I need to prove this and I'm pretty sure this is straight forward.

$f[A\cup{B}]=f\{x|x\in{A} \text{ or } x\in{B}\}=\{f(x)|x\in{A} \text{ or } x\in{B}\}=f[A]\cup{f[B]}$

I'm writing this because of a slight confusion with the notation. should the $f$ distribute over the entire set builder notation? In other words, $f\{x|x\in{A} \text{ or } x\in{B}\}=\{f(x)|f(x)\in{A} \text{ or } f(x)\in{B}\}$? I don't think this is the case since the right hand side are simply the conditions imposed upon $x$...

I need to prove this and I'm pretty sure this is straight forward.

$$f[A\cup{B}]=f\{x|x\in{A} \text{ or } x\in{B}\}=\{f(x)|x\in{A} \text{ or } x\in{B}\}=f[A]\cup{f[B]}$$

I'm writing this because of a slight confusion with the notation. should the $f$ distribute over the entire set builder notation? In other words, $$f\{x|x\in{A} \text{ or } x\in{B}\}=\{f(x)|f(x)\in{A} \text{ or } f(x)\in{B}\}\;?$$ I don't think this is the case since the right hand side are simply the conditions imposed upon $x$...

I need to prove this and I'm pretty sure this is straight forward.

let $x\in{A\cup{B}}$. Then $f[A\cup{B}]=f\{x|x\in{A} \text{ or } x\in{B}\}=\{f(x)|x\in{A} \text{ or } x\in{B}\}=f[A]\cup{f[B]}$

I'm writing this because of a slight confusion with the notation. should the $f$ distribute over the entire set builder notation? In other words, $f\{x|x\in{A} \text{ or } x\in{B}\}=\{f(x)|f(x)\in{A} \text{ or } f(x)\in{B}\}$? I don't think this is the case since the right hand side are simply the conditions imposed upon $x$...

I need to prove this and I'm pretty sure this is straight forward.

$f[A\cup{B}]=f\{x|x\in{A} \text{ or } x\in{B}\}=\{f(x)|x\in{A} \text{ or } x\in{B}\}=f[A]\cup{f[B]}$

I'm writing this because of a slight confusion with the notation. should the $f$ distribute over the entire set builder notation? In other words, $f\{x|x\in{A} \text{ or } x\in{B}\}=\{f(x)|f(x)\in{A} \text{ or } f(x)\in{B}\}$? I don't think this is the case since the right hand side are simply the conditions imposed upon $x$...