Proof verification: $f(S\cup T) = f(S) \cup f(T)$, where $S$ and $T$ are subsets of the set $A$ and $f$ is a function from $A$ to $B$.

Question

Problem: Let $f$ be a function from set $A$ to set $B$. Let $S$, $T$ be subsets of $A$.
Show: $f( S\cup T ) = f(S)\cup f(T)$

My proof:
$f( S\cup T ) \subseteq f(S) \cup f(T)$

Let $b \in f( S\cup T )$
$\rightarrow$ $\exists a \in S\cup T$, such that $f(a) = b$,
therefore,
case 1: $a \in S \rightarrow f(a) \in f(S) \rightarrow b \in f(S)$
or
case 2: $a \in T \rightarrow f(a) \in f(S) \rightarrow b \in f(T)$
case 1 & case 2 $\rightarrow b \in f(S) \cup f(T)$

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

Let $b \in f(S) \cup f(T)$
$\rightarrow b \in f(S) \rightarrow \exists a_1 \in S$ such that $f(a_1) = b \rightarrow a_1 \in S\cup T$
or
$\rightarrow b \in f(T) \rightarrow \exists a_2 \in T$ such that $f(a_2) = b \rightarrow a_2 \in S\cup T$
$\rightarrow b \in f(S\cup T)$

I am pretty sure about the first half, but I also think the second half is alright. Anyways, thanks in advance.

Answer 1

Your proof fine, but I'd write it a bit differently (avoiding all those arrows).

1. If $X\subseteq Y\subseteq A$, then $f(X)\subseteq f(Y)$ (easy proof).

2. Since $S\subseteq S\cup T$ and $T\subseteq S\cup T$, we have $f(S)\cup f(T)\subseteq f(S\cup T)$ (by 1).

3. Suppose $b\in f(S\cup T)$, then there exists $a\in S\cup T$ with $f(a)=b$. In the case $a\in S$, we have $b=f(a)\in f(S)$; in the case $a\in T$, we have $b=f(a)\in f(T)$. Therefore $b\in f(S)\cup f(T)$.