# boundary of a closed set is nowhere dense

Let $A$ be closed. In general, the boundary of a set is closed. So it suffices to show that $\partial A$ has empty interior.

Let $U\subset \partial A$ be open. Since $\partial A\subset \overline{A}=A$, this implies that $U\subset A$. Since $\mathrm{int}(A)$ is the largest open subset of $A$, we must have $U\subset \mathrm{int}(A)$. Therefore $U\subset \partial A\cap \mathrm{int}(A)$. But $\partial A\cap \mathrm{int}(A)=(\overline{A}-\mathrm{int}(A))\cap \mathrm{int}(A)=\mathrm{\varnothing}$, so $U=\mathrm{\varnothing}$.

Title | boundary of a closed set^{} is nowhere dense |
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Canonical name | BoundaryOfAClosedSetIsNowhereDense |

Date of creation | 2013-03-22 18:34:01 |

Last modified on | 2013-03-22 18:34:01 |

Owner | neapol1s (9480) |

Last modified by | neapol1s (9480) |

Numerical id | 4 |

Author | neapol1s (9480) |

Entry type | Derivation |

Classification | msc 54A99 |