# 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

## 1 Answer

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 $\{1,2\}\cap\{3,4\}=\emptyset$).

Whenever you are writing down a proof, re-read it when you are done and see if actually proved what was asked, and if you are convinced by it (from the actual words wrote down, not your thoughts on it).

Can you try b) yourself?

Tip: don't stop using mathjax when typing up your question (you stopped for some reason when writing down your own work).