Thread: Equality of two sets (using Boolean algebra)

1. Equality of two sets (using Boolean algebra)

Hello. I recently started my course, so this is pretty trivial.

Prove:
$\displaystyle $f\left( {\bigcup\limits_{i \in \ell } {{A_i}} } \right) = \bigcup\limits_{i \in \ell } {f\left( {{A_i}} \right)}$$

Well, analogous examples demonstrated in my university begin with

$\displaystyle $\forall y \in f\left( {\bigcup\limits_{i \in \ell } {{A_i}} } \right) \Rightarrow \left( {\exists x:f\left( x \right) = y} \right) \wedge \left( {x \in \bigcup\limits_{i \in \ell } {{A_i}} } \right)$$

The problem is that I am not sure how to represent this highly abstract union of sets since
$\displaystyle $x \in \bigcup\limits_{i \in \ell } {{A_i}} \Rightarrow x \in {A_i} \wedge i \in \ell$$
I guess wouldn't make enough sense (I could tell the same about intersection of those very same sets).

So what is missing on my mind?
Thanks for help.

2. Originally Posted by Pranas
The problem is that I am not sure how to represent this highly abstract union of sets since $\displaystyle $x \in \bigcup\limits_{i \in \ell } {{A_i}} \Rightarrow x \in {A_i} \wedge i \in \ell$$
I guess wouldn't make enough sense (I could tell the same about intersection of those very same sets
This is one standard way: $\left( {\exists j \in \ell } \right)\left[ {x \in A_j } \right]$

3. Originally Posted by Plato
This is one standard way: $\left( {\exists j \in \ell } \right)\left[ {x \in A_j } \right]$
I believe it's all I needed. Thanks again

And an intersection would be $\left( {\forall j \in \ell } \right)\left[ {x \in A_j } \right]$ right?