# Index Family Subset Proof: Union subset Intersection

1. May 2, 2013

### mliuzzolino

1. The problem statement, all variables and given/known data

Let ${B_j: j \in J}$ be an indexed family of sets. Show that $\bigcup_{i \in J} B_i \subseteq \bigcap_{j \in J} B_j$ iff for all i, j, $\in$ J, Bi = Bj.

2. Relevant equations

3. The attempt at a solution

First show that $\bigcup_{i \in J} B_i \subseteq \bigcap_{j \in J} B_j \Rightarrow$ for all i, j, $\in$ J, Bi = Bj.

By contrapositive, $B_i \neq B_j \Rightarrow \bigcup_{i \in J} B_i \not\subset \bigcap_{j \in J} B_j$

Suppose $B_i \neq B_j$.

Let $x \in \bigcup_{i \in J} B_i$. So there exists an i in J such that x in Bi. But since Bi $\neq$ Bj, $i \neq j$ and there exists an $i \in J \ni x \notin B_j.$.

I know that the definition of the index family of intersections is for all j in J, x in Ej. But I'm not sure how to say that this isn't the case in the above proof...

Any guidance for a lost soul?

2. May 2, 2013

### LCKurtz

The contrapositive would be if there exist p and q such that $B_p\neq B_q$ then $\bigcup_{i \in J} B_i \not\subset \bigcap_{j \in J} B_j$. Don't use i and j for the indices and also the particular values.

Why don't you start with with x in one of $B_p,B_q$ and not the other? Then look at where it fits in the result.

Last edited: May 2, 2013
3. May 2, 2013

### mliuzzolino

Maybe I entirely mistook your guidance, but this is where I ended up going with it...not sure it's logically consistent though?

$\exists p, q \in J \ni B_p \neq B_q \Rightarrow \bigcup_{i_J} B_i \not\subset \bigcap_{j \in J} B_j$

Suppose $\exists p, q \in J \ni B_p \neq B_q$. Let $x \in \bigcup_{i \in J} B_i$, then $\exists i \in J \ni x \in B_i$. Then suppose $x \in B_p$ then $x \notin B_q$.

Therefore, for all $j \in J$, namely j = q, $x \notin B_j$.

So $\bigcup_{i \in J} B_i \not\subset \bigcap_{j \in J} B_j$.

A bit of these steps seem dodgy to me. Your input is greatly appreciated! :)

4. May 2, 2013

### LCKurtz

No. Don't start with saying x is in the union. Without loss of generality you can say there is an x that is $B_p$ and not $B_q$. Start with that x and show it is in the left side but not the right side.

5. May 2, 2013

### mliuzzolino

So, again my copious stupidity may be valiantly working against my attempts to understand this, but...

By supposition $B_p \neq B_q$, $x \in B_p$ AND $x \notin B_q$. Since there exists $p \in J \ni x \in B_p$, then $x \in \bigcup_{i \in J} B_i.$

Then, since there exists $j \in J,$ namely j = q, $\ni x \notin B_j$, then $x \notin \bigcap_{j \in J} B_j$.

Therefore, $\bigcup_{i \in J} B_i \not\subset \bigcap_{j \in J} B_j$.

6. May 2, 2013

### LCKurtz

Good. You are now halfway done with the original problem. The other direction is even easier

7. May 2, 2013

Thanks!!