# Beginning Rigorous Mathematics/Sets and Functions

We will now discuss some important set operations. Recall from the preliminaries that we only intuitively define a set to be a collection of distinct mathematical objects. There are much better and rigorous definitions of what sets are and form a subject for study all by itself, into which we will not delve.

Recall the meaning of the symbol "$\in$", which reads "is an element of" so that if $A$ is a set and $x$ an object, then $x\in A$ a is a statement (which is either true of false, depending on whether $x$ is an element of $A$).

# Sets and Set operations

In the following discussion, $A$ and $B$ denote any sets.

## Definitions

### The empty set: $\emptyset$

We will assume that the empty set exists, and is denoted by $\emptyset$. As the name suggests, the empty set contains no elements so that for any object $x$ the statement $x\in \emptyset$ is false.

### Set equality

Usually there is no ambiguity when we use the symbol "=" to refer to equality between sets. It is important that equality between sets is completely different to equality between numbers.

We define the logical statement "$A = B$" to be true by definition when the statement "$x\in A \Leftrightarrow x\in B$" (which reads "$x$ is contained in $A$ if and only if $x$ is contained in $B$") is true, and false otherwise. Intuitively this means that sets are equal if and only if they contain exactly the same elements. For example, $\{1,2,3\}=\{1,2\}$ is false since "$3\in \{1,2\}$" is false. It might be helpful to check the truth table to see that "$3\in \{1,2,3\} \Leftrightarrow 3\in \{1,2\}$" is a false statement. It should then be clear that "$\{5,6,7\}=\{5,6,7\}$" is true.

### Subsets

If every element of the set $A$ is an element of $B$, we then say that $A$ is a subset of $B$.

Rigorously, we say that the statement "$A\subset B$" is true by definition when the statement "$x\in A \Rightarrow x\in B$" (which reads "If $x$ is contained in $A$ then $x$ is contained in $B$") is true.

We have seen previously that the statement $\{1,2\}=\{1,2,3\}$ is false, however the statement $\{1,2\}\subset \{1,2,3\}$ is true. It should be clear that $\{1,2,3,5\}\subset \{1,2,3,4\}$ is false, since the statement "$5\in \{1,2,3,5\} \Rightarrow 5 \in \{1,2,3,4\}$" is false.

### Intersection

We define the intersection of sets by the symbol "$\cap$". Rigorously we write "$A\cap B := \{x|x\in A \land x\in B\}$" which reads "The intersection of the sets $A$ and $B$ is by definition equal to the set which contains exactly the elements which are contained in both $A$ and $B$".

For example "$\{1,2,3,4\}\cap \{1,4,5\}:=\{1,4\}$".

We say that $A$ and $B$ are disjoint when $A\cap B = \emptyset$.

### Union

We define the union of sets by the symbol "$\cup$". Rigorously we write "$A\cup B := \{x|x\in A \lor x\in B\}$" which reads "The union of the sets $A$ and $B$ is by definition equal to the set which contains exactly the elements which are contained in either one of $A$ and $B$".

For example "$\{1,2,3,4\}\cup \{1,4,5\}:=\{1,2,3,4,5\}$".

### Complement

To define the complement of the set $A$ we assume that the set $A$ is a subset of some universal set $X$. We say "$A$ lives in $X$". Often the universal set is implicitly clear, for example when we are studying real analysis we often just assume $X=\mathbb{R}$ or when studying complex analysis we assume $X=\mathbb{C}$.

We define the complement of a set by the superscript "$c$". Rigorously "$A^c := \{x \in X | \lnot x\in A\}$" which reads "the complement of $A$ in $X$ is the set of all elements which are contained in $X$ and not in $A$".

For example, if we assume $X=\{1,2,3,4,5\}$ then $\{1,5\}^c:=\{2,3,4\}$.

### Relative complement

We define the relative complement of sets by the symbol "$\backslash$". Rigorously, "$A\backslash B := \{x\in A | \lnot x\in B\}$" which reads "the relative complement of $B$ in $A$ is by definition equal to the set containing all the elements contained in $A$ and is not contained in $B$".

For example "$\{1,2,3,4\}\backslash \{1,4,5\} := \{2,3\}$".

## Some basic results

### Lemma 1

$(A = B) \Leftrightarrow (A\subset B) \land (B\subset A)$. Which reads "A equals B if and only if A is a subset of B AND B is a subset of A"

proof As explained in the previous chapter, $(A = B) \Leftrightarrow (A\subset B) \land (B\subset A)$ will be true by adjunction when both $(A = B) \Rightarrow (A\subset B) \land (B\subset A)$ and $(A\subset B) \land (B\subset A) \Rightarrow (A = B)$ are true.

We prove first $(A = B) \Rightarrow (A\subset B) \land (B\subset A)$.

Let $A = B$, then by definition we have $x\in A \Leftrightarrow x\in B$, which is logically equivalent to $(x\in A \Rightarrow x\in B)\land (x\in A \Rightarrow x\in B)$. By simplification we have that $x\in A \Rightarrow x\in B$ is true, and that $x\in A \Rightarrow x\in B$ is true. Therefore by definition $A \subset B$, and $B\subset A$ are both true. By adjunction $(A\subset B)\land(B\subset A)$ is true. Therefore $(A = B) \Rightarrow (A\subset B) \land (B\subset A)$ is true.

Conversely, we prove $(A\subset B) \land (B\subset A) \Rightarrow (A = B)$.

Let $(A\subset B) \land (B\subset A)$. Then, by simplification, both $A\subset B$ and $B\subset A$ are true. By definition both $x\in A \Rightarrow x\in B$ and $x\in B \Rightarrow x\in A$ are true. By adjunction, $(x\in A \Rightarrow x\in B) \land (x\in B \Rightarrow x\in A)$ is true, which is logically equivalent to $x\in A \Leftrightarrow x\in B$. Then by definition $A=B$ is true. Therefore $(A\subset B) \land (B\subset A) \Rightarrow (A = B)$ is true. QED.

### Lemma 2

$A\subset A\cup B$

### Lemma 3

$A\cap B\subset A$ and $A\cap B\subset B$