Defined Piece by Piece - Proper Subset
Definition
If $A$ and $B$ are sets, then $A$ is called a proper subset of $B$, if, and only if, every element of $A$ is also an element of $B$, and $B$ contains at least one element that is not contained in $A$.
Notation
\(A \subset B \Longleftrightarrow \forall \; x_{1}\; (x_{1} \in A \Longrightarrow x_{1} \in B) \; \wedge \; \exists \; x_{2}\; (x_{2} \in B \wedge x_{2} \notin A)\).
Meaning of the used symbols
| Symbol | Meaning |
|---|---|
| $ \subset $ | is a proper subset of |
| $ \Longleftrightarrow $ | if, and only if |
| $ \forall $ | for all (universal quantifier) |
| $ \in $ | is an element of |
| $ \Longrightarrow $ | if, then |
| $ \wedge $ | and |
| $ \exists $ | it exists (existential quantifier) |
| $ \notin $ | is not an element of |
Example
Let $A$ and $B$ be the following finite sets: $A = \{1, 2, 3\}$, $B = \{1, 2, 3, 4, 5\}$. All three elements of $A$ are in $B$, but there are also the elements $4$ and $5$ that are containted in B, but not in $A$. Therefore, $A$ is a proper subset of $B$ and we can write $A \subset B$.
As we have seen in the definition of subsets (see link below), the symbol that expresses “is a subset of” $\subseteq$ contains also some sort of equals sign. Due to the fact that a proper subset $A$ by definition does not contain all the element of set $B$, $A$ being a proper subset of $B$ is also denoted as $A \subsetneq B$.
Negation of the definition of proper subset
Starting from he formal definition, a set $A$ is not a proper subset of set $B$, if there is at least an element in set $A$ that is not contained in set $B$ or in case every element of $B$ is also contained in set $A$ (see the non-examples section below).
Notation of the negation
\[A \not\subset B \Longleftrightarrow \exists \; x_{1}\; (x_{1} \in A \wedge x_{1} \notin B) \; \vee \; \forall \; x_{2}\; (x_{2} \in B \Longrightarrow x_{2} \in A).\]Meaning of the used symbols
| Symbol | Meaning |
|---|---|
| $ \not\subset $ | is not a proper subset of |
| $ \Longleftrightarrow $ | if, and only if |
| $ \exists $ | it exists (existential quantifier) |
| $ \in $ | is an element of |
| $ \wedge $ | and |
| $ \notin $ | is not an element of |
| $ \vee $ | (not exclusive) or |
| $ \forall $ | for all (universal quantifier) |
| $ \Longrightarrow $ | if, then |
Non-examples of a proper subset
Let $A$ and $B$ be the following sets: $A = \{1, 2, 3, 4\}$, $B = \{1, 2, 3\}$. In this case, set $A$ contains an element, that is not contained in set $B$, and is therefore not even a subset of $B$. This is an example for the first part of the expression above: $\exists \; x_{1}\; (x_{1} \in A \wedge x_{1} \notin B)$.
Let $A$ and $B$ be the following sets: $A = \{1, 2, 3\}$, $B = \{1, 2, 3\}$. In this case, every element that is contained in set $B$ is also contained in set $A$. But by definition of proper subset, set $B$ must contain at least one element that is not contained in $A$. This is an example for the second part of the expression above: $\forall \; x_{2}\; (x_{2} \in B \Longrightarrow x_{2} \in A).$
It is clear that these two cases cannot occur simultaneously.
Jargon-free explanation
Given two boxes, if all the objects of the first box are also contained within the second box, and the second box contains at least one additional object that is not contained the first box, then the first box is said to be a proper subset of the second box.