# Notes I: Sets and Functions: Solutions

Problem 1:

If $\{ A_{i}\}$ and $\{ B_{j}\}$ are two classes of sets such that $\{ A_{i}\} \subseteq \{ B_{j}\}$, prove that $\bigcup_{i}A_{i} \subseteq \bigcup_{i}B_{i}$ and $\bigcap_{j}B_{j} \subseteq \bigcap_{i}A_{i}$.

The first part is clear as the hypothesis says the set of sets $\{ A_{i}\}$ is a subset of the set of sets $\{ B_{j}\}$. So, quite clearly $\bigcup_{i}A_{i} \subseteq \bigcup_{j}A_{j}$. For the second part, we can use the generalized De Morgan’s laws. (Note: if you try to derive it straight, you will prove exactly what is opposite of the required result :-)). For the second part note that $X \subseteq Y \Longleftrightarrow Y^{'} \bigcap X^{'}$. Using the result in the first part, that is, consider $\bigcup_{i}A_{i} \subseteq \bigcup_{j}A_{j}$ and now take the complement of both sides so we get: $\bigcup_{i}A_{i} \subseteq \bigcup_{j}B_{j} \Longleftrightarrow (\bigcup_{j}B_{j})^{'} \subseteq (\bigcup_{i}A_{i})^{'}$, that is, $\bigcap_{j}B_{j} \subseteq \bigcap_{i}A_{i}$. QED.

Problem 2:

The difference between two sets A and B, denoted by $A-B$, is the set of all elements in A and not in B, thus $A-B = A \bigcap B^{'}$. Show the following:

(a)$A-B = A - (A \bigcap B) = (A \bigcup B) -B$.

Answer (a): $A-B$ is the set of elements $x \in A$, but not in B. Hence, $A-B = A-(A \bigcap B)$, simply by definition. Again, if we consider elements in A or B or both, and then take away elements of B, we are just left with elements of A only, but not of B. That is, $A-B = A-(A \bigcap B) = (A \bigcup B) - B$. QED. ps: A venn diagram may help visualize v well.

(b) TPT: $(A-B) - C = A - (B \bigcup C)$ (PS: a Venn diagram can help visualize and perhaps, guide the writing of the proof also).

Let $x \in A-B$, but $x \notin C$. Then, $x \in A$, but $x \notin B$, or $x \notin C$, or not in both B and C. That is, $x \in A-(B \bigcup C)$. By reversing the argument, we get the other subset relationship. Hence, $(A-B)-C = A- (B \bigcap C)$. QED.

(c) TPT: $A - (B-C) = (A-B) \bigcup (A \bigcap C)$.

Answer (c): Let $x \in A-(B-C)$. That is, $x \in A$ and $x \notin (B-C)$. That is, $x \in A, x \notin B, x \in C$. That is, $x \in A, x \notin B$, or x could be both in A and C. That is, $x \in (A-B) \bigcup (A \bigcap C)$. Simply reversing the arguments, we get the reverse subset relation. Hence, the two sets are equal. Hence, $A-(B-C) = (A-B) \bigcup (A \bigcap C)$. QED.

(d) TPT: $(A \bigcup B) - C = (A-C) \bigcup (B-C)$.

answer (d): let $x \in A\bigcup B$, but $x \notin C$. Then, $x \in A$, or $x \in B$, or x is an element of both A and B, but x is not an element of C. Which clearly means, x is an element of A but not C, OR x is an element of B but not C. That is, $x \in (A-C) \bigcup (B-C)$ (upon reversing the arguments, we get the other subset relations. Hence, the two sets are equal). QED.

problem (e): TPT: $A-(B \bigcup C) = (A-B) \bigcap (A-C)$ (once again, note that for set theoretic relations with up to three sets, venn diagrams are helpful to visualize and guide the proofs..)

answer (c): let $x \in A$, but not in $B \bigcup C$. That is, x is an element of A, but not of B, not of C, or not both of B and C. That is, x is an element of A not of B, and x is an element of A not of C. That is, $x \in (A-B) \bigcap (A-C)$. Reversing the arguments, we get the other subset relationship. Hence, the two sets are equal. Hence, $A - (B \bigcup C) = (A-B) \bigcap (B-C)$. QED.

Problem 3:

The symmetric difference of two sets A and B, denoted by $A \triangle B$, is defined by $A \triangle B = (A-B) \bigcup (B-A)$; it is the union of the difference of two sets in opposite orders. Prove the following:

3(a) Symmetric difference is associative: $A \triangle (B \triangle C) = (A \triangle B) \triangle C$.

3(b): TPT: $A \triangle \phi = A$; and $A \triangle A = \phi$. (this is some sort of ‘existence of additive inverse of ‘in a symmetric relation’).

Answer 3b: both results are obvious from definition.

3(c): TPT: Symmetric difference is commutative operation: $A \triangle B = B\triangle A$.

answer 3c: By definition, LHS is $(A-B)\bigcup (B-A) = (B-A) \bigcup (A-B)$, which is RHS. Hence, done.

3(d): TPT: Some sort of distributive law: $A \bigcap (B \triangle C) = (A \bigcap B) \triangle (A \bigcap C)$.

answer 3d: Let $x \in A$, and also simultaneously $x \in (B \triangle C)$. That is, $x \in A$, and $x \notin (B \bigcap C)$. That is, $x \in A, x \notin (A \bigcap B)\bigcap (B \bigcap C)$. That is, $x \in RHS$. Reversing the arguments, we get the other subset relation. Hence, the two sets are equal. Hence, we can say “intersection distributes over symmetric difference.” QED.

Problem 4:

A ring of sets is a non-empty class A of sets such that if A and B are in A then $A \triangle B$ is in A and $A \bigcap B$ is also in A. Show that A must contain the empty set, $A \bigcup B$ and $A-B$. Also, show that if a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets. Also, prove that a Boolean algebra of sets is a ring of sets.

Answer 4(i): TPT: A must also contain the empty set, the set $A \bigcup B$ and the set $A-B$. As $A \triangle B$ is in A, so also $A \triangle A = \phi \in A$; now, it is already given that the ring of sets is non-empty, so it contains a non empty universal set also, call it U. We know from problem 2 in this blog that $A - B = A \bigcap B^{'}$. As U exists, so does $B^{'}$ by definition and so $A-B$ is in A. Also, as $A-B$ exists, $A-B^{'}$ exists in A, but $A-B^{'}=A\bigcap B$ also exists in A. QED.

Answer 4(ii): TPT: If a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets.

Let A be any non-empty class of sets. Clearly, it therefore has a non empty universal set. As $\phi \in U$, so also $\phi$ is contained in A. Hence, the complement of a set A is also in A. Now, $A \bigcup B \in A$, so $U - (A \bigcup B)$ is in A, that is, by De Morgan law, $B^{'} \bigcap A^{'}$ is in A. That is, again, $A \bigcap B$ is in A. Now, since $A-B$ is in A, hence, $B-A$ is in A, and since $A \bigcup B$ is in A, then $(A-B) \bigcup (B-A)$ is in A also. Hence, A is a ring of sets.

4(iii) TPT: Show that a boolean algebra of sets is a ring of sets.

Answer 4(iii); Firstly, as a boolean algebra of sets is non-empty, and it also contains the empty set and the Universal set. As A is in A implies that $A^{'}$ is in A, so $A \bigcup B^{'}$ is also in A (by definition of Boolean algebra), but this is precisely that $A-B$ is in A. Now, as $A \triangle B = (A-B) \bigcup (B-A)$ clearly is also in A. Hence, a Boolean algebra of sets is a ring of sets.

Problem 5:

Show that the class of all finite subsets (including the empty set) of an infinite set is a ring of sets but is not a Boolean algebra.

In problem 4 above, we showed that if a non empty class of sets contains the union and difference of any pair of the sets, then it is a ring of sets. The given class is a subclass of all an infinite set such that it contains finite subsets only; that is, again, we know that finite unions of finite subsets and finite intersections of finite subsets also lie in that subclass, but the complements might be infinite sets so that it need not always be a Boolean algebra of sets.

Problem 6:

Show that the class of all finite unions of closed-open intervals on the real line is a ring of sets but is not a Boolean algebra of sets.

Same reasoning as above tells us that it is indeed a ring of sets; but consider for example the closed-open interval $[a,b)$, the complement of this is not in that class as it is an infinite interval so that it is not a Boolean algebra of sets.

Problem 7:

Assuming that the Universal set U is non empty, show that Boolean algebras of sets can be described as rings of sets which contain U.

As $U \neq \phi$, by definition of Boolean algebras, $U^{'}$ is in this class of sets, so that the empty set is also a part of this class of sets. Again, by definition of Boolean algebras, the union of two sets of this class is also in this set so also the intersection of any two sets of this class is also in this class. So, for any two sets A and B, $A-B$ is also in this set (as complement of B exists), and so also, $A \triangle B = (A-B) \bigcup (B-A)$ is also in this set. In other words, this is also a ring of sets.