Completeness is the property that Cauchy sequences do not escape the space. It separates spaces with no missing limit points from spaces with gaps. The real numbers are complete under the usual metric, while the rational numbers are not. Completeness is one of the central structural properties in metric space theory and functional analysis. :::definition[Complete Metric Space] Let $(X,d)$ be a metric space. The space $(X,d)$ is called $complete$ if every Cauchy sequence in $X$ converges to a point of $X$. ::: :::example The spaces $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{R}^n$ with their usual metrics are complete. ::: :::example The metric space $\mathbb{Q}$ with the usual metric is not complete. ::: :::proof Given that $X=\mathbb{Q}$ with $d(x,y)=|x-y|$. To prove that $\mathbb{Q}$ is not complete. The decimal approximations $1.4,1.41,1.414,\ldots$ form a Cauchy sequence in $\mathbb{Q}$, but their real limit is $\sqrt{2}$. Since $\sqrt{2}\notin\mathbb{Q}$, the sequence does not converge in $\mathbb{Q}$. Hence $\mathbb{Q}$ is not complete. $\square$ ::: :::example The rational sequence $$\begin{equation*} x_n=1+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!} \end{equation*}$$ is Cauchy in $\mathbb{Q}$ and converges in $\mathbb{R}$ to $e$. Since $e$ is irrational, it does not converge in $\mathbb{Q}$. ::: :::theorem[Completeness of Discrete Metric Spaces] Every nonempty set with the discrete metric is complete. ::: :::proof Given that $X$ has the discrete metric. To prove that $X$ is complete. Let $\{x_n\}$ be Cauchy. For $\epsilon=1/2$, there exists $n_0$ such that $d(x_n,x_m)<1/2$ whenever $m,n\geq n_0$. Since the discrete metric takes only the values $0$ and $1$, this implies $d(x_n,x_m)=0$ for all $m,n\geq n_0$. Hence the sequence is eventually constant and therefore convergent in $X$. $\square$ ::: :::example Let $X=\mathbb{N}$ and define $d(m,n)=|1/m-1/n|$. Then $(X,d)$ is not complete. ::: :::proof Given that $X=\mathbb{N}$ and $d(m,n)=|1/m-1/n|$. To prove that $(X,d)$ is not complete. The sequence $x_n=n$ is Cauchy because $|1/n-1/m|\to 0$ as $m,n\to\infty$. If it converged to $k\in\mathbb{N}$, then $d(n,k)=|1/n-1/k|\to 1/k\neq 0$, a contradiction. Hence $(X,d)$ is not complete. $\square$ ::: :::example Let $X=\mathbb{R}$ and define $d(x,y)=|x-y|/(\sqrt{1+x^2}\sqrt{1+y^2})$. Then $(X,d)$ is not complete because the sequence $x_n=n$ is Cauchy but has no finite real limit in this metric. ::: :::exercise [1] Define completeness. [2] Why is $\mathbb{Q}$ incomplete? [3] Prove that a discrete metric space is complete. ::: :::answer [1] Every Cauchy sequence converges inside the space. [2] It has rational Cauchy sequences whose real limits are irrational. [3] A Cauchy sequence must be eventually constant when distances below $1/2$ are forced. ::: :::call-to-action[Continue to Standard Complete Spaces] subtitle: Next, prove completeness for common finite-dimensional, sequence, and function spaces. button: Next Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/completeness-of-standard-metric-spaces button-ghost: Previous Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/cauchy-sequences-in-metric-spaces :::