What mathematics is built on

Mathematics begins with set theory. And set theory begins with one question: does an object belong to the chosen set? This is where equality, inclusion, and all the basic operations begin. To understand mathematics, we first need to learn the language of set theory.

We will use one simple example throughout: the sets $A = \{1, 2, 3\}$ and $B = \{3, 4\}$. The numbers are chosen only to illustrate the definitions. Three has no secret life, even if it must belong to two sets at once.

Two outlined regions, elements, and mutual inclusion directions connected by one continuous flow

The foundation: $\in$ and $=$

How sets are denoted

Sets are denoted by uppercase Latin letters: $A$, $B$, and so on. Their elements are denoted by lowercase letters: $a$, $b$, and so on.

The membership symbol $\in$

An object may belong to a set—that is, be one of its elements—or it may not. If an abstract object $a$ is an element of a set $A$, we write $a \in A$; otherwise, we write $a \notin A$.

In our example, $1 \in A$, $3 \in A$ and $3 \in B$, while $4 \notin A$. The notation does not explain what an object is; it only records its position relative to a set.

The equality symbol $=$

Sets $A$ and $B$ are equal if they contain the same elements; we then write $A = B$.

The empty set $\varnothing$

A set with no elements is called empty and is denoted by $\varnothing$. It is a set in its own right, with a remarkably short inventory: there is nothing in it.

The subset symbol $\subset$

If every element of $A$ belongs to $B$, then $A$ is called a subset of $B$, written $A \subset B$. The same relation can be read from the other side: $B$ is a superset of $A$, so $B \supset A$.

The criterion for equality of sets

$A \subset B,\quad B \subset A \Longleftrightarrow A = B$

In other words, two sets are equal if and only if each is a subset of the other.

Consider the direction from inclusion to equality. From $A \subset B$, every element of $A$ belongs to $B$; from $B \subset A$, every element of $B$ belongs to $A$. The sets therefore contain the same elements, so $A = B$.

Three exact diagrams highlight the union, the shared intersection lens, and the left difference of two sets

Union, intersection, and set difference

Three operations on two sets

The union collects elements that belong to at least one of the two sets:

$A \cup B = \{x : x \in A \lor x \in B\}.$

The intersection keeps the elements that belong to both sets:

$A \cap B = \{x : x \in A \land x \in B\}.$

The difference keeps the elements of the first set that do not belong to the second:

$A \setminus B = \{x : x \in A \land x \notin B\}.$

In our example, $A \cup B = \{1, 2, 3, 4\}$, $A \cap B = \{3\}$, and $A \setminus B = \{1, 2\}$.

Decomposing a union

$A \cup B = (A \setminus B) \cup (B \setminus A) \cup (A \cap B).$

The formula divides the union into three parts: elements found only in $A$, elements found only in $B$, and elements shared by both. To prove equality of the sets, we check both inclusions.

Proof of the decomposition

We will prove it using the criterion for equality of sets.

Every element of $A \setminus B$ belongs to $A$, every element of $B \setminus A$ belongs to $B$, and every element of $A \cap B$ belongs to both $A$ and $B$. Each of the three parts therefore lies in $A \cup B$.

Now take $x \in A \cup B$. If $x$ belongs to both sets, then $x \in A \cap B$. If it belongs to $A$ but not $B$, then $x \in A \setminus B$. If it belongs to $B$ but not $A$, then $x \in B \setminus A$.

Each set is included in the other, so the equality is proved.

The same family of three regions branches into the full covered union and the common central intersection

Existence and universality

The union of a family of sets

Let the sets $A_\alpha$ form a family indexed by $\alpha \in \mathfrak{U}$. An element belongs to the union of the family if there is at least one index for which the element belongs to the corresponding set:

$\underset{\alpha \in \mathfrak{U}}{\cup} A_{\alpha} = \{x : \exists \alpha \in \mathfrak{U}: x \in A_{\alpha}\}.$

The key phrase is “the existential quantifier $\exists$.” One suitable set is enough to place the element in the whole union.

The intersection of a family of sets

Here the key phrase is “the universal quantifier $\forall$.” The element must belong to every set in the family:

$\underset{\alpha \in \mathfrak{U}}{\cap} A_{\alpha} = \{x : \forall \alpha \in \mathfrak{U}: x \in A_{\alpha}\}.$

The union asks, “Can we find at least one witness?” The intersection asks, “Does everyone confirm the claim?” One vote is enough in the first case; in the second, the vote must be unanimous.

Two pairs of diagrams in a universal frame show the complement of a union and of an intersection

Complements of unions and intersections

Fix a set $X$ and a family $A_\alpha \subset X$ for every $\alpha \in \mathfrak{U}$. The difference $X \setminus A_\alpha$ contains the elements of $X$ that are absent from $A_\alpha$. Negating the membership condition changes the quantifier and, with it, the operation.

The complement of a union

$X \setminus \underset{\alpha \in \mathfrak{U}}{\cup} A_{\alpha} = \underset{\alpha \in \mathfrak{U}}{\cap} (X \setminus A_{\alpha}).$

Test an arbitrary element. The condition $x \in X \setminus \underset{\alpha \in \mathfrak{U}}{\cup} A_{\alpha}$ means that $x \in X$ and $x$ does not belong to the union. Therefore, $x \notin A_\alpha$ for every $\alpha$, so $x \in X \setminus A_\alpha$ for every $\alpha$. This is exactly membership in the intersection of the complements.

The complement of an intersection

$X \setminus \underset{\alpha \in \mathfrak{U}}{\cap} A_{\alpha} = \underset{\alpha \in \mathfrak{U}}{\cup} (X \setminus A_{\alpha}).$

Here $x \notin \underset{\alpha \in \mathfrak{U}}{\cap} A_\alpha$ means that there is an index $\alpha$ for which $x \notin A_\alpha$. For that index, $x \in X \setminus A_\alpha$, and therefore $x$ belongs to the union of the complements.

Both formulas change two things at once: “there exists” becomes “for every,” or vice versa, while union becomes intersection.

Three regions branch and reconnect to show two equivalent routes through the same flow

Distributivity of union and intersection

Two distributive laws

$(A \cup B) \cap C = (A \cap C) \cup (B \cap C),$

$(A \cap B) \cup C = (A \cup C) \cap (B \cup C).$

In the first law, an element must belong to $C$ and to at least one of $A$ or $B$. It therefore lies in either $A \cap C$ or $B \cap C$. In the second law, an element belongs to both $A$ and $B$, or it belongs to $C$; equivalently, it belongs to both $A \cup C$ and $B \cup C$.

One way to read every proof

Every lemma above follows the same route: choose an arbitrary $x$, expand the definitions of the operations, transform the membership conditions, and assemble them back into the target set. To establish equality, check both inclusions or build a chain of equivalent conditions.