Equality of two sets (using Boolean algebra)

**Pranas** · April 2nd 2011, 11:10 AM

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.

**Plato** · April 2nd 2011, 11:19 AM

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: $\left( {\exists j \in \ell } \right)\left[ {x \in A_j } \right]$

**Pranas** · April 2nd 2011, 11:25 AM

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?

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