Real Analysis: Sequences
Cauchy Sequences
Cauchy Sequences
Given a metric space $(X, d),$ a Cauchy sequence is a sequence $\{a_n\}$ with the property that for every $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $d(a_n, a_m) < \varepsilon$ for all $n, m > N$. This definition is different from that of a convergent sequence because it requires that all terms in the sequence be arbitrarily close to one another as opposed to a fixed value. However, there is a major theorem that shows that the two definitions are equivalent in Euclidean space.
Cauchy sequences have many applications. For example, if you can prove that a computer algorithm outputs a Cauchy sequence, you need not solve for the limit explicitly to know that it converges. In fact, finding that limit may be the point of the algorithm!
Cauchy Criterion for $\mathbb{R}$
If the terms in a Cauchy sequence can get closer together within an arbitrary degree of precision, it stands to reason that they should converge to a single point. However, this is not always so. For example, take the sequence of rational numbers where the $n$th element in the sequence has the first $n$ digits of $\sqrt{2}.$ The sequence is a Cauchy sequence in both $\mathbb{R}$ and $\mathbb{Q},$ but it only converges in $\mathbb{R}.$
The Cauchy Criterion for a metric space $(X, d)$ is the property that every Cauchy sequence in $X$ converges to a point in $X.$ A metric space that fulfills the Cauchy Criterion is said to be Cauchy complete, or simply complete. Notably, the real numbers are complete, and the rational numbers are not.
Topological Completeness and the Construction of $\mathbb{R}$
It is not a coincidence that the term "complete" coincides with the one used to attribute the least upper bound property to $\mathbb{R}.$ The least upper bound property is in fact equivalent to the Cauchy Criterion for $\mathbb{R},$ as each property implies the other. However, because metric spaces have no intrinsic notion of order, the least upper bound property cannot be used to generalize completeness. Cauchy sequences, though, are readily constructible for every metric space. This generalization of completeness to arbitrary metric spaces is covered in the Completeness section of Global Metric Topology.
Problems
Show that $\left\{\frac{1}{n}\right\}$ is a Cauchy sequence.
Pick $\varepsilon > 0.$ By the Archimedean property, there exists an $N \in \mathbb{N}$ such that $N \varepsilon > 1.$ For all $n > N$ it then follows that $\varepsilon > \frac{1}{N} > \frac{1}{n}.$ Next, note that $0 < \frac{1}{m} < \frac{1}{n}$ for all $m > n.$ Therefore $\frac{1}{n} - \frac{1}{m} < \varepsilon.$ Therefore $\left\{\frac{1}{n}\right\}$ is a Cauchy sequence.
Show that every convergent sequence in $\mathbb{R}$ with the Euclidean metric is a Cauchy sequence.
Assume $\{a_n\}$ is a convergent sequence in $\mathbb{R}$ under the Euclidean metric. Then there is some $L \in \mathbb{R}$ such that for every $\dfrac{\varepsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that $|a_n - L| < \dfrac{\varepsilon}{2}$ for all $n > N.$ It follows that $a_n \in \left(L - \dfrac{\varepsilon}{2}, L + \dfrac{\varepsilon}{2}\right)$ whenever $n > N.$ Thus for any $n, m > N,$ we see that $|a_n - a_m| < \varepsilon.$ Therefore $\{a_n\}$ is a Cauchy sequence.
Show that every convergent sequence in any metric space $(X, d)$ is Cauchy.
Let $\{a_n\}$ be a Cauchy sequence in $(X, d).$ Then there is some $L \in X$ such that for every $\frac{\varepsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that $d(a_n, L) < \frac{\varepsilon}{2}$ for every $n > N.$ By the triangle inequality, $d(a_n, a_m) \leq d(a_n, L) + d(L, a_m) < \varepsilon$ for all $n, m > N.$ It follows that $\{a_n\}$ is a Cauchy sequence.
Show that every Cauchy sequence is bounded.
Let $\{a_n\}$ be a Cauchy sequence in a metric space $(X, d).$ Then there exists an $N \in \mathbb{N}$ such that $d(a_n, a_m) < 1$ for all $n, m \in \mathbb{N}$ where $n > N$ and $m > N.$ It follows that $\{a_n\}$ is bounded for $n > N.$ Likewise, since the set $\{a_1, \ldots, a_N\}$ is finite, it is bounded. Since the union of two bounded sets is bounded, it follows that $\{a_n\}$ is bounded.
Cauchy Criterion for $\mathbb{R}$: Show that every Cauchy sequence in $\mathbb{R}$ converges.
Let $\{a_n\}$ be a Cauchy sequence in $\mathbb{R}.$ It follows that $\{a_n\}$ is bounded. By the Bolzano-Weierstrass theorem, $\{a_n\}$ has a convergent subsequence, call it $\{a_{n_k}\}.$ Let $\lim\limits_{n_k \rightarrow \infty} a_{n_k} = L.$ We show that $\lim\limits_{n \rightarrow \infty} a_n = L.$
Since $\{a_n\}$ is Cauchy, for any $\frac{\varepsilon}{2} > 0,$ there exists an $N \in \mathbb{N}$ such that $d(a_n, a_m) < \frac{\varepsilon}{2}$ for all $n \geq N$ and $m \geq N.$ Since $\{a_{n_k}\}$ converges to $L,$ it follows that there exists an $M \in \mathbb{N}$ such that $d(a_{n_k}, L) < \frac{\varepsilon}{2}$ whenever $n_k > M.$ Take $n$ and $n_k$ to be greater than $\max\{N, M\}.$ By the triangle inequality,
$d(a_n, L) < d(a_n, a_{n_k}) + d(a_{n_k}, L)\\d(a_n, L) < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2}\\d(a_n, L) < \varepsilon$
Therefore $\lim\limits_{n \rightarrow \infty} a_n = L.$
Show that a Cauchy sequence in a metric space $(X, d)$ converges if and only if it has a subsequence that converges in $X.$
Let $\{a_n\}$ be a Cauchy sequence in a metric space $(X, d).$ If $\{a_n\}$ converges, then it has a convergent subsequence, namely itself. Conversely, assume $\{a_n\}$ has a convergent subsequence, $\{a_{n_k}\}$ whose limit is $L \in X.$ Then for any $\frac{\varepsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that $d(a_{n_k}, L) < \frac{\varepsilon}{2}$ whenever $n_k > N.$ Also, for any $\frac{\varepsilon}{2} > 0$ there exists an $M \in \mathbb{N}$ such that $d(a_n, a_{n_k}) < \frac{\varepsilon}{2}$ whenever $n > M.$ Take $n$ and $n_k$ to be greater than $\max\{N, M\}.$ It follows from the triangle inequality that
$d(a_n, L) < d(a_n, a_{n_k}) + d(a_{n_k}, L)\\d(a_n, L) < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2}\\d(a_n, L) < \varepsilon$
Therefore $\{a_n\}$ converges.
Cauchy Criterion for $\mathbb{R}^n$: Show that every Cauchy sequence in $\mathbb{R}^n$ converges under the Euclidean metric.
Let $\{\mathbf{a}_n\}$ be a Cauchy sequence in $\mathbb{R}^k$ under the Euclidean metric. Then for every $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $d(\mathbf{a}_n, \mathbf{a}_m) < \varepsilon$ for all $n \geq N$ and $m \geq N.$ It follows that $d(a_{n, i}, a_{m, i}) < \varepsilon$ for every $1 \leq i \leq k.$ Therefore the $i$th components of $\{\mathbf{a}_n\}$ form Cauchy sequences $\{a_{n,i}\}$ in $\mathbb{R}.$ By the Cauchy criterion on $\mathbb{R},$ each converges to a value $L_i.$ It follows by the convergence criterion for $\mathbb{R}^n$ that $\{\mathbf{a}_n\}$ converges to $\mathbf{L} = \{L_1, \ldots, L_k\}.$