# Help with proof.. No idea how to do second part. Suppose $\{A_i\mid i\in I\}$, $\{B_i\mid i\in I\}$....

I absolutely don't understand the second part of this question:

Suppose $\{A_i\mid i\in I\}$, $\{B_i\mid i\in I\}$ are indexed families of sets and $I\ne\varnothing$. Prove that if $\bigcup_{i\in I}A_i\nsubseteq\bigcup_{i\in I}B_i$, then there exists some $j\in I$ such that for all $i\in I$, $A_j\nsubseteq B_i$.

How do I prove this?? Here's my work so far. I tried 'translating' the statement.

$$(\exists i\in I:x\in A_i\to x\in B_i)\\\downarrow\\(\exists i\in I:x\notin A_i\lor x\in B_i)\\(\exists i\in I:x\in A_i\land x\notin B_i)$$ Suppose $\bigcup_{i\in I}A_i\not\subset\bigcup_{i\in I}B_i$. This means there is some $x\in\bigcup_{i\in I}A_i$ s.t. $x\notin\bigcup_{i\in I}B_i$. Since $x\in\bigcup_{i\in I}A_i$, then $x\in A_i$. Similarly, $x\notin B_i$.

A scan of the handwritten work is here.

• Try to write your question. we cannot trust links to pictures on other sites. Mar 15, 2016 at 9:30
• @MartinSleziak Great edit, it is a pity I cannot give you more credits than a simple comment. Mar 15, 2016 at 22:05
• @AD I have just added a tag, the post was edited mostly by Brian M. Scott: math.stackexchange.com/posts/1698372/revisions Mar 15, 2016 at 22:37

Suppose $\bigcup A_i\not\subset\bigcup B_i$. Then there is $x\in \bigcup A_i$ such that $x\not\in \bigcup B_i$.
This means that $x$ belongs to some $A_i$ and that this $x$ is in no $B_j$. Hence $A_i$ cannot be part of any $B_j$.
You are on the right track. Suppose $\bigcup_i A_i \nsubseteq \bigcup_i B_i$: then there exists some $x \in \bigcup_i A_i$ such that $x \notin \bigcup_i B_i$.
Now, apply the definition of union: $x \in \bigcup_i A_i$ is equivalent to $$\exists i \in I \mbox{ such that } x \in A_i$$ fix such an index, and give it a name, for example $i_0$.
Going on, $x \notin \bigcup_i B_i$ is equivalent to $$\neg (\exists i \in I \mbox{ such that } x \in B_i)$$ or, equivalently, $$\forall i \in I \ \ \ \ x \notin B_i$$
so, necessarily $A_{i_0} \nsubseteq B_i$ for all $i$. This was exactly what you needed to show.