2.2 Subsequences, Cauchy sequences and completeness

Subsequences

Definition

Subsequence

A subsequence of $(x_n)$ is $(x_{n_k})$ where $n_1<n_2<\cdots$ are positive integers. If $x_n\to L$, then every subsequence converges to $L$ because $n_k\ge k$ and therefore eventually lies beyond any prescribed index.

Theorem

Bolzano–Weierstrass

Every bounded real sequence has a convergent subsequence.

Proof

Nested-interval proof

Place every term in a closed bounded interval $I_1$. Bisect it. At least one half contains infinitely many terms; choose that half as $I_2$. Continue, obtaining nested closed intervals $I_k$ of lengths tending to zero, each containing infinitely many terms. Choose indices $n_k$ strictly increasing with $x_{n_k}\in I_k$; this is possible because each interval contains infinitely many terms. Completeness gives a common point $L$ in all intervals: the supremum of their left endpoints lies in each interval because every left endpoint is at most every right endpoint. Since $x_{n_k},L\in I_k$, their distance is at most the interval length and tends to zero.

The conclusion is about a subsequence, not necessarily the whole sequence. For example, $(-1)^n$ has constant subsequences converging to $1$ and $-1$, but does not converge.

Cauchy criterion

Definition

Cauchy sequence

A sequence $(x_n)$ is Cauchy if for every $\varepsilon>0$ there exists $N$ such that $|x_m-x_n|<\varepsilon$ whenever $m,n\ge N$. Unlike the limit definition, no candidate limit is named.

Theorem

Cauchy completeness of $\mathbb R$

A real sequence converges if and only if it is Cauchy.

Proof

Both directions

If $x_n\to L$, choose a tail within $\varepsilon/2$ of $L$ and apply the triangle inequality to $x_m-L$ and $L-x_n$. Conversely, a Cauchy sequence is bounded: choose a tail within $1$ of one fixed tail term, then include the finite initial terms. By Bolzano–Weierstrass it has a subsequence $x_{n_k}\to L$. Given $\varepsilon>0$, choose $N$ so all pairs of terms beyond $N$ differ by less than $\varepsilon/2$, and choose $k$ with $n_k\ge N$ and $|x_{n_k}-L|<\varepsilon/2$. For every $n\ge N$, $|x_n-L|\le|x_n-x_{n_k}|+|x_{n_k}-L|<\varepsilon$.

This equivalence uses completeness of $\mathbb R$. It fails as a statement that every rational Cauchy sequence converges to a rational number. For instance, decimal truncations of $\sqrt2$ are rational and Cauchy but have no rational limit.

Worked example

Convergence without knowing the sum

For $s_n=\sum_{k=1}^n2^{-k}$ and $m>n$,

$$|s_m-s_n|\le\sum_{k=n+1}^\infty2^{-k}=2^{-n}.$$

Given $\varepsilon>0$, choose $N$ with $2^{-N}<\varepsilon$. If $m,n\ge N$, the tail between them is at most $2^{-N}$, so $(s_n)$ is Cauchy and converges. The exact sum is $1$, but the Cauchy argument establishes existence without guessing it. The geometric-tail identity can also be verified by a finite geometric sum and taking the elementary limit $2^{-j}\to0$.

Method comparison

Use a subsequence to extract a limit from boundedness, not to prove the full sequence converges. Use two subsequences with different limits to disprove convergence. Use the Cauchy criterion when differences of tails are easier to control than the unknown limit. A bounded Cauchy sequence converges because of real completeness; boundedness by itself is insufficient.

Practice

1. Show every subsequence of a convergent sequence has the same limit.
2. Find two convergent subsequences of $x_n=(-1)^n+1/n$ and conclude whether $(x_n)$ converges.
3. Prove $x_n=1/n$ is Cauchy.
4. Explain why existence of one convergent subsequence does not imply convergence of the whole sequence.

Full solutions

1. Given $\varepsilon>0$, choose $N$ for the original sequence. Since $n_k\ge k$, for $k\ge N$ we have $n_k\ge N$ and $|x_{n_k}-L|<\varepsilon$.
2. $x_{2k}=1+1/(2k)\to1$ and $x_{2k-1}=-1+1/(2k-1)\to-1$. Their limits differ, so the original sequence cannot converge.
3. Given $\varepsilon>0$, choose $N>2/\varepsilon$. For $m,n\ge N$, $|1/m-1/n|\le1/m+1/n\le2/N<\varepsilon$.
4. The sequence $(-1)^n$ has a subsequence constantly $1$, yet its odd subsequence is constantly $-1$, so the whole sequence diverges.

Synthesis

Boundedness gives a convergent subsequence, while the Cauchy property controls the entire tail. Completeness upgrades that internal tail control to a real limit.