Index Family Subset Proof: Union subset Intersection

[mliuzzolino]

Homework Statement

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, B_i = B_j.

Homework Equations

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, B_i = B_j.

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 B_i. But since B_i $\neq$ B_j, $i \neq j$ 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 E_j. But I'm not sure how to say that this isn't the case in the above proof...


Any guidance for a lost soul?

 

[LCKurtz]

Homework Statement

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, B_i = B_j.

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, B_i = B_j.

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

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.

Suppose $B_i \neq B_j$. ($\color{red}{B_p\ne B_q}$)

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

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.

 

[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! :)

 

[LCKurtz]

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 $\exists p, q \in J \ni B_p \neq B_q$. [STRIKE]Let $x \in \bigcup_{i \in J} B_i$, then $\exists i \in J \ni x \in B_i$.[/STRIKE]

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.

 

[mliuzzolino]

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.

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

 

[LCKurtz]

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

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

 

[mliuzzolino]

Thanks!