How to prove union and intersection of classes if sets? Im working on some book exercises and struggle with a proof.
If $B_i$ and $C_j$ are two classes of sets such that $B_i \subseteq  C_j $.
How can I show that $\bigcup\limits_i B_{i}  \subseteq \bigcup\limits_{j} C_{j}$ and $\bigcap\limits_j C_{j} \subseteq \bigcap\limits_{i} B_{i}$
So far I have tried for $\bigcup\limits_i B_{i}  \subseteq \bigcup\limits_{j} C_{j}$:
Let $x  \in \bigcup\limits_i B_{i} \subseteq \bigcup\limits_j C_{j} \iff x \in B_i \cup C_j$
By the definition of union, it follows that $x \in B_i$ or  $x \in C_j$
By definition of subset it follows that $x \in B_i \implies x \in C_j$
So far I have tried for $\bigcap\limits_j C_{j} \subseteq \bigcap\limits_{i} B_{i}$:
Take any subset $x \subseteq B_i \cap C_j$. 
$x \subseteq B_i \wedge x \subseteq C_j$, meaning that $x \in \bigcup\limits_i B_{i} \wedge x \in \bigcup\limits_j C_{j}$. 
However not sure whether that is correct. Appreciate any help.
 A: You misstated the problem for  Simmons Introduction to Topology and Analysis.
The problem was:
Let $\{A_i\}, \{B_j\}$ be to classes of sets. so that $\{A_i\} \subset \{B_i\}$ (in other words it is the classes of sets that are subsets of each other and NOT the individual sets)
Now show that $\cup_i A_i \subset \cup_j B_j$ and $\cap B_j \subset \cap_i A_i$.
Again, my advice is element chase:
Let $x \cup_i A_i$ then $x \in A_i$ for some $A_i$.  Now $\{A_i\} \subset \{B_i\}$ so $A_i = B_k$ for some $B_k \in \{B_j\}$ so $x \in B_k$. so $x \in \cup_j B_j$.
So $\cup_i A_i \subset \cup_j B_j$.
If $x \in \cap_j B_j$ then $x \in B_j$ for all the $B_j$.  Let $A_i$ be any set in $\{A_i\} \subset \{B_j\}$.  Then $A_i = B_k$ for $B_k \in \{B_j\}$.  So $x \in B_k = A_i$ because $x$ is in all $B_j$.  Thus it is in all the $A_i \in \{A_i\} \subset \{B_j\}$ so it is in $\cap_i A_i$.
$\cap B_j \subset \cap_i A_i$.
A: For $\bigcup_i B_i \subseteq \bigcup_i C_i$, take an element $x \in B_i \subseteq C_i$. We then conclude $\bigcup_i B_i \subseteq \bigcup_i C_i$, because $x \in B_i \subseteq \bigcup_i B_i$ and because for all $i$, $B_i \subseteq C_i$, we have $x \in B_i \subseteq C_i \subseteq \bigcup_i C_i$.
For $\bigcap_i C_i \subseteq \bigcap B_i$, use element-chasing again by taking an element in all $C_i$, that is, an element in $\bigcap_i C_i$, than it is in all $B_i$, that is, an element in $\bigcap_i B_i$, because $B_i \subseteq C_i$ for all $i$. A hint: look at the complements of $B_i$ and $C_i$ for all $i$.
Make sure you state well from which set you are taking an arbitrary element. Show that that arbitrary element is also in the second set and you have proven the set inclusion.
