# $f[A\cup{B}]=f[A]\cup{f[B]}$ [duplicate]

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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 $x$...

## marked as duplicate by Nick Peterson, Brian Rushton, Olivier Bégassat, Dan Rust, AlexRJan 5 '14 at 18:12

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• Indeed, it's not the case : your second line is completely true. – D.L. Sep 17 '13 at 14:52
• You sort of misuse $x$ in that first line. "Let $x\in A\cup B$" means "Let $x$ be a particular element of $A\cup B$." Then the next sentence you write, $f[A\cup B]=f\{x\mid x\in A\text{ or } x\in B\}$, which actually is a different usage of $x$. – Thomas Andrews Sep 17 '13 at 14:55
• If you're worried about how $f$ interacts with the set builder notation, you can avoid it completely by proving $f(A\cup B)\subseteq f(A)\cup f(B)$ and $f(A)\cup f(B)\subseteq f(A\cup B)$. But as has already been said, your existing argument is correct. – mdp Sep 17 '13 at 14:57
• I was speaking of the sentence after the point. I don't understand the utility of the "let $x\in A\cup B$". – D.L. Sep 17 '13 at 14:59
• In hindsight I thought it was needed but I see what @ThomasAndrews is saying in that, mentioning that i'm letting $x\in{A\cup{B}}$ implies that I'll be using a particular $x$ for a proof...I will remove that. Thanks everyone! – Iceman Sep 17 '13 at 15:03

## 1 Answer

Your initial argument is correct.

You are correct that your second statement, the one followed with the question mark, is NOT true:

$$f\{x|x\in{A} \text{ or } x\in{B}\}\neq \{f(x)|f(x)\in{A} \text{ or } f(x)\in{B}\}$$

We are interested in the $x \in A$ or the $x \in B$. Very often times, we have $f: A \to X$, where $x \in A$, but $f(x) \notin A$, but rather $f(x) \in X$.

Perhaps you were thinking of the set $\{f(x) \mid f(x) \in f[A] \text { or } f(x) \in f[B]\}$

• so the $f$ can distribute so long as its over both the elements of the sets and the sets themselves. That was another thought i had... – Iceman Sep 17 '13 at 15:36