Cauchy sequences

Cauchy sequences

Definition 1(Definition).

A sequence $(a_n)$ in $\mathbb{R}$ is Cauchy if, for every $ϵ>0$, there exists an $N\in \mathbb{N}$ such that whenever $m,n\ge N$ it follows that $|a_n-a_m|<ϵ$.

A sequence $(a_n)$ in a metric space $(X,d)$ is Cauchy if, for every $ϵ>0$, there exists an $N\in \mathbb{N}$ such that whenever $m,n\ge N$ it follows that $d(a_n,a_m)<ϵ$.

Notice that this definition bears striking resemblance to the definition of convergence for a sequence. The motivation for this definition is that to test for convergence, you need a candidate limit value. However, you can test whether a sequence is Cauchy without such an entity.

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Cauchy Criterion

Our interest in Cauchy sequences is motivated by the definition of Cauchy sequences having no mention of a limit, and thus possibly allowing us to test for convergence without a candidate limit. We will now prove that this is in fact possible in $\mathbb{R}$.

Theorem 2(Theorem).

A sequence in $\mathbb{R}$ converges if and only if it is a Cauchy sequence.

Proof of convergent $⟹$ Cauchy

Proof
Let $(a_n)\to a$ . Let $ϵ>0$ be arbitrary. Then, there must exist an $N\in \mathbb{N}$ such that $|a_n-a|<\frac{ϵ}{2}$ for all $n\ge N$. Let $m\ge N$ too. Now, (surprise) triangle inequality:

$$\begin{array}{rl}|a_n-a_m|& \le |a_n-a|+|a-a_m|\\ & <\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ.\end{array}$$

❏

Proof can be adapted to show the same for metric spaces. Thus, in any metric space $X$, every convergent sequence is a Cauchy sequence.

Proof of Cauchy $⟹$ convergent

Proof in $\mathbb{R}$

Here’s how we’ll go about proving this:

1. We will show that being Cauchy implies being bounded.
2. We will show that given any sequence, a monotone subsequence can be extracted.
3. Thus, we should be able to extract a monotone subsequence from a Cauchy sequence, which will be bounded since Cauchy sequences are bounded. From the monotone convergence theorem, the subsequence must converge.
4. We will then we will show that the limit of the subsequence must equal the limit of the Cauchy sequence.

Step 1: Cauchy $⟹$ bounded

Lemma 3(Lemma).

Cauchy sequences are bounded.

Proof
Given $ϵ>0$, there exists an $N$ such that $|x_m-x_n|<ϵ$ for all $m,n\ge N$. Thus, the sequence is bounded by $|x_N|+ϵ$ for all $n\ge N$. It follows that

$$M=\max \{|x_1|,|x_2|,\dots ,|x_{N-1}|,|x_N|+ϵ\}$$

is a bound for the sequence $(x_n)$. ❏

This can be proved any metric space.

Step 2: Pervasive monotonicity

Lemma 4(Lemma).

Every sequence in $\mathbb{R}$ has a monotone subsequence.

Proof
Consider a sequence $(a_n)$. We say $a_k$ is a peak if for all $n>k$, we have $a_k\ge a_n$. Now, we can either have infinitely many peaks, or finitely many peaks.

If we have infinitely many peaks, the subsequence of peaks forms a monotone decreasing subsequence.

If we have finitely many peaks, let the last peak $a_k$. Consider $a_{k+1}$ (if there are no peaks, make this $a_1$), which is not a peak. It is guaranteed that $a_m$ exists such that $m>k+1$ and $a_m\ge a_{k+1}$. The same is guaranteed for $a_m$. Thus, we can construct a monotone increasing subsequence. ❏

Recall that a monotone sequence has no meaning in a metric space. Thus, this sequence of proofs cannot to used to show that Cauchy sequences in metric spaces converge.

Step 3: Persuasive subsequence

Lemma 5(Lemma).

If a subsequence of a Cauchy sequence converges to $l$, the Cauchy sequence also converges to $l$.

Proof
Let $(a_{i_k})\to l$ be a subsequence of a Cauchy sequence $(a_n)$. Let $ϵ>0$ be arbitrary. We have $|a_{i_k}-l|<\frac{ϵ}{2}$ for all $k\ge N_1$, for some $N_1$. Also, we have $|a_{i_k}-a_n|<\frac{ϵ}{2}$, for all $n,k\ge N_2$ (since $i_k\ge k$). As usual, let $N=\max \{N_1,N_2\}$. Triangle inequality gives us $|a_n-l|<ϵ$ for all $n\ge N$. ❏

Again, this argument can be made for any metric space.

Proof in ${\mathbb{R}}^k$

It can be shown that if a sequence is Cauchy in ${\mathbb{R}}^k$, it must be Cauchy slot-wise too (very similar argument to how we proved slot-wise convergence). Since Cauchy sequences in $\mathbb{R}$ converge, every slot converges, and thus the vector sequence converges.

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Complete metric space

Definition 6(Definition).

A metric space $X$ is called a complete metric space if every Cauchy sequence in $X$ converges (in $X$).

Note the overload of the term “complete”, which is also used in the context of ordered sets to signify the presence of the LUB property.

Examples: $\mathbb{R}$, ${\mathbb{R}}^k$, $[a,b]$, closed box in ${\mathbb{R}}^k$.

Relation with sequential compactness

It is clear that being complete does not imply being sequentially compact (example: $\mathbb{R}$).

However, being sequentially compact does imply being complete (Since any sequence must have a convergent subsequence, due to the existence of which a Cauchy sequence must converge).