# a complete subspace of a metric space is closed

Let $X$ be a metric space, and let $Y$ be a complete subspace^{} of $X$. Then $Y$ is closed.

Proof

Let $x\in \overline{Y}$ be a point in the closure^{} of $Y$. Then by the definition of closure, from each ball $B(x,\frac{1}{n})$ centered in $x$, we can select a point ${y}_{n}\in Y$. This is clearly a Cauchy sequence^{} in $Y$, and its limit is $x$, hence by the completeness of $Y$, $x\in Y$ and thus $Y=\overline{Y}$.

Title | a complete subspace of a metric space is closed |
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Canonical name | ACompleteSubspaceOfAMetricSpaceIsClosed |

Date of creation | 2013-03-22 16:31:29 |

Last modified on | 2013-03-22 16:31:29 |

Owner | ehremo (15714) |

Last modified by | ehremo (15714) |

Numerical id | 5 |

Author | ehremo (15714) |

Entry type | Result |

Classification | msc 54E50 |