Prove $f(A \cup B) = f(A) \cup f(B)$ where $f: X \rightarrow Y$ Can anyone verify my proof of the above equality?
Define 
\begin{align*}
f(A) &= \left\{ f(x) : x \in A \right\} \\
f(B) &= \left\{ f(x) : x \in B \right\} 
\end{align*}
Then
\begin{align*}
f(A) \cup f(B) &= \left\{ f(x) : x \in A \lor B \right\} \\
f(A) \cup f(B) &= \left\{ f(x) : x \in A \cup B \right\} \\
\end{align*}
Since $A \cup B \in X$,
\begin{align*}
f(A) \cup f(B) = f(A \cup B)
\end{align*}
 A: It looks like you have the right idea but your notation could use a wee bit of help:


*

*When you say $x \in A \lor B$ this should really be $x \in A \lor x \in B$. At least in my opinion $A \lor B$ means nothing (nor does $(x \in A) \lor B$), but of course a true logician can correct me here.

*When you say $A \cup B \in X$ this should really be $A \cup B \subset X$. The first statement says that $A \cup B$ is an element of $X$, but from how you used $A,B$ this is not true.


I would write out the proof in the following way:

Claim: Let $f : X \to Y$ and $A, B \subset X$ then $f(A) \cup f(B) = f(A \cup B)$
Proof:
  $$
\begin{eqnarray}
f(A) \cup f(B) & = & \{f(x) : x \in A \} \cup \{f(x) : x \in B\} \\
& = & \{f(x) : x \in A \lor x \in B\} \\
& = & \{f(x) : x \in A \cup B\} \\
& = & f(A \cup B)
\end{eqnarray}
$$

Hope this helps!

Of course if one is persuaded by the proof above they can do double inclusion; that is, prove that $f(A) \cup f(B) \subset f(A \cup B)$ and $f(A) \cup f(B) \supset f(A \cup B)$.
Really this depends on how rigorous of a proof you (or your teacher) is looking for. If this is a more intense set theory class then you may be required to write out a few more steps between the first and second lines above.

In general I personally prefer the double inclusion route (it may take more writing but it tends to be a bit easier / cleaner from my experience). Here is that form of the proof:

Let $y \in f(A) \cup f(B)$ then $y \in f(A) \lor y \in f(B) \implies \exists x \in A : f(x) = y \lor \exists x \in B : f(x) = y$ so $\exists x \in A \lor \exists x \in B : f(x) = y \implies \exists x \in A \cup B : f(x) = y \implies y \in f(A \cup B)$. 
Now let $y \in f(A \cup B)$ then $\exists x \in A \lor \exists x \in B : f(x) = y$ so $\exists x \in A : f(x) = y \lor \exists x \in B : f(x) = y \implies y \in f(A) \lor y \in f(B) \implies y \in f(A) \cup f(B)$.

A: You start with the definition of $f(A)$ and $f(B)$, which is good (I would remove the word “Define,” because it is not you who is defining anything here, though.)
And you have as your conclusion the correct statement. However, the way you get from the beginning to the end needs improvement.
Your equation $$f(A) \cup f(B) = \left\{ f(x) : x \in A \lor B \right\}$$ isn’t a clear statement. The expression $x\in A\lor B$ doesn’t have a meaning, because $\lor$ only makes sense when it’s between two true-false statements. ($B$ is not a true-false statement.) What you have to do to fix this depends on what definitions you can use, but the likely first step in looking at $f(A)\cup f(B)$ is to use the definition of $\cup$, which is not what you’ve done. An example of a correct statement about $f(A) \cup f(B)$ using $\lor$ is this one:
$$f(A) \cup f(B) = \left\{ y\in Y : y \in f(A) \lor y \in f(B) \right\}.$$
Also, you say that the final result is true “since $A\cup B\in X$,” but you don’t have any reason to know that $A\cup B\in X$ (did you mean $A\cup B\subseteq X$?), nor would it (with either $\in$ or $\subseteq$ lead to your conclusion.
You haven’t looked at the set $f(A\cup B)$ at all, and that would be useful, since you’re trying to show it equals something.
The original question is to show that one set equals another set. What do you need to do to show that? You can either show each is equal to the same set, or you can show that each is a subset of the other (and you need to know exactly what subset means to do that).
