Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f(S \cup T) = f(S) \cup f(T)$

f(S) encompasses all x that is in S f(T) encompasses all x that is in T

thus the domain being the same, both the LHS and RHS map to the same ys, since the function f(x) is the same for both.

Can you post the solution?

share|improve this question
1  
You have correct intuition about why the result is true. The language you use is a bit fuzzy. At this stage of the game, I suggest you divide the argument into two parts: (1) if $y\in f(S\cup T)$ then $y\in f(S) \cup f(T)$; (2) the other direction. –  André Nicolas Nov 6 '12 at 3:17

3 Answers 3

up vote 2 down vote accepted

$$y\in f(S\cup T)\Longrightarrow \exists\,x\in S\cup T\,\,s.t.\,\,f(x)=y$$

and now:

$$x\in S\Longrightarrow\,y=f(x)\in f(S)\;\;;\;\;x\in T\Longrightarrow\,y=f(x)\in T$$

so that anyway $\,y=f(x)\in f(S)\cup f(T)\,\Longrightarrow f(S\cup T)\subset f(S)\cup f(T)$

Now you try to do the other way around: $\,f(S)\cup f(T)\subset f(S\cup T)$

share|improve this answer
    
this one is easier to understand –  George Milton Nov 6 '12 at 20:15

Let $x\in f(S\cup T)$. Then there is a $y\in S\cup T$ such that $f(y) = x$. Assume without loss of generality that $y\in S$. Then $x = f(y)\in f(S) \subseteq f(S)\cup f(T)$. Hence you have proved on of the directions of your inclusion.

For the other one you do similarly. Hence start with $x\in f(S)\cup f(T)$. Say that $x\in f(S)$. Then there is a $y\in S \subseteq S\cup T$ ... (you can probably finish the argument).

share|improve this answer
    
I don't really understand honestly –  George Milton Nov 6 '12 at 20:13

Here is how I would do this, with a slightly different notation to prevent confusion: start with the most complex side $\;f[S] \cup f[T]\;$, and determine which elements $\;y\;$ are in that set by expanding the definitions and simplifying using predicate logic.

So we calculate, for every $\;y\;$: \begin{align} & y \in f[S] \cup f[T] \\ \equiv & \;\;\;\;\;\text{"definition of $\;\cup\;$"} \\ & y \in f[S] \;\lor\; y \in f[T] \\ \equiv & \;\;\;\;\;\text{"definition of $\;\cdot[\cdot]\;$, twice"} \\ & \langle \exists x :: x \in S \land f(x) = y \rangle \;\lor\; \langle \exists x :: x \in T \land f(x) = y \rangle \\ \equiv & \;\;\;\;\;\text{"logic: simplify, using the fact that $\;\lor\;$ distributes over $\;\exists\;$"} \\ & \langle \exists x :: (x \in S \land f(x) = y) \;\lor\; (x \in T \land f(x) = y) \rangle \\ \equiv & \;\;\;\;\;\text{"logic: simplify by extracting common conjunct"} \\ & \langle \exists x :: (x \in S \;\lor\; x \in T) \land f(x) = y \rangle \\ \equiv & \;\;\;\;\;\text{"reintroduce $\;\cup\;$ using its definition"} \\ & \langle \exists x :: x \in S \cup T \land f(x) = y \rangle \\ \equiv & \;\;\;\;\;\text{"reintroduce $\;\cdot[\cdot]\;$ using its definition"} \\ & y \in f[S \cup T] \\ \end{align}

By set extensionality, this proves the statement.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.