Proving that $f(W\cap X) \subseteq f(W) \cap f(X)$

I am trying to write the proofs and/or counterexamples to these problems, but I'm not sure if my proofs are right; I have trouble trying to write proofs.

Theorem $5.4.2.$ Suppose $f: A \rightarrow B$ and $W$ and $X$ are subsets of $A$. Then $f (W \cap X) \subseteq f(W) \cap f(X)$. Furthermore, if $f$ is one-to-one, then $f (W \cap X)= f(W) \cap f(X)$.

Now, here are some questions, in each case, justify your answers with proofs and counterexamples.

Suppose $f: A \rightarrow B$

1) Suppose $W$ and $X$ are subsets of $A$

a)Will it always be true that $f (W \cup X)= f(W) \cup f(X)$?

This is what I have..

Let $a\in f(W\cup X)$, then there exists some $y\in W \cup X$ such that $f(y)=a$. Suppose that $y\in W$, then $a=f(y)\in f(W)$. Suppose that $y\in X$, then $a=f(y)\in f(X)$. Take $a\in f(W) \cup f(X)$. Then $a\in f(W)$, that is there exists some $y\in f(W)$ such that $a=f(y)$. But $y\in W \rightarrow y\in W \cup X$. Thus, $a\in f(W \cup X)$. Therefore $f(W \cup X) = f(W) \cup f(X)$.

b)Will it always be true that $f (W \setminus X)= f(W) \setminus f(X)$?

c)Will it always be true that $W \subseteq X \iff f(W)\subseteq f(x)$?

This is what I have...

Let $W \subseteq X$. Let $t\in F(W)$, then $t=f(Z)$. Therefore $Z\in W \subset X$. If $t = f(Z)$ then $Z\in X$ which means $t\in F(X)$. Hence $f(W) \subset f(x)$.

I'm really unsure if what I have done is correct, but if anybody could help/explain, that would be much appreciated.

• Why this edit to the title? That's not what the OP is trying to prove, right? – B. Pasternak May 16 '16 at 9:10

For a), the inclusion $f(W\cup X)\subset f(W)\cup f(X)$ is allright, I would just conclude with, "...and thus $a\in f(W)\cup f(X)$, so $f(W\cup X)\subset f(W)\cup f(X)$ holds. The other inclusion is not quite done. Take $a\in f(W)\cup f(X)$, and do the same thing as before. Then (non-exclusive or) $a\in f(W)$ or $a\in f(X)$; if $a\in f(W)$ then (as you say) there exists $y\in W$ such that $a=f(y)$, and $y\in W\cup X$, so $a\in f(W\cup X)$. The case $a\in f(X)$ goes of course exactly in the same way, but you have to mention it, else your proof is not complete.
It's really unclear what you write for c). Suppose $W\subset X$, then for all $w\in W$ it holds that $w\in X$. Let $a\in f(W)$, then there exists $w\in W$ such that $a=f(w)$; as $w\in X$, $a=f(w)\in f(X)$, so $f(W)\subset f(X)$. Conversely, suppose $f(W)\subset f(X)$. Let $a\in f(W)$, then $a\in f(X)$ and there exists $w\in W$ such that $a=f(w)$. Now if $f$ is injective, then $w\in W\subset A$ is the unique element such that $f(w)=a$, so necessarily $w\in X$ and thus $W\subset X$. If $f$ is however not injective, then there might be some other element $x'\in A$ such that $a=f(x')$, and it needn't hold that $x'\in W$ or $x'\in X$, so in fact the other implication only holds when $f$ is injective (as a counterexample consider the function $f:\{1,2,3,4\}\to\{5\}$ sending every element to $5$, then $f(\{1,2\})=f(\{3,4\})$ but $\{1,2\}\cap\{3,4\}=\emptyset$).