The most basic object

In mathematics, a set constitutes the most basic object. We define a mathematical object by the properties of the operations we can perform on it.

In the case of sets, we only have a single operation: the membership predicate takes two sets and maps them to a truth value. In infix notation:

x \in A \\ x \notin A \iff \neg (x \in A)

Finite sets can be written by enumerating their elements:

\{ A, B, C \}

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When enumerating sets, the order of elements does not matter. Elements can be repeated without an effect: $\{ A, B, C \} = \{ A, C, B \} = \{ A, B, C, A \}$

In the following, we define some shorthand notation and some properties of the membership predicate. If you want a more formal treatment of set theory have a look at the Zermelo-Fraenkel set theory (opens in a new tab).

Empty set: There exists an empty set, usually denoted as $\emptyset$ . $\forall x \neg (x \in \emptyset)$
Subset: $\begin{align*} A \subseteq B &\iff \forall x (x \in A \to x \in B) \\ A \subset B &\iff A \subseteq B \land A \neq B \\ \end{align*}$
Equality: $\begin{align*} A = B &\iff \forall x (x \in A \leftrightarrow x \in B) \\ &\iff A \subseteq B \land B \subseteq A \end{align*}$
Axiom of pairing: For any sets $A, B$ there exists a set $\{A, B\}$ .
Union: For any sets $A, B$ there exists a set $A \cup B$ . Defined as: $x \in A \cup B \iff x \in A \lor x \in B$
Intersection: For any sets $A, B$ there exists a set $A \cap B$ . Defined as: $x \in A \cap B \iff x \in A \land x \in B$
Difference: For any sets $A, B$ there exists a set $A \setminus B$ . Defined as: $x \in A \setminus B \iff x \in A \land x \notin B$
Filter: For any predicate $P$ and set $A$ there exists a set containing all the elements of $A$ for which $P$ holds. In set notation: $\{ x \in A \mid P(x) \}$
Power set: For any set $A$ there exists a set $\mathcal{P}(A)$ containing all subsets of $A$ . $x \in \mathcal{P}(A) \iff x \subseteq A$

Existance of SingletonsSource: Max Obreiter

Prove that for any set $A$ there exists a set $\{ A \}$ containing only $A$ .

Proving Consistency (Theorem 3.4)Source: Noah Tittelbach

Prove that: $A\subseteq B \iff A = A\cap B$ (The lemma also includes $B = A\cup B$ , but we will skip that here.)

Proof Patterns Relations