# Unions and Functions on Sets

Given these conditions, I seek a proof.

Let $f: A \rightarrow B$ be a function, and let $X$ and $Y$ be subsets of $A$.

Prove that $f(X \cup Y) = f(X) \cup f(Y)$.

I can't seem to figure it out. It appears obvious, but materializing a proof is troubling me. What is the best method of proof for this?

• It is a very strange statement. Consider: $X=\{ab\}, Y=\{bc\}$ and let $f$ be constructed as follows: $f: abc\rightarrow c$, $ab\rightarrow a$, $bc\rightarrow b$. In this case $f(X\cup Y) = c$, while $f(X)\cup f(Y)=ab$. – Andremoniy Oct 29 '15 at 11:19
• In your example as it is, $f$ is not a proper function. For something close to that, let $S=\{a,b,c\}$ and $f: P(S) \rightarrow S$ such that $f(\{a, b, c\}) = c$, $f(\{a, b\}) = a,$ and $f(\{b, c\}) = b$. Then if $X=\{\{a, b\}\}$ and $Y=\{\{b, c\}\}$, $f(X) = a$ and $f(Y)=b$ indeed. However, $X \cup Y = \{\{a, b\}, \{b, c\}\}$, not $\{a, b, c\},$ and in fact $f(X \cup Y) = \{a, b\}$. Note that $X$ and $Y$ are elements of the powerset of $S$ and not of $S$ itself as per definition of $f$. If we had $f: S \rightarrow S$ instead, then $\{a,b,c\} \notin S$ and $f(\{a,b,c\})$ would not be defined. – AlCorreia Mar 17 at 9:14

I suppose the best way to do this is to show the two inclusions (namely $f(X\cup Y)\subseteq f(X)\cup f(Y)$ and $f(X)\cup f(Y)\subseteq f(X\cup Y)$), and to use definitions. For the first inclusion, this gives :

Let $x\in f(X\cup Y)$. This means that there exists a $y\in X\cup Y$ such that $f(y)=x$. Now $y\in X\cup Y$ means that either $y\in X$ or $y\in Y$. Now if $y\in X$, then $x\in f(X)$, and if $y\in Y$, then $x\in f(Y)$. In any instance, $x\in f(X)\cup f(Y)$. We proved that any $x$ in $f(X\cup Y)$ is also in $f(X)\cup f(Y)$, that is precisely $f(X\cup Y)\subseteq f(X)\cup f(Y)$. The other inclusions proceeds similarly.

• $\LaTeX$ tip: use \subset or \subseteq for set inclusion, not \in. We want to show $f(X\cup Y)\subseteq f(X)\cup f(Y)$, not $f(X\cup Y)\in f(X)\cup f(Y)$ (which could be true, but is unlikely) – Arturo Magidin Mar 23 '12 at 15:17
• @Dimitri: It would have been good for the OP to make the correction himself, given that he's new and this is his first post. – Arturo Magidin Mar 23 '12 at 15:33
• @ArturoMagidin, I could revert it, if you'd like. – sxd Mar 23 '12 at 15:43
• @Dimitri: that's okay; just keep it in mind for next time. – Arturo Magidin Mar 23 '12 at 15:44
• @Arturo: Sorry, I don't know what crossed my mind at the time. Thank you Dimitri for the edit. – Samuel T Mar 26 '12 at 11:14

The simplest method two prove two sets are equal is to show that each one is contained in the other.

The simplest method to show that one set is contained in the other is to show that any element in the one set is also an element in the other.

Here you will want to show that $$f(X\cup Y)\subseteq f(X)\cup f(Y)\quad\text{and}\quad f(X)\cup f(Y)\subseteq f(X\cup Y)$$ both hold.

For example, to show that $f(X\cup Y)\subseteq f(X)\cup f(Y)$, let $b\in f(X\cup Y)$. We need to show that $b\in f(X)\cup f(Y)$; that is, we need to show that either $b\in f(X)$, or $b\in f(Y)$.

Since $b\in f(X\cup Y) = \{f(a)\mid a\in X\cup Y\}$, there exists $a\in X\cup Y$ such that $b=f(a)$. Since $a\in X\cup Y$, either $a\in X$ or $a\in Y$. If $a\in X$, then $b=f(a)\in f(X)\subseteq f(X)\cup f(Y)$, and we are done. If $a\in Y$, then $b=f(a)\in f(Y)\subseteq f(X)\cup f(Y)$, and we are done. Since these are the only cases, we see that if $b\in f(X\cup Y)$, then $b\in f(X)\cup f(Y)$. This proves the first inclusion.

I'll let you work out the second inclusion.

To get started, write out the definitions of $f(X\cup Y)$, $f(X)$, and $f(Y)$: $$f(X\cup Y)=\{b\in B\mid \text{there exists an }a\in X\cup Y\text{ such that }b=f(a)\}$$ $$f(X)=\{b\in B\mid \text{there exists an }a\in X\text{ such that }b=f(a)\}$$ $$f(Y)=\{b\in B\mid \text{there exists an }a\in Y\text{ such that }b=f(a)\}$$ Also, remember that for two sets $C$ and $D$, $$z\in C\cup D\iff z\in C\;\text{ or }\,z\in D.$$ Do you see how to go the rest of the way?