how to prove $pr_i(\alpha \setminus \beta) \supseteq pr_i\alpha \setminus pr_i\beta$

Question

For those who are not familiar with the syntax

$pr_i \alpha = \{ pr_i(a,b) / a \alpha b \} \text{ for }\alpha \subseteq A \times B$

which is same as $\begin{cases} (x= pr_1 \alpha) \Leftrightarrow \exists b(b \in B)(x \alpha b)\\ (y= pr_2 \alpha) \Leftrightarrow \exists a(a \in A)(a \alpha y) \end{cases}$

I coundn't prove the statement because I'm not familiar with such proof writing, I can only provide a counterexample.

$A=B=\{ a,b,c,d,e \}$

$\alpha = \{ (a,b) , (c,d) , (a,e) \}$, $\beta = \{ (c,d), (b,c), (e,a) (a,d)\}$

$pr_1 (\alpha \setminus \beta) = pr_1 \{ (a,b), (a,e) \} = a$

$pr_1 (\alpha) \setminus pr_1 (\beta) = \{ a,c \} \setminus \{ a,b,c,e \} = \varnothing $

Note that the proof should be provided for both $i=1$ and $i=2$, Thank you.

Answer 1

Let’s look at the first projection: you want to show that $\operatorname{pr}_1(\alpha\setminus\beta)\supseteq\operatorname{pr}_1\alpha\setminus\operatorname{pr}_1\beta$.

Suppose that $a\in\operatorname{pr}_1\alpha\setminus\operatorname{pr}_1\beta$; you need to show that $a\in\operatorname{pr}_1(\alpha\setminus\beta)$. What does it mean for $a$ to be an element of the set $\operatorname{pr}_1\alpha\setminus\operatorname{pr}_1\beta$? It means that $a\in\operatorname{pr}_1\alpha$, but $a\notin\operatorname{pr}_1\beta$. Break it down a step further: what does it mean for $a$ to be an element of $\operatorname{pr}_1\alpha$? It means that there is some $b_0\in B$ such that $a\mathrel{\alpha}b_0$, i.e., such that $\langle a,b_0\rangle\in\alpha$. On the other hand, $a\notin\operatorname{pr}_1\beta$, so no matter what $b\in B$ you choose, $a\not\mathrel{\beta}b$, i.e., $\langle a,b\rangle\notin\beta$.

Now let’s look at what we want: $a$ is an element of $\operatorname{pr}_1(\alpha\setminus\beta)$ if and only if there is some $b\in B$ such that $a\mathrel{(\alpha\setminus\beta)}b$, i.e., such that $\langle a,b\rangle\in\alpha\setminus\beta$, which in turn means that $\langle a,b\rangle\in\alpha$, but $\langle a,b\rangle\notin\beta$. Do we know of such a $b$? Yes: the $b_0$ that we found above works, as you should now verify.

The argument for the second projection map, $\operatorname{pr}_2$, is very similar; you should try to carry it out on your own with the argument for $\operatorname{pr}_1$ as a model.