# subset

Given two sets $A$ and $B$, we say that $A$ is a subset of $B$ (which we denote as $A\subseteq B$ or simply $A\subset B$) if every element of $A$ is also in $B$. That is, the following implication^{} holds:

$$x\in A\Rightarrow x\in B.$$ |

The relation^{} between $A$ and $B$ is then called set inclusion.

Some examples:

The set $A=\{d,r,i,t,o\}$ is a subset of the set $B=\{p,e,d,r,i,t,o\}$ because every element of $A$ is also in $B$. That is, $A\subseteq B$.

On the other hand, if $C=\{p,e,d,r,o\}$, then neither $A\subseteq C$ (because $t\in A$ but $t\notin C$) nor $C\subseteq A$ (because $p\in C$ but $p\notin A$). The fact that $A$ is not a subset of $C$ is written as $A\u2288C$. Similarly, we have $C\u2288A$.

If $X\subseteq Y$ and $Y\subseteq X$, it must be the case that $X=Y$.

Every set is a subset of itself, and the empty set^{} is a subset of every other set. The set $A$ is called a proper subset^{} of $B$, if $A\subset B$ and $A\ne B$. In this case, we do not use $A\subseteq B$.

Title | subset |

Canonical name | Subset |

Date of creation | 2013-03-22 11:52:38 |

Last modified on | 2013-03-22 11:52:38 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 13 |

Author | Wkbj79 (1863) |

Entry type | Definition |

Classification | msc 03-00 |

Classification | msc 00-02 |

Related topic | EmptySet |

Related topic | Superset^{} |

Related topic | TotallyBounded |

Related topic | ProofThatAllSubgroupsOfACyclicGroupAreCyclic |

Related topic | Property2 |

Related topic | CardinalityOfAFiniteSetIsUnique |

Related topic | CriterionOfSurjectivity |

Defines | set inclusion |