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
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