Proving or Disproving Completeness in Metric Spaces

Hello! In the previous lesson, convergence meant that terms eventually stayed close to a specified point in the space. The sequence $2-1/n$, for example, converges to $2$ in $\mathbb{R}$ but has no limit in $(0,2)$. Completeness asks what we can conclude when the terms become close to one another, even before we know whether a suitable limit exists.

This lesson develops that distinction, proves completeness for familiar spaces, and constructs Cauchy sequences that reveal when a space is incomplete. The aim is not just to recognize the examples, but to see the proof pattern you can reuse.

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Cauchy sequences: closeness without a proposed limit

A sequence $(x_n)$ in a metric space $(X,d)$ is Cauchy if

$$\text{for every }\epsilon >0,\text{ there is an }N\text{ such that }n,m\ge N⟹d(x_n,x_m)<\epsilon .$$

Compare this with convergence: instead of comparing each late term with one fixed limit $x$, we compare every pair of sufficiently late terms. The condition says nothing directly about where that tail might converge.

Watch the selected passages of Dr. Bevin Maultsby’s Cauchy sequences and complete metric spaces, Real Analysis II. They establish the definition, the fundamental one-way implication, and the meaning of completeness.

Here is the implication at the heart of that proof. If $x_n\to x$, choose $N$ so that $d(x_k,x)<\epsilon /2$ whenever $k\ge N$. For any $n,m\ge N$, the triangle inequality gives

$$d(x_n,x_m)\le d(x_n,x)+d(x,x_m)<\epsilon .$$

Thus every convergent sequence is Cauchy, in every metric space. A space is complete if the converse holds there: every Cauchy sequence of its points converges to a point in that same space.

A quantifier trap is worth avoiding. It is not enough that consecutive terms become close. For the harmonic partial sums

$$h_n=\sum _{k=1}^n\frac{1}{k},$$

we have $|h_{n+1}-h_n|=1/(n+1)\to 0$. Yet

$$|h_{2n}-h_n|=\sum _{k=n+1}^{2n}\frac{1}{k}\ge \frac{n}{2n}=\frac{1}{2}.$$

The pairs $h_n,h_{2n}$ occur arbitrarily far along the sequence while remaining at least $1/2$ apart. The Cauchy definition requires control of all pairs beyond one cutoff.

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Disproving completeness by finding a missing limit

To show that a space is not complete, one counterexample suffices: find a sequence whose terms all lie in the space, prove it is Cauchy in its metric, and prove it does not converge to any point of that space.

Consider $X=(0,1]$ with the usual distance inherited from $\mathbb{R}$. Let $x_n=1/n$. Every term belongs to $X$, and for $n,m\ge N$,

$$|x_n-x_m|\le \frac{1}{n}+\frac{1}{m}\le \frac{2}{N}.$$

Given $\epsilon >0$, choose $N>2/\epsilon$; this proves that $(x_n)$ is Cauchy in $X$. But its limit in $\mathbb{R}$ is $0\notin X$. It cannot have some other limit in $X$: convergence in the inherited metric would also be convergence in $\mathbb{R}$, where limits are unique. Hence $X$ is incomplete.

The same strategy works when the missing limit is less visible. Set $X=\mathbb{Q}$ with its usual metric and, for $n\ge 1$, define the rational decimal truncations

$$q_n=\frac{\lfloor 10^n\sqrt{2}\rfloor }{10^n}.$$

The defining property of the floor function gives

$$0\le \sqrt{2}-q_n<10^{-n}.$$

Consequently, for $n,m\ge N$,

$$|q_n-q_m|\le |q_n-\sqrt{2}|+|\sqrt{2}-q_m|<2\cdot 10^{-N}.$$

So $(q_n)$ is Cauchy in $\mathbb{Q}$, but it converges in $\mathbb{R}$ to the irrational number $\sqrt{2}$. Uniqueness again rules out a rational limit. Thus $\mathbb{Q}$ is incomplete.

Watch how the video finds analogous missing-limit examples, then contrasts them with a complete discrete space.

“Missing points” is useful intuition, but the actual test remains the Cauchy definition plus a proof that no limit lies in the stated space. In particular, we used the larger space $\mathbb{R}$ to locate each candidate limit, not to change the distances between the sequence terms.

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Proving completeness: start with an arbitrary Cauchy sequence

A positive completeness proof has a different logical shape. Examples of convergent sequences are not enough: we must begin with an arbitrary Cauchy sequence and show that it converges in the space.

For the discrete metric, this proof is short and illuminating. Let $(x_n)$ be Cauchy and apply its definition with $\epsilon =1/2$. There is an $N$ such that $d(x_n,x_m)<1/2$ for all $n,m\ge N$. Distinct points have distance $1$, so every term $x_n$ with $n\ge N$ equals $x_N$. The sequence is eventually constant and therefore converges to $x_N$. Every discrete metric space is complete, regardless of how many points it has.

For Euclidean space, we build a limit coordinate by coordinate. Read Proposition 7.4.4 in Section 7.4.1, “Cauchy sequences and completeness,” of RA Completeness and compactness. Its proof reduces completeness of ${\mathbb{R}}^r$ to the familiar completeness of $\mathbb{R}$.

Let us make the last step of that proof explicit. Suppose $(z_j)$ is Cauchy in Euclidean ${\mathbb{R}}^r$, with

$$z_j=(z_{j,1},\dots ,z_{j,r}).$$

For each coordinate $i$,

$$|z_{j,i}-z_{k,i}|\le \Vert z_j-z_k{\Vert }_2.$$

Thus $(z_{j,i}{)}_j$ is Cauchy in $\mathbb{R}$, and real completeness supplies a limit $a_i$. The proposed vector limit is $a=(a_1,\dots ,a_r)$, which belongs to ${\mathbb{R}}^r$. Given $\epsilon >0$, choose one cutoff large enough that

$$|z_{j,i}-a_i|<\frac{\epsilon }{\sqrt{r}}\text{for every }i=1,\dots ,r$$

whenever $j$ exceeds that cutoff. We can use one cutoff because there are only finitely many coordinates. Then

$$\Vert z_j-a{\Vert }_2={\left(\sum _{i=1}^r|z_{j,i}-a_i{|}^2\right)}^{1/2}<\epsilon .$$

This proves that Euclidean ${\mathbb{R}}^r$ is complete. Notice where the underlying fact that $\mathbb{R}$ is complete enters; coordinatewise Cauchy behavior alone does not manufacture real coordinate limits.

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Closed subspaces: a reusable completeness test

There is a convenient way to extend these proofs. A closed subset of a complete metric space is complete under the inherited metric. Suppose $F$ is closed in a complete space $M$, and let $(x_n)$ be Cauchy in $F$. The inherited distances are unchanged, so the sequence is Cauchy in $M$. Completeness of $M$ gives a limit $x\in M$. Every ball around $x$ contains a term of the sequence and hence meets $F$; therefore $x$ belongs to the closure of $F$. Since $F$ is closed, $x\in F$, as required.

For a walkthrough of this proof, watch the final selected passage of the video. It also points out a subtlety: a limit of points in $F$ need not be an accumulation point of $F$ if, for example, the sequence is constant. Saying that the limit lies in the closure handles both cases at once.

Since $\mathbb{R}$ is complete, this immediately proves that $[0,1]$, and indeed any closed subset of $\mathbb{R}$, is complete with the usual inherited metric. It also explains the contrast with $(0,1]$: adding its missing endpoint $0$ changes the answer.

There is a converse when the ambient space $M$ is complete: if $F\subseteq M$ is complete under the inherited metric, then $F$ is closed in $M$. To see why, take $x$ in the closure of $F$ and choose $x_n\in F$ with $d(x_n,x)<1/n$. This sequence converges to $x$ in $M$, so it is Cauchy in $F$. Completeness of $F$ gives a limit $y\in F$. Both are limits in $M$, hence $x=y\in F$. We have proved the useful criterion:

$$\boxed{\text{Within a complete ambient metric space, a subspace is complete exactly when it is closed.}}$$

The ambient assumption matters. A space is always closed in itself, but $\mathbb{Q}$, for example, is not complete.

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Takeaways

Completeness closes the gap between two statements: convergence always implies Cauchy, while Cauchy implies convergence only in a complete space. To disprove completeness, exhibit one Cauchy sequence with no limit in the space. To prove it, start with an arbitrary Cauchy sequence and find its limit—directly, coordinatewise, or by placing it inside a known complete space and using closedness.

The proofs also show why naming the metric and the space is indispensable. The same rational truncations are Cauchy whether viewed in $\mathbb{Q}$ or $\mathbb{R}$, but only the latter contains their limit. In the next lesson, we will use sequences in a different role: testing whether a function between metric spaces is continuous.