Subsequences select terms from a sequence without changing their order. They help locate possible limiting behavior. Cauchy sequences measure internal stabilization: late terms become close to one another. A fundamental result says that if a Cauchy sequence has even one convergent subsequence, then the whole sequence converges to the same limit. :::definition[Subsequence] Let $\{x_n\}$ be a sequence in a metric space $(X,d)$. Let $n_1<n_2<n_3<\cdots$ be a strictly increasing sequence of positive integers. Then $\{x_{n_k}\}_{k\geq 1}$ is called a $subsequence$ of $\{x_n\}$. ::: :::definition[Subsequential Limit] Let $\{x_{n_k}\}$ be a subsequence of $\{x_n\}$. If $x_{n_k}\to x$, then $x$ is called a $subsequential limit$ of $\{x_n\}$. ::: :::definition[Cauchy Sequence] Let $(X,d)$ be a metric space. A sequence $\{x_n\}$ is called a $Cauchy sequence$ if for every $\epsilon>0$, there exists $n_0\in\mathbb{N}$ such that $$\begin{equation*} d(x_n,x_m)<\epsilon,\qquad m,n\geq n_0. \end{equation*}$$ ::: :::theorem[Cauchy Sequence with a Convergent Subsequence] Let $(X,d)$ be a metric space. If a Cauchy sequence $\{x_n\}$ has a convergent subsequence, then $\{x_n\}$ is convergent and has the same limit. ::: :::proof Given that $\{x_n\}$ is Cauchy and $x_{n_k}\to x$. To prove that $x_n\to x$. Let $\epsilon>0$. Choose $n_1$ such that $d(x_n,x_m)<\epsilon/2$ whenever $m,n\geq n_1$. Choose $k$ so large that $n_k\geq n_1$ and $d(x_{n_k},x)<\epsilon/2$. Then for $n\geq n_1$, $$\begin{align*} d(x_n,x)&\leq d(x_n,x_{n_k})+d(x_{n_k},x)\\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &=\epsilon. \end{align*}$$ Hence $x_n\to x$. $\square$ ::: :::theorem[Convergence from Three Subsequences] Let $\{x_n\}$ be a sequence in a metric space. Suppose $\{x_{2n}\}$, $\{x_{2n+1}\}$, and $\{x_{3n}\}$ are convergent. Then $\{x_n\}$ is convergent. ::: :::proof Given that the three subsequences are convergent. To prove that $\{x_n\}$ is convergent. Let $x_{2n}\to a$, $x_{2n+1}\to b$, and $x_{3n}\to c$. The subsequence $\{x_{6n}\}$ is common to $\{x_{2n}\}$ and $\{x_{3n}\}$, so $a=c$. The subsequence $\{x_{6n+3}\}$ is common to $\{x_{2n+1}\}$ and $\{x_{3n}\}$, so $b=c$. Hence $a=b=c$. Since the even and odd subsequences converge to the same limit, the whole sequence converges. $\square$ ::: :::exercise [1] Define a subsequence. [2] Define a subsequential limit. [3] Prove that a Cauchy sequence with a convergent subsequence converges. ::: :::answer [1] A subsequence is $\{x_{n_k}\}$ with $n_1<n_2<\cdots$. [2] It is a limit of some subsequence. [3] Use the Cauchy condition to compare $x_n$ with a late subsequence term and then compare that term with the limit. ::: :::call-to-action[Continue to Cauchy Sequences] subtitle: Next, focus directly on Cauchy examples and nonexamples. button: Next Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/cauchy-sequences-in-metric-spaces button-ghost: Previous Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/limits-of-sequences-in-metric-spaces :::