Complete Metric Spaces

When Cauchy Sequences Must Have Limits

A Cauchy sequence has terms that become arbitrarily close to one another. The previous tutorial showed that this condition does not always produce a limit in the space: the sequence $1/(n+1)$ is Cauchy in $(0,1)$, but its limit in $\mathbb{R}$ is not in $(0,1)$. Completeness describes precisely the spaces where this obstruction never occurs.

The key phrase is “in the space.” A sequence may converge in a larger space while failing to converge in the space where it was originally defined. We will define completeness, prove how it passes to closed subsets, and establish two ways to recognize when a change in the space or metric preserves completeness.

Definition of Completeness

Definition (Complete Metric Space): A metric space $(X,d)$ is complete if every Cauchy sequence in $X$ converges to a point of $X$.

Completeness is a property of the whole metric space, not of an individual sequence. To establish it, one must show that an arbitrary Cauchy sequence has a limit and that the limit belongs to the space. To show a space is not complete, it is enough to find one Cauchy sequence that does not converge there.

Completeness is distinct from boundedness. The previous tutorial established that every Cauchy sequence is bounded, but that does not ensure a limit exists in the space. Nor does boundedness alone imply completeness. The issue is whether the space contains the limits required by its own Cauchy sequences.

Definition (Banach Space): A Banach space is a normed vector space that is complete with respect to the metric induced by its norm, $d(x,y)=\|x-y\|$.

The real line with its usual metric is complete, as follows from the Completeness of Finite-Dimensional Normed Spaces Theorem in Tutorial 625. The definition above extends the same idea to general metric spaces, where there may be no vector operations or coordinates.

Worked Example: A Closed Interval Is Complete

Consider $X=[1,4]$ with the usual metric. The set $[1,4]$ is closed in the complete metric space $\mathbb{R}$. The Closed Subset Theorem proved below will imply that $X$ is complete. In particular, a Cauchy sequence of points in $[1,4]$ has a real limit, and closedness ensures that this limit still lies between $1$ and $4$, inclusive.

This illustrates why the endpoints matter. The interval $[1,4]$ includes its boundary points, so limits approaching either endpoint remain in the space. The interval $(1,4)$, by contrast, omits both endpoints and is not closed in $\mathbb{R}$; the next example shows directly how this can lead to incompleteness.

Closed Subsets of Complete Spaces

A Cauchy sequence in a subset $A$ of a metric space is Cauchy in the ambient space as well, because both spaces use the same distances between points of $A$. If the ambient space is complete, that sequence has a limit there. Closedness is what guarantees that the limit remains in $A$.

Theorem (Closed Subset of a Complete Space): Let $(X,d)$ be a complete metric space, and let $A$ be a closed subset of $X$. Equipped with the restricted metric, $A$ is complete.

Proof. If $A$ is empty, there are no sequences in $A$, so the statement that every Cauchy sequence in $A$ converges in $A$ holds vacuously. Now suppose $A$ is nonempty, and let $(a_n)$ be a Cauchy sequence in $A$. Since the metric on $A$ is the restriction of $d$, the same pairwise distance estimates show that $(a_n)$ is Cauchy in $X$. Completeness of $X$ gives a point $x\in X$ such that $a_n\to x$ in $X$. Since every $a_n$ belongs to the closed set $A$, the Sequential Characterization of Closed Sets implies $x\in A$. Thus the sequence converges to a point of $A$ in its restricted metric, proving that $A$ is complete. $\square$

The theorem uses both hypotheses for distinct purposes. Completeness of $X$ supplies an ambient limit; closedness of $A$ keeps that limit inside $A$. Without the first hypothesis, a Cauchy sequence may lack any ambient limit. Without the second, its ambient limit may fall outside the subset.

Worked Example: The Open Interval Is Not Complete

Let $X=(1,4)$ with the usual metric, and set $x_n=4-1/n$ for positive integers $n$. Each term belongs to $X$: $x_1=3$, and for every $n\geq1$, $3\leq x_n<4$. For positive integers $m,n$, $$ |x_m-x_n|=\left|\frac{1}{n}-\frac{1}{m}\right| \leq \frac{1}{n}+\frac{1}{m}. $$ If $m,n\geq N$, this is at most $2/N$. Given $\varepsilon>0$, choose a positive integer $N>2/\varepsilon$. Then $|x_m-x_n|\leq2/N<\varepsilon$, so $(x_n)$ is Cauchy in $X$.

As a sequence in $\mathbb{R}$, it converges to $4$, since $|x_n-4|=1/n\to0$. It cannot converge to a point $x\in(1,4)$: convergence to $x$ in $X$ would also be convergence in $\mathbb{R}$, and the Uniqueness of Metric Limits Theorem would then give $x=4$, contrary to $x\in(1,4)$. Therefore $X$ is not complete. The omitted endpoint is exactly where this Cauchy sequence’s limit lies.

Completeness and Closedness in a Complete Ambient Space

The Closed Subset Theorem gives one direction: a closed subset of a complete space is complete. In a complete ambient space, the converse holds too. If a subset is complete, a sequence in it that approaches one of its ambient limit points must have a limit in the subset; uniqueness of limits then forces that point to be the ambient limit point.

Theorem (Complete Subsets of a Complete Space Are Closed): Let $(X,d)$ be complete and $A\subseteq X$, equipped with the restricted metric. If $A$ is complete, then $A$ is closed in $X$. Consequently, for subsets of a complete metric space, completeness is equivalent to being closed.

Proof. If $A$ is empty, it is closed. Otherwise, take any $x$ in the closure of $A$. If $x\in A$, there is nothing to prove. If $x\notin A$, every open ball centered at $x$ meets $A$. For each positive integer $n$, choose $a_n\in A$ such that $d(a_n,x)<1/n$. For any positive integers $m,n$, the triangle inequality gives $$ d(a_m,a_n)\leq d(a_m,x)+d(x,a_n)<\frac{1}{m}+\frac{1}{n}. $$ Given $\varepsilon>0$, choose $N>2/\varepsilon$. If $m,n\geq N$, then $d(a_m,a_n)<2/N<\varepsilon$, so $(a_n)$ is Cauchy in $A$. By completeness of $A$, there is $a\in A$ such that $a_n\to a$ in $A$, hence also in $X$. The construction gives $a_n\to x$ in $X$, since $d(a_n,x)<1/n$. By the Uniqueness of Metric Limits Theorem, $a=x$. Thus $x\in A$, so $A$ contains every point of its closure and is closed. Combining this with the Closed Subset Theorem proves the stated equivalence. $\square$

The equivalence depends on the ambient space being complete. It does not say that every complete metric space, viewed as a subset of an arbitrary larger metric space, must be closed. Rather, it characterizes complete subsets when the surrounding space is already known to be complete.

Completeness Is Preserved by Isometries

A distance-preserving bijection changes the names of points but not the distances between them. It therefore preserves the Cauchy condition and convergence, which together determine completeness.

Theorem (Completeness Is Invariant Under Surjective Isometries): Let $F:X\to Y$ be a bijection between metric spaces $(X,d_X)$ and $(Y,d_Y)$, and suppose $d_Y(F(x),F(x'))=d_X(x,x')$ for every $x,x'\in X$. Then $X$ is complete if and only if $Y$ is complete.

Proof. Suppose first that $X$ is complete, and let $(y_n)$ be a Cauchy sequence in $Y$. Surjectivity gives $x_n\in X$ with $F(x_n)=y_n$ for each $n$. For all $m,n$, $$ d_X(x_m,x_n)=d_Y(F(x_m),F(x_n))=d_Y(y_m,y_n). $$ Thus $(x_n)$ is Cauchy in $X$, so it converges to some $x\in X$. Distance preservation gives $$ d_Y(y_n,F(x))=d_Y(F(x_n),F(x))=d_X(x_n,x)\longrightarrow 0. $$ Hence $(y_n)$ converges in $Y$, proving $Y$ complete. Conversely, the inverse bijection $F^{-1}:Y\to X$ also preserves distances: for $y=F(x)$ and $y'=F(x')$, $d_X(F^{-1}(y),F^{-1}(y'))=d_Y(y,y')$. Applying the same argument to $F^{-1}$ shows that completeness of $Y$ implies completeness of $X$. $\square$

Worked Example: A Relabeling Does Not Change Completeness

Let $X=\mathbb{R}$ with its usual metric and $Y=\{(t,0):t\in\mathbb{R}\}$, with the metric inherited from the Euclidean plane. Define $F:X\to Y$ by $F(t)=(t,0)$. This map is bijective, and for $s,t\in\mathbb{R}$, $$ d_Y(F(s),F(t))=\sqrt{(s-t)^2+(0-0)^2}=|s-t|=d_X(s,t). $$ It is a surjective isometry. Since $\mathbb{R}$ is complete, the theorem shows that $Y$ is complete as well. No new analysis of Cauchy sequences is needed: the isometry transfers their distances exactly.

A Bounded Transform Also Preserves Completeness

Tutorial 651 introduced the bounded metric transform $\rho=d/(1+d)$, which retains the same convergent sequences as $d$. For completeness, it is useful to know more: the transform retains exactly the same Cauchy sequences. This is stronger than agreement about convergence, since completeness is defined using Cauchy sequences.

Theorem (Completeness Under the Bounded Metric Transform): Let $d$ be a metric on $X$, and define $\rho(x,y)=d(x,y)/(1+d(x,y))$. A sequence in $X$ is Cauchy with respect to $d$ if and only if it is Cauchy with respect to $\rho$. Consequently, $(X,d)$ is complete if and only if $(X,\rho)$ is complete.

Proof. For $t\geq0$, the function $t/(1+t)$ is increasing. If a sequence is Cauchy for $d$, let $\varepsilon>0$ and choose $\delta=\min\{\varepsilon,1\}/2$. Whenever $d(x_m,x_n)<\delta$, $$ \rho(x_m,x_n)=\frac{d(x_m,x_n)}{1+d(x_m,x_n)} \leq d(x_m,x_n)<\delta<\varepsilon. $$ Thus the sequence is Cauchy for $\rho$. In the other direction, fix $\varepsilon>0$ and put $\eta=\varepsilon/(1+\varepsilon)$. If $\rho(x_m,x_n)<\eta$, then $$ \frac{d(x_m,x_n)}{1+d(x_m,x_n)}<\frac{\varepsilon}{1+\varepsilon}, $$ which, after multiplying by the positive denominators and cancelling the common term, gives $d(x_m,x_n)<\varepsilon$. The Cauchy condition for $\rho$ therefore implies the Cauchy condition for $d$. Finally, $d$ and $\rho$ have the same convergent sequences by the bounded metric result from Tutorial 651. Since they also have the same Cauchy sequences, every Cauchy sequence converges for one metric exactly when it converges for the other. This proves the completeness equivalence. $\square$

Worked Example: A Bounded Complete Metric on the Real Line

On $\mathbb{R}$, let $d(s,t)=|s-t|$ and define $\rho(s,t)=|s-t|/(1+|s-t|)$. The bounded metric transform theorem from Tutorial 651 ensures that $\rho$ is a metric, and $0\leq\rho(s,t)<1$ for all $s,t$. The real line is complete under $d$; by the theorem just proved, it is also complete under $\rho$.

This example distinguishes boundedness of a metric from boundedness of its space in the original metric. The $\rho$-distances are all less than $1$, but the same set $\mathbb{R}$ is unbounded under the usual metric $d$. Changing the metric changes the numerical distances, while the bounded transform preserves the Cauchy sequences and therefore completeness.

Why Completeness Matters

Completeness guarantees that a process whose successive approximations become mutually close does not run out of points in the space. This is why the property is important in analysis: arguments often first establish that a sequence is Cauchy, then use completeness to obtain a limit that remains available for further reasoning.

The closed-subset results provide a practical test. If a space is presented as a subset of a known complete space, proving that it is closed is enough to establish completeness. Conversely, if the subset is complete, the ambient completeness theorem forces it to be closed. The isometry result lets us transfer completeness across exact distance-preserving identifications, while the bounded-transform result permits a change of metric without changing the Cauchy sequences.

A common pitfall is to confuse “the sequence has a limit in some larger space” with “the sequence converges in this space.” Convergence in a metric space requires the limit to be a point of that space. The example $(1,4)$ makes the distinction explicit: the Cauchy sequence approaches $4$ in $\mathbb{R}$, but $4\notin(1,4)$. Completeness is precisely the guarantee that such a missing limit cannot occur.

Takeaway: A metric space is complete when every Cauchy sequence converges to a point in that space. Closed subsets of complete spaces are complete, and subsets of a complete space are complete exactly when they are closed. Surjective isometries and the bounded metric transform preserve completeness.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What additional requirement does completeness impose beyond the Cauchy condition itself?
In the Closed Subset Theorem, where is completeness used, and where is closedness used?
Why does the sequence $x_n=4-1/n$ show that $(1,4)$ is not complete?
Why does the characterization of complete subsets as closed require the ambient space to be complete?
Which distance identity allows a surjective isometry to transfer Cauchy sequences?
How does the bounded metric transform preserve the Cauchy condition in both directions?

What You'll Learn