Introduction to Set Theory

In set theory there are two wings of study, the axiomatic and the naive theory. The axiomatic wing relies on an informal understanding of sets as collections of objects. The naïve approach, well developed at the end of the 19th century, treats sets as demonstrable forms with propositions, lemmas, theorems and corollaries. Set theory is very important in mathematics, especially for modern theory such as:

• numbers
• relations
• functions

Specification axiom

Axioms are fundamental to mathematics. However, when we assume that any operations on sets can lead to paradoxes such as Russell’s paradox. Axiomatic theory was fundamental in determining which operations are allowed for sets.

The specification axiom says that if a set $A$ exists and we can describe elements of this set through a property, then there is a set $B$, a subset of $A,$ that contains these elements.

Specification Axiom: Let $A$ be a set and $p(x)$ a proposition in $x$ and must be expressed in terms of quantifiers and variables as:

• quantifiers: $\wedge, \vee$ among others
• variables: $x, y, A, B, dots$

Then there is a set that generates a consistency of all elements $x$ of $A$ that make $p(x)$ true. More specifically

$$\\{x \in A; ~p(x) ~\text{é verdadeira}\\}$$

This axiom is very important to avoid semantic paradoxes, such as the Richard paradox. The set described in Richard’s paradox cannot be constructed from the specification axiom, as the axiom’s property is restricted to logical operators and some quantifiers.

Russell’s paradox

Another more interesting application of this axiom is that it guarantees that there is no set that contains all sets. . Thus, we reach the conclusion that B /∈ U, that is, there is no a universe set. The argument that led to this conclusion is called the paradox by Russel,

Russell’s paradox is very important for understanding some properties of set theory. From it we can apply the axiom of specification and show that there is no set that contains all sets. Note Russell’s argument.

Russell’s Paradox: In a certain city there is a barber who only shaves men who do not shave themselves. Who shaves the barber?

How do we respond to this?

Let us suppose that there is a set whose elements are all sets, that is, the universe set.

Let $U$ be the universe set.

Therefore, from the specification axiom, we can construct the set:

The objective is to check if$B\in U.$ With this, we have two possibilities:

• $B\in B$ or
• $B\notin B$

1. If $B\in B$, by specifying $B,$ we have $B\notin B$
2. If $B\notin B$, then $B\in B$

Therefore, we have a contradiction.

Construction of Sets

From what has been seen so far, the specification axiom is very important for set theory, as from it we can construct several important sets such as:

• empty
• unity
• intersection

Empty Set

Definition: the set is said to be empty, denoted by $\emptyset$, the set that does not have any elements, where the existence of this set is guaranteed by the specification axiom.

In fact, given any set $A,$ we have

$$\emptyset = \\{x \in A; x \neq x\\}.$$

Proposition For every set $A,$ we have $\emptyset \subseteq A.$

Demo:

We already know that either $\emptyset \subseteq A$ or $\emptyset \cancel{\subseteq} A.$, and that these possibilities are mutually exclusive.

So, we just need to show that it is not possible to have $\emptyset \cancel{\subseteq} A.$

For $\emptyset \cancel{\subseteq} A$ to occur, there must exist an element $x$ such that $x \in \emptyset$ and $x \notin A.$. But since $\meptyset$ has no elements, such an $x$ cannot exist.

Therefore, the possibility $\emptyset \cancel{\subseteq} A$ does not occur, and only the possibility $\emptyset \subseteq A$ remains, which must be true.

$\square$

Unity

Definition: Let $A$ and $B$ be any two sets, we define the union of $A$ and $B,$ denoted by $A\cup B,$ the set formed by the elements $x$ such that $x$ is in one of the two sets $A$ or $B.$ More specifically,

$$A \cup B = \\{x; ~x \in A \vee x \in B\\}$$

Intersection

Definition: Let $A$ and $B$ be any two sets, the intersection of $A$ and $B,$ denoted by $A\cap B,$ is the set formed by the elements $x$ such that $x$ is in both sets $A$ and $B.$

$$A \cap B = \\{x; ~ x \in A \wedge x \in B\\}$$

Important results

I, Euler as a mathematician, need to guarantee the existence of the sets $A\cap B$ and $A\cup B$ and these sets are guaranteed by the specification axiom. From the definitions, we derive the following logical equivalences:

Below we present some important results.

Theorem 1 Let $A$ and $B$ be any sets. Then:

• a) $A \subseteq A \cup B \text{ e } B \subseteq A \cup B$
• b) $A \cap B \subseteq A \text{ e } A \cap B \subseteq B$
• c) $A \subseteq B \Leftrightarrow A \cup B = B \text{ e } A \subseteq B \Leftrightarrow A \cap B = A$
• d) $A \cup (B \cap A) = A \text{ e } A \cap (B \cup A) = A.$

Demo:

(a) By definition, this item is equivalent to showing that

In fact, this result is equivalent to

This sentence is a tautology, as it is an implication of the type

(b) is constructed in a similar way, just change a few points. Here’s an exercise 😂😂😂

(c) Of this item I will only show $A \subseteq B \Leftrightarrow A \cup B = B$ the other is as an exercise 😂😂😂

Note that the tautology $(p \Leftrightarrow q) \quad \Leftrightarrow \quad (p \Rightarrow q) \wedge (q \Rightarrow p),$ is valid

Before showing the result, it is worth remembering that $\Leftrightarrow$ is an if, and only if, in which you have to show the round trip, that is:

• departure: $\Rightarrow$
• return: $\Leftarrow$

Go $(\Rightarrow)$ $A \subseteq B,$ then $A \cup B = B$

Note that for equality of sets, we have to show that $A \cup B \subseteq B$ and $B \subseteq A \cup B.$

The second inclusion follows from (a).

For the first, let $x \in A \cup B,$ then $x \in A \wedge x \in B.$

If $x \in A,$ by hypothesis we have $A \subseteq B,$ in this way $x \in B.$

So $x \in B,$ in both cases, as we wanted.

Volta $(\Leftarrow)$ If $A \cup B = B,$ then, we have:

As $A \subseteq A \cup B = B,$ we clearly have that $A \subseteq B.$

(d) Exercise 😂😂😂

$\square$

Disjoint Sets: We say that two sets $A$ and $B$ are disjoint if they have no elements in common, that is, if $A \cap B = \emptyset.$

Theorem 2: Let $X, A$ and $B$ be sets. So we have:

• (a) $\emptyset \subseteq A,\quad A \cup \emptyset = A$ e $A \cap \emptyset = \emptyset$
• (b)
  • $X \subseteq A \cup B \quad \Leftrightarrow \quad X \subseteq A \vee X \subseteq B$ e
  • $X \subseteq A \cap B \quad \Leftrightarrow \quad X \subseteq A \wedge X \subseteq B.$

(a) For this item we will show that $\emptyset \subseteq A.$

By definition, we have to show that $x \in \emptyset \Rightarrow x \in A.$

As the proposition $p : x \in \emptyset$ is always false, then $p \Rightarrow q$ is true for any proposition $q,$ which shows the inclusion.

(b) In this item we show the equivalence $X ⊆ A ∩ B ⇐⇒ X ⊆ A ∧ X ⊆ B,$ the other is as an exercise 😂😂😂.

(Ida)

• Hypothesis: $X \subseteq A \cap B$
• Thesis $X \subseteq A \wedge X \subseteq B.$

Let $x ∈ X,$ as by hypothesis $X \subseteq A \cap B,$ we have $x \in A \cap B$ and, by the definition of intersection, we have $x \in A \wedge x \in B.$

Therefore $X \subseteq A \wedge X \subseteq B.$

(Back)

• Hypothesis: $X \subseteq A \wedge X \subseteq B$
• Thesis $X \subseteq A \cap B.$

Let $x ∈ X,$ by hypothesis be $x \in A \wedge x \in B$ and, by the definition of intersection, we have $x \in A \cap B.$

Therefore $X \subseteq A \cap B$

$\square$

This is the summary of the introduction to sets. Wait for the next posts.