# discontinuity of characteristic function

Theorem. For a subset $A$ of ${\mathbb{R}}^{n}$, the set of the
discontinuity (http://planetmath.org/Continuous^{})
points of the characteristic function^{} ${\chi}_{A}$ is the
boundary (http://planetmath.org/BoundaryFrontier) of $A$.

Proof. Let $a$ be a discontinuity point of ${\chi}_{A}$. Then any
neighborhood^{} (http://planetmath.org/Neighborhood) of $a$
contains the points $b$ and $c$ such that ${\chi}_{A}(b)=1$ and ${\chi}_{A}(c)=0$. Thus
$b\in A$ and $c\notin A$, whence $a$ is a boundary point of $A$.

If, on the contrary, $a$ is a boundary point of $A$ and $U(a)$ an arbitrary neighborhood of $a$, it follows that $U(a)$ contains both points belonging to $A$ and points not belonging to $A$. So we have in $U(a)$ the points $b$ and $c$ such that ${\chi}_{A}(b)=1$ and ${\chi}_{A}(c)=0$. This means that ${\chi}_{A}$ cannot be continuous at the point $a$ (N.B. that one does not need to know the value ${\chi}_{A}(a)$).

Title | discontinuity of characteristic function |
---|---|

Canonical name | DiscontinuityOfCharacteristicFunction |

Date of creation | 2015-02-03 21:23:33 |

Last modified on | 2015-02-03 21:23:33 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 3 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 03-00 |

Classification | msc 26-00 |

Classification | msc 26A09 |