Hints to exercises
The power set of a set always includes the empty set as well as the whole set that we are forming the power set of. In this case they happen to coincide so \({\mathcal P}(\emptyset) = \{ \emptyset \}\). Notice that \(2^0 =1\).
I won’t spoil your fun, but as a check \({\mathcal P}({\mathcal P}(\emptyset))\) should have \(2\) elements, and \({\mathcal P}({\mathcal P}({\mathcal P}(\emptyset)))\) should have \(4\).
Three and one.
A contradiction.
The universe of discourse.
The set of all multiples of \(6\).
Many answers are possible. Perhaps the easiest is \(\exists y \in {\mathbb Z}, x = y^2\).
5, 10
\(\binom{16}{8} = 12870\)
\(n\)
\(\in\), \(\subseteq\), \(\in\), \(\subseteq\)
When \(p=2\) we have seen these sets. \(P_1\) is the even numbers, \(P_2\) is the doubly-even numbers, etc.
A subset is called proper if it is neither empty nor equal to the superset. If we are talking about finite sets then the proper subsets do indeed have fewer elements than the supersets. Among infinite sets it is possible to have proper subsets having the same number of elements as their superset, for example there are just as many even natural numbers as there are natural numbers all told.
Turn “logical negation” into “set complement” and reverse the direction of the inclusion.
The smallest example I can think of would be \(A=\emptyset\) and \(B=\{\emptyset\}\). You should come up with a different example.
It would probably be helpful to have precise definitions of the sets described in the problem.
The fourth powers are \[ F = \{x {\,:\,}\exists y \in {\mathbb Z}, x=y^4 \}. \]
The squares are \[ S = \{x {\,:\,}\exists z \in {\mathbb Z}, x=z^2 \}. \]
To show that one set is contained in another, we need to show that the first set’s membership criterion implies that of the second set.
- \(\{ b, \{1, 2\} \}\)
- \(\{1, 2, a, b, \{1, 2\} \}\)
- \(\{ 1, 2 \}\)
- \(\{ a \}\)
- \(\{ 1, 2, a \}\)
- This is just the ace of hearts.
- All of the hearts and the other three aces.
- These two cards are known as the one-eyed jacks.
- The king of hearts, a.k.a. the suicide king.
- \(\emptyset\)
- Eight cards: all four kings and all four aces.
You’re on your own for this one.
Here’s the first one (although I’m omiting justifications for each step).
\[\begin{align*} & x \in \overline{A\cap B} \\ \iff \; & {\lnot}(x \in A\cap B) \\ \iff \; & {\lnot}(x \in A \; \land \; x \in B) \\ \iff \; & {\lnot}(x \in A) \; \lor \; {\lnot}(x \in B) \\ \iff \; & x \in \overline{A} \; \lor \; x \in \overline{B} \\ \iff \; & x \in \overline{A} \cup \overline{B} \end{align*}\]To better understand what is going on, first figure out what the first three or four intervals actually are. \[\begin{align*} I_1 & = \underline{~~~~~~~~~~~~~~~~~~~~~~~} \\ I_2 & = \underline{~~~~~~~~~~~~~~~~~~~~~~~} \\ I_3 & = \underline{~~~~~~~~~~~~~~~~~~~~~~~} \\ I_4 & = \underline{~~~~~~~~~~~~~~~~~~~~~~~} \end{align*}\]
Any negative real number \(r\) will be in the intersection only if \(r \geq -1\). Certainly \(0\) is in the intersection since it is in each of the intervals. Are there any positive numbers in the intersection?
In order to be in the union a real number just needs to be in one of the intervals.
You’re on your own for this one.
One of the answers to the last two questions is \(\emptyset\) and the other is \(U\). Decide which is which.
You’re on your own for this one.
This exercise, as well as the previous one, is really just about converting set-theoretic statements into their logical equivalents, applying some rules of logic that we’ve already verified, and then returning to a set-theoretic version of things.
Try to model make use of the De Morgan’s laws in logic.
The definition of \(A \triangle B\) is \((A\setminus B) \cup (B\setminus A)\). The definition of \(X \setminus Y\) is \(X \cap \overline{Y}\). Restating things in terms of \(\cap\) and \(\cup\) (and complementation) should help. So your first few lines should be:
Suppose \(x \in A \triangle B\).
Then, by definition, \(x \in (A\setminus B) \cup (B\setminus A)\). So, \(x \in (A \cap \overline{B}) \cup (B \cap \overline{A})\). \[\vdots\]
The center region contains \(2\) and \(4\).
I found it easier to experiment by making my drawings on graph paper. I never did anage to draw the \(5\)-set Venn diagram with just rectangles…probably just a lack of persistence.
Of course, rectangles are rectilinear, so one could use the solution from the previous problem (if, unlike me, you were persistant enough to find it). Otherwise, start with the \(4\) set diagram made with rectangles and use your \(5\)th (rectilinear) curve to split each region into \(2\) — don’t forget to split the region on the outside too.
Fortunately the instructions don’t say to prove that rectilinear curves will always suffice, so we can be less rigorous. Try to argue as to why it will alway be possible to add one more rectilinear curve to an existing Venn diagram and split every region into two.
One might also argue that any continuous curve can be approximated using rectilinear curves. So if a Venn diagram can be constructed using continuous curves we can also get the job done with rectilinear curves.}
\((A \cap B \cap \overline{C}) \cup (A \cap \overline{B} \cap C)\)
It is \((A \cap \overline{B} \cap \overline{C}) \cup (\overline{A} \cap B \cap \overline{C}) \cup (\overline{A} \cap \overline{B} \cap C)\). Now find the disjunctive normal form of \(A \triangle (B \triangle C)\).
The statement “All men are mortal” would be interpreted on a Venn diagram by showing the set of “All men” as being entirely contained within the set of “mortal beings.” Socrates is an element of the inner set. Zeus, on the other hand, lies outside of the outer set.
Obviously we’ll need one of the \(4\)-set Venn diagrams.
After constructing Venn diagrams for both sets you should be able to see that there are \(4\) regions where they differ. One is \(A \cap B \cap \overline{C} \cap \overline{D}\). What are the other three?
In order to get started on this you’ll need to convert the conditionals into equivalent disjunctions. Recall that \(X \implies Y \; \equiv \; {\lnot}X \lor Y\).}
If it’s not a set then it doesn’t necessarily have to have the property that we can be sure whether an element is in it or not.