# How do I prove the following: $f(S\cup T) = f(S) \cup f(T)$ [duplicate]

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I have a question.

How do I prove the following identity? $$f(S\cup T) = f(S) \cup f(T)$$

## marked as duplicate by Joel Reyes Noche, user147263, Hanul Jeon, Paul, Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 3 '14 at 7:11

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• Prove both inclusions. Go step by step. If $y$ belongs to the lhs, there is $x$ in $S\cup T$ such that $y= f(x)$. If $x$ belongs to $S$ in the first place, then $y=f(x)$ belongs to $f(S)\subseteq f(S)\cup f(T)$. Etc... – Julien Mar 23 '13 at 12:09
• it's not clear. Are you saying that given two sets s and t, the set of values of f on the union of s and t is the same as the union of the set of values of f(s) and f(t)? – mau Mar 23 '13 at 12:10
• 1. please use latex for math notation. 2. what are your own thoughts? what is your definition of them being equal? – akkkk Mar 23 '13 at 12:10
• Please notice for further reference the $\LaTeX$ editing that has been made to your post. – Andreas Caranti Mar 23 '13 at 12:13

## 4 Answers

Element chasing is a promising method here.

$y\in f(s\cup t)$ if and only if there is some $x\in s\cup t$ such that $f(x)=y$. If $x\in s$ then $y\in f(s)$, if $x\in t$ then $y\in f(t)$. Therefore $y\in f(s)\cup f(t)$.

I leave the second inclusion to you.

• Apparently something is wrong in the argument above? – Asaf Karagila Mar 23 '13 at 13:56

You know that if $R$ is a relation an d $A$ is a set then $$R(A)=\{y\mid\exists x(x\in A\wedge xRy)\}$$ now try to show that $$R(A\cup B)=R(A)\cup R(B)$$

Begin your argument as follows. Can you move it further? $$y\in f(s\cup t) \iff \exists x\in s\cup t \hbox{s.t.} f(x) = y$$

You don't prove a function, but an identity which states that $$f(S \cup T) = f(S) \cup f(T)$$ When $x\in f(S\cup T)$ it means there is a $y \in S \cup T$ such that $f(y)=x$ make the same on the right hand side and compare both.

• thanks first wanted to write $y\in S \vee y \in T$ but then i messed up – Dominic Michaelis Mar 23 '13 at 12:14
• Hi Dominic. I've noticed you often use the verb "to proof". It exists, but it does not mean "to prove". So I thought I'd let you know. I hope you don't mind. – Julien Mar 23 '13 at 12:16
• @julien oh thanks i hope i won't forget it – Dominic Michaelis Mar 23 '13 at 12:24
• @julien: Perhaps Dominic was pointing out that we are proofing the proof against mistakes? :-) – Asaf Karagila Mar 23 '13 at 12:25
• @AsafKaragila He, he! That was funny. And indeed a possibility. Although it's been ruled out by Dominic's comment above. – Julien Mar 23 '13 at 12:32