Cauchy Sequences

A Cauchy sequence \(\{a_n\}_{n=1}^\infty\) is one which has the following property:

\[{}\forall \epsilon > 0 \ \exists N \in {\mathbb N} \text{ such that }{}\]

\[{}n,m > N \Rightarrow |a_n - a_m | < \epsilon.{}\]

In other words, for any threshold \(\epsilon\), there is a point beyond which all terms in the tail of the sequence are \(\epsilon\)-close to each other. This is closely related to the definition of convergence but distinct. It's generally easier to show a sequence is Cauchy than it is to show that it converges because the definition of “Cauchyness” does not make reference to any limit.

Warning

It is extremely important to observe that all terms in the tail of a Cauchy sequence are close to each other. This is a much stronger condition than merely saying that adjacent terms are close to one another. This distinction must be kept in mind at all times.

It's not so difficult to prove that all convergent sequences are Cauchy.

Theorem 1 (Convergent Sequences are Cauchy)

If \(\{a_n\}_{n=1}^\infty\) is a convergent sequence of real numbers, then it must be a Cauchy sequence.

Proof

Meta (Main Idea)

Apply the definition of convergence at threshold \(\epsilon/2\) combined with the triangle inequality

\[{}|a_n - a_m|{}\]

\[{}\leq |a_n - L| + |a_m - L|.{}\]

The really remarkable thing is that the converse is true:

Theorem 2 (Cauchy Sequences Converge)

If \(\{a_n\}_{n=1}^\infty\) is a Cauchy sequence of real numbers, then it must be convergent.

The proof of this theorem, like Bolzano-Weierstrass, is a bit involved. The first step is the following intermediate result, which is often useful in its own right.

Proposition

If \(\{a_n\}_{n=1}^\infty\) is Cauchy, then it must be bounded.

Proof

Meta (Main Idea)

There is some threshold \(N\) such that \(|a_n - a_m| < 1\) for \(n,m > N\). In particular, fixing \(m = N+1\) gives \(|a_n| < |a_{N+1}| + 1\) by the triangle inequality. This bounds \(|a_n|\) for all but finitely many \(N\). Then show that \(|a_n| \leq |a_1| + \cdots + |a_{N+1}| + 1\) must hold for actually all values of \(n\) and you're done.

Now by the Bolzano-Weierstrass Theorem, there is a subsequence \(\{a_{n_j}\}_{j=1}^\infty\) which converges to some limit \(L\). From here we use the definition of Cauchyness to show that any Cauchy sequence with a convergent subsequence (which happens to be all Cauchy sequences) must actually just converge.

To see this, fix \(\epsilon > 0\), and let \(J\) be some threshold such that \(|a_{n_j} - L| < \epsilon/2\) when \(j > J\). Next let \(N\) be some threshold such that \(|a_{n} - a_m| < \epsilon/2\) when \(n,m > N\). Notice that, since \(n_j \rightarrow \infty\) as \(j \rightarrow \infty\), there is always some choice of \(j\) such that \(j > J\) and \(n_j > N\). Fix \(m := n_j\) for any such \(j\). Then when \(n > N\),

\[{}|a_n - L| \leq |(a_n - a_m) + (a_m) - L| {}\]

\[{}\leq |a_n - a_m| + |a_m - L|{}\]

\[{}\leq \frac{\epsilon}{2} + |a_{n_j} - L|{}\]

\[{}< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.{}\]

Exercise

Prove directly from the definitions that a subsequence of a Cauchy sequence is necessarily also Cauchy.