UPSC 2025 Maths Optional Paper 2 Q2a — Step-by-Step Solution

15 marks · Section A

Cauchy sequences; completeness of R · Real Analysis · asked 3× in 13 yrs · Read the full method →

Question

Define Cauchy sequence and prove that every convergent sequence of real numbers is a Cauchy sequence. What is the importance of Cauchy condition?

Technique

An $\varepsilon/2$ -argument using the triangle inequality, followed by a discussion of the completeness of $\mathbb{R}$ .

Solution

Definition (Cauchy sequence). A sequence $(a_n)_{n\ge1}$ of real numbers is a Cauchy sequence if for every $\varepsilon > 0$ there exists a positive integer $N$ (depending on $\varepsilon$ ) such that

|a_m - a_n| < \varepsilon \quad\text{for all } m,n \ge N.

Intuitively, the terms eventually get arbitrarily close to one another, without reference to any limit value.

Theorem. Every convergent sequence of real numbers is a Cauchy sequence.

Proof. Suppose $(a_n)$ converges to a limit $L\in\mathbb{R}$ , i.e.

\forall\,\varepsilon>0,\ \exists\, N\in\mathbb{N}\ \text{ s.t. } |a_n - L| < \tfrac{\varepsilon}{2}\ \text{ for all } n\ge N.

Let $\varepsilon > 0$ be given. Choose $N$ as above for the value $\varepsilon/2$ . Then for any $m,n \ge N$ , both $|a_m - L| < \varepsilon/2$ and $|a_n - L| < \varepsilon/2$ . By the triangle inequality,

|a_m - a_n| = |(a_m - L) - (a_n - L)| \le |a_m - L| + |a_n - L| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.

Thus for every $\varepsilon>0$ there is $N$ with $|a_m - a_n| < \varepsilon$ for all $m,n\ge N$ . Hence $(a_n)$ is a Cauchy sequence. $\qquad\blacksquare$

Importance of the Cauchy condition.

Intrinsic criterion for convergence. A Cauchy sequence is defined entirely in terms of its own terms — no prior knowledge of the limit is needed. This lets us establish convergence even when the limit is unknown or hard to compute.
Completeness of $\mathbb{R}$ . The converse — every Cauchy sequence of reals converges — is the completeness property of $\mathbb{R}$ (Cauchy’s general principle of convergence). Together with the theorem above, this gives

\text{a real sequence converges} \iff \text{it is Cauchy}.

This equivalence characterises $\mathbb{R}$ as a complete metric space. The rationals $\mathbb{Q}$ are not complete: e.g. the decimal truncations of $\sqrt2$ form a Cauchy sequence in $\mathbb{Q}$ with no rational limit.

Foundation for analysis. Completeness underlies the convergence of series, uniform convergence (via the Cauchy criterion), the construction of $\mathbb{R}$ from $\mathbb{Q}$ , the Banach fixed-point theorem, and the theory of complete (Banach) spaces.

Answer

A Cauchy sequence satisfies: $\forall \varepsilon>0\ \exists N$ with $|a_m-a_n|<\varepsilon$ for $m,n\ge N$ . Every convergent real sequence is Cauchy (proved by the $\varepsilon/2$ triangle-inequality argument). The Cauchy condition is important because in $\mathbb{R}$ it is equivalent to convergence (completeness), giving an intrinsic test for convergence that needs no knowledge of the limit.