Once distance is available, convergence can be defined in any metric space. A sequence converges when its distance from a fixed point becomes arbitrarily small. This generalizes ordinary convergence of real sequences and becomes the foundation for closed sets, continuity, completeness, and compactness in metric spaces. :::definition[Sequence in a Metric Space] Let $(X,d)$ be a metric space. A $sequence$ in $X$ is a function $$\begin{equation*} f:\mathbb{N}\to X. \end{equation*}$$ If $f(n)=x_n$, then the sequence is denoted by $\{x_n\}_{n\geq 1}$. ::: :::definition[Limit of a Sequence in a Metric Space] Let $(X,d)$ be a metric space and let $\{x_n\}$ be a sequence in $X$. A point $x\in X$ is called a $limit$ of $\{x_n\}$ if for every $\epsilon>0$, there exists $n_0\in\mathbb{N}$ such that $$\begin{equation*} d(x_n,x)<\epsilon,\qquad n\geq n_0. \end{equation*}$$ In this case, we write $x_n\to x$. ::: :::theorem[Metric Convergence Criterion] Let $(X,d)$ be a metric space. Then $$\begin{equation*} x_n\to x \iff d(x_n,x)\to 0. \end{equation*}$$ ::: :::proof Given that $(X,d)$ is a metric space. To prove that $x_n\to x$ if and only if $d(x_n,x)\to 0$. The definition of $x_n\to x$ says precisely that for every $\epsilon>0$, $d(x_n,x)<\epsilon$ eventually. This is exactly the definition that the real sequence $d(x_n,x)$ converges to $0$. Hence the equivalence holds. $\square$ ::: :::theorem[Uniqueness of Limit] Let $(X,d)$ be a metric space. If $x_n\to x$ and $x_n\to y$, then $x=y$. ::: :::proof Given that $x_n\to x$ and $x_n\to y$. To prove that $x=y$. Let $\epsilon>0$. Choose $n_0$ such that $d(x_n,x)<\epsilon/2$ and $d(x_n,y)<\epsilon/2$ for $n\geq n_0$. Then $$\begin{align*} d(x,y)&\leq d(x,x_n)+d(x_n,y)\\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &=\epsilon. \end{align*}$$ Since $\epsilon>0$ is arbitrary, $d(x,y)=0$, and therefore $x=y$. $\square$ ::: :::example In $\mathbb{R}$ with the usual metric, $x_n\to x$ if and only if $|x_n-x|\to 0$. Thus real convergence is metric convergence for the usual distance. ::: :::solved-problem Let $(X,d)$ be a metric space and let $x_n=x$ for every $n$. Prove that $x_n\to x$. ::: :::solution Let $x_n=x$ for all $n$. For every $\epsilon>0$, $$\begin{equation*} d(x_n,x)=d(x,x)=0<\epsilon. \end{equation*}$$ Thus the convergence condition holds with $n_0=1$. Hence $x_n\to x$. $\square$ ::: :::exercise [1] Define convergence of a sequence in a metric space. [2] State the metric convergence criterion. [3] Prove uniqueness of limits in a metric space. ::: :::answer [1] $x_n\to x$ if for every $\epsilon>0$, $d(x_n,x)<\epsilon$ eventually. [2] $x_n\to x$ if and only if $d(x_n,x)\to 0$. [3] Use the triangle inequality to show $d(x,y)<\epsilon$ for every $\epsilon>0$. ::: :::call-to-action[Continue to Subsequences] subtitle: Next, study subsequences and the Cauchy criterion. button: Next Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/subsequences-and-cauchy-criteria button-ghost: Previous Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/metrics-on-function-spaces :::