1.9: Hints to get started on some exercises

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Bob Dumas and John E. McCarthy
University of Washington and Washington University in St. Louis

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Exercise 1.2. You could do this with a Venn diagram. However, once there are more than three sets (see Exercise 1.13), this approach will be difficult. An algebraic proof will generalize more easily, so try to find one here. Argue for the two inclusions \[\begin{aligned} (X \cup Y)^{c} & \subseteq X^{c} \cap Y^{c} \\ X^{c} \cap Y^{c} & \subseteq(X \cup Y)^{c} \end{aligned}\] separately. In the first one, for example, assume that \(x \in(X \cup Y)^{c}\) and show that it must be in both \(X^{c}\) and \(Y^{c}\).

Exercise 1.13. Part of the problem here is notation - what if you have more sets than letters? Start with a finite number of sets contained in \(U\), and call them \(X_{1}, \ldots, X_{n}\). What do you think the complement of their union is? Prove it as you did when \(n=2\) in Exercise 1.2. (See the advantage of having a proof in Exercise \(1.2\) that did not use Venn diagrams? One of the reasons mathematicians like to have multiple proofs of the same theorem is that each proof is likely to generalize in a different way). Can you make the same argument work if your sets are indexed by some infinite index set?

Now do the same thing with the complement of the intersection.

Exercise 1.14. Again there is a notational problem, but while \(Y\) and \(Z\) play the same rôle in Exercise 1.3, \(X\) plays a different rôle. So rewrite the equations as \[\begin{aligned} &X \cap\left(Y_{1} \cup Y_{2}\right)=\left(X \cap Y_{1}\right) \cup\left(X \cap Y_{2}\right) \\ &X \cup\left(Y_{1} \cap Y_{2}\right)=\left(X \cup Y_{1}\right) \cap\left(X \cup Y_{2}\right), \end{aligned}\] and see if you can generalize these.

Exercise 1.35. (i) Again, this reduces to proving two containments. If \(y\) is in the left-hand side, then there must be some \(x_{0}\) in some \(U_{\alpha_{0}}\) such that \(f(x)=y\). But then \(y\) is in \(f\left(U_{\alpha_{0}}\right)\), so \(y\) is in the right-hand side.

Conversely, if \(y\) is in the right-hand side, then it must be in \(f\left(U_{\alpha_{0}}\right)\) for some \(\alpha_{0} \in A\). But then \(y\) is in \(f\left(\cup_{\alpha \in A} U_{\alpha}\right)\), and so is in the left-hand side.