. Then $\bigcup_i X_i \subset Y$, and $\bigcup_i X_i$ is the unique set satisfying this minimality condition

Proof:

Suppose $z \in \bigcup_i X_i$. Then there is some $i,$ such that $z\in X_i$ but since $X_i \subset Y$ for each index, it follows that $z\in Y$. Therefore $\bigcup_i X_i \subset Y$.

Uniqueness: ...

Does somebody have a hint? Thanks as usual.

In particular let $(P,\leq)$ be a poset and let $Q\subseteq P$. If there is a $p\in Q$ such for all $q\in Q$ we have that $p\leq q$ then $p$ is unique: Assume $p'$ is a least element of $Q$. Then by definition $p'\leq p$ and $p\leq p'$, i.e. $p=p'$.
In your case $Q=\{Y : (\forall i\in I)[ X_i\subseteq Y]\}$.
• @JoseAntonio: Yes. If you assume that $Y$ is such as set, then by the minimality of both, $Y\subseteq\bigcup_iX_i$ and $\bigcup_iX_i\subseteq Y$, and hence they are equal. – Apostolos Aug 18 '13 at 20:39
If $Y$ fulfills the condition, then $\bigcup_i X_i\subseteq Y$, so if $\bigcup_i X_i\ne Y$ we have the smaller set $\bigcup_i X_i\subsetneq Y$ that fulfills the same condition, i.e. $Y$ is not minimal.
• But why not in the book $\subset$ is used either as proper subset or as subset with the possibility of equality. So, I've not shown yet that $\bigcup_i X_i\ne Y$. Or I misunderstood something? – Jose Antonio Aug 18 '13 at 20:04