# Subset

In mathematics, a set A is a subset of a set B if all elements of A are also elements of B; B is then a superset of A. It is possible for A and B to be equal; if they are unequal, then A is a proper subset of B. The relationship of one set being a subset of another is called inclusion (or sometimes containment). A is a subset of B may also be expressed as B includes (or contains) A or A is included (or contained) in B. Euler diagram showing A is a proper subset of B,  A⊂B,  and conversely B is a proper superset of A.

The subset relation defines a partial order on sets. In fact, the subsets of a given set form a Boolean algebra under the subset relation, in which the join and meet are given by intersection and union, and the subset relation itself is the Boolean inclusion relation.

## Definitions

If A and B are sets and every element of A is also an element of B, then:

• A is a subset of B, denoted by $A\subseteq B,$ or equivalently
• B is a superset of A, denoted by $B\supseteq A.$ If A is a subset of B, but A is not equal to B (i.e. there exists at least one element of B which is not an element of A), then:

• A is a proper (or strict) subset of B, denoted by $A\subsetneq B$ (or $A\subset B$ [circular reporting?][better source needed]). Or equivalently,
• B is a proper (or strict) superset of A, denoted by $B\supsetneq A$ (or $B\supset A$ [circular reporting?]).
• The empty set, written $\{\}$ or $\varnothing ,$ is a subset of any set X and a proper subset of any set except itself.

For any set S, the inclusion relation $\,\subseteq \,$ is a partial order on the set ${\mathcal {P}}(S)$ (the power set of S—the set of all subsets of S) defined by $A\leq B\iff A\subseteq B$ . We may also partially order ${\mathcal {P}}(S)$ by reverse set inclusion by defining $A\leq B{\text{ if and only if }}B\subseteq A.$ When quantified, $A\subseteq B$ is represented as $\forall x\left(x\in A\implies x\in B\right).$ We can prove the statement $A\subseteq B$ by applying a proof technique known as the element argument:

Let sets A and B be given. To prove that $A\subseteq B,$ 1. suppose that a is a particular but arbitrarily chosen element of A,
2. show that a is an element of B.

The validity of this technique can be seen as a consequence of Universal generalization: the technique shows $c\in A\implies c\in B$ for an arbitrarily chosen element c. Universal generalisation then implies $\forall x\left(x\in A\implies x\in B\right),$ which is equivalent to $A\subseteq B,$ as stated above.

## Properties

• A set A is a subset of B if and only if their intersection is equal to A.
Formally:
$A\subseteq B{\text{ if and only if }}A\cap B=A.$ • A set A is a subset of B if and only if their union is equal to B.
Formally:
$A\subseteq B{\text{ if and only if }}A\cup B=B.$ • A finite set A is a subset of B, if and only if the cardinality of their intersection is equal to the cardinality of A.
Formally:
$A\subseteq B{\text{ if and only if }}|A\cap B|=|A|.$ ## ⊂ and ⊃ symbols

Some authors use the symbols $\,\subset \,$ and $\,\supset \,$ to indicate subset and superset respectively; that is, with the same meaning and instead of the symbols, $\,\subseteq \,$ and $\,\supseteq .$ For example, for these authors, it is true of every set A that $A\subset A.$ Other authors prefer to use the symbols $\,\subset \,$ and $\,\supset \,$ to indicate proper (also called strict) subset and proper superset respectively; that is, with the same meaning and instead of the symbols, $\,\subsetneq \,$ and $\,\supsetneq .\,$ This usage makes $\,\subseteq \,$ and $\,\subset \,$ analogous to the inequality symbols $\,\leq \,$ and $\,<.\,$ For example, if $x\leq y,$ then x may or may not equal y, but if $x then x definitely does not equal y, and is less than y. Similarly, using the convention that $\,\subset \,$ is proper subset, if $A\subseteq B,$ then A may or may not equal B, but if $A\subset B,$ then A definitely does not equal B.

## Examples of subsets

• The set A = {1, 2} is a proper subset of B = {1, 2, 3}, thus both expressions $A\subseteq B$ and $A\subsetneq B$ are true.
• The set D = {1, 2, 3} is a subset (but not a proper subset) of E = {1, 2, 3}, thus $D\subseteq E$ is true, and $D\subsetneq E$ is not true (false).
• Any set is a subset of itself, but not a proper subset. ($X\subseteq X$ is true, and $X\subsetneq X$ is false for any set X.)
• The set {x: x is a prime number greater than 10} is a proper subset of {x: x is an odd number greater than 10}
• The set of natural numbers is a proper subset of the set of rational numbers; likewise, the set of points in a line segment is a proper subset of the set of points in a line. These are two examples in which both the subset and the whole set are infinite, and the subset has the same cardinality (the concept that corresponds to size, that is, the number of elements, of a finite set) as the whole; such cases can run counter to one's initial intuition.
• The set of rational numbers is a proper subset of the set of real numbers. In this example, both sets are infinite, but the latter set has a larger cardinality (or power) than the former set.

Another example in an Euler diagram:

## Other properties of inclusion A ⊆ B {\displaystyle A\subseteq B} and B ⊆ C {\displaystyle B\subseteq C} implies A ⊆ C . {\displaystyle A\subseteq C.}

Inclusion is the canonical partial order, in the sense that every partially ordered set $(X,\preceq )$ is isomorphic to some collection of sets ordered by inclusion. The ordinal numbers are a simple example: if each ordinal n is identified with the set $[n]$ of all ordinals less than or equal to n, then $a\leq b$ if and only if $[a]\subseteq [b].$ For the power set $\wp {P}(S)$ of a set S, the inclusion partial order is—up to an order isomorphism—the Cartesian product of $k=|S|$ (the cardinality of S) copies of the partial order on $\{0,1\}$ for which $0<1.$ This can be illustrated by enumerating $S=\left\{s_{1},s_{2},\ldots ,s_{k}\right\},$ , and associating with each subset $T\subseteq S$ (i.e., each element of $2^{S}$ ) the k-tuple from $\{0,1\}^{k},$ of which the ith coordinate is 1 if and only if $s_{i}$ is a member of T.

## References

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7. Subsets and Proper Subsets (PDF), archived from the original (PDF) on 2013-01-23, retrieved 2012-09-07