Proof of Uniqueness of Metric Limits

Why a Metric Limit Cannot Split in Two

The definition of convergence permits the index after which a sequence is close to its limit to depend on the chosen tolerance. It does not permit the limit to depend on that choice. The Uniqueness of Metric Limits Theorem was established earlier in this course. Here we examine its proof closely: the key idea is to compare two proposed limits through the same sequence term and use the triangle inequality to bound their separation.

The argument is useful beyond confirming that a limit is well-defined. It gives a practical test for rejecting a proposed limit: if a sequence converges to $x$, then it must eventually stay a definite distance away from any different point $y$. A related estimate will also show that two sequences whose corresponding terms become arbitrarily close cannot have different limits.

The Epsilon Argument for Uniqueness

Suppose a sequence $(x_n)$ converges both to $x$ and to $y$ in the same metric space $(X,d)$. If $x\ne y$, their separation $D=d(x,y)$ is positive. Convergence lets us choose one sufficiently late term $x_n$ that is within $D/3$ of both proposed limits. The triangle inequality would then make the distance between the limits less than $2D/3$, contradicting its value $D$.

Recall (Uniqueness of Metric Limits): A sequence in a metric space has at most one limit.

Proof. Suppose $x_n\to x$ and $x_n\to y$. Assume, for contradiction, that $x\ne y$, and set $D=d(x,y)>0$. Apply convergence to $x$ with tolerance $D/3$: there is an index $N_1$ such that $d(x_n,x)<D/3$ whenever $n\geq N_1$. Apply convergence to $y$ with the same tolerance: there is an index $N_2$ such that $d(x_n,y)<D/3$ whenever $n\geq N_2$. Choose $n\geq\max\{N_1,N_2\}$. The triangle inequality gives $$ D=d(x,y)\leq d(x,x_n)+d(x_n,y)<\frac{D}{3}+\frac{D}{3}=\frac{2D}{3}. $$ Since $D>0$, we have $2D/3<D$, which contradicts the displayed inequality. Thus $x=y$. $\square$

There are two details worth keeping visible. First, the same index $n$ must satisfy both estimates; choosing $n$ at least as large as the two starting indices guarantees this. Second, the contradiction depends on $D$ being positive, which follows from $x\ne y$ and the defining property of a metric. If the proposed limits are already equal, there is nothing to prove.

Worked Example: Two Proposed Limits in the Euclidean Plane

Use the Euclidean metric on $\mathbb{R}^2$, and let $$ x_n=\left(2+\frac1n,\,-3+\frac{(-1)^n}{n}\right). $$ The distance from $x_n$ to $x=(2,-3)$ is $$ d_2(x_n,x) =\sqrt{\left(\frac1n\right)^2+\left(\frac{(-1)^n}{n}\right)^2} =\frac{\sqrt{2}}{n}, $$ because $((-1)^n)^2=1$. This tends to zero, so $x_n\to(2,-3)$.

Could the same sequence converge to $y=(2,-2)$? The distance between the proposed limits is $d_2(x,y)=1$. For every $n$, the second-coordinate difference between $x_n$ and $y$ is $$ \left|-3+\frac{(-1)^n}{n}-(-2)\right| =\left|-1+\frac{(-1)^n}{n}\right|. $$ If $n\geq2$, then $(-1)^n/n\leq1/2$, so this absolute value is at least $1/2$. The Euclidean distance is at least the absolute difference in any one coordinate; hence $d_2(x_n,y)\geq1/2$ for every $n\geq2$. In particular, the sequence cannot eventually lie within $1/3$ of $y$, as convergence to $y$ would require. The example exhibits the separation argument concretely: the sequence approaches one point while staying away from the other.

A Quantitative Test for a Wrong Limit

The uniqueness proof can be used in a one-sided way. Instead of assuming convergence to two points and deriving a contradiction, fix the actual limit $x$ and a different point $y$. The distance $d(x,y)$ provides a radius around $y$ that the sequence must eventually avoid.

Theorem (Eventual Exclusion of a Different Point): Let $(X,d)$ be a metric space, suppose $x_n\to x$, and let $y\ne x$. Then, for all sufficiently large $n$, $$ x_n\notin B_{d(x,y)/2}(y). $$

Proof. Put $D=d(x,y)$, so $D>0$. By convergence to $x$, there is an index $N$ such that $d(x_n,x)<D/2$ whenever $n\geq N$. The triangle inequality, applied to the path from $x$ to $y$ through $x_n$, gives $$ D=d(x,y)\leq d(x,x_n)+d(x_n,y). $$ Consequently, for $n\geq N$, $$ d(x_n,y)\geq D-d(x,x_n)>D-\frac{D}{2}=\frac{D}{2}. $$ Thus $x_n$ is outside the open ball of radius $D/2$ centered at $y$ for every $n\geq N$, as claimed. $\square$

This radius is a convenient explicit choice, not a uniquely determined one. The important point is that a positive separation between $x$ and $y$ leaves a positive margin: once the terms are sufficiently close to $x$, they cannot also be close to $y$. This provides a useful way to test a candidate limit. If terms keep entering every neighborhood of $y$ arbitrarily late, then the sequence cannot converge to a different point $x$.

Worked Example: Excluding a Point in the Discrete Metric

Let $X=\{a,b,c\}$ have the discrete metric $\delta$, where $\delta(u,v)=0$ if $u=v$ and $\delta(u,v)=1$ otherwise. Consider a sequence that is eventually equal to $a$. It converges to $a$: for any $\varepsilon>0$, all sufficiently late terms have distance $0<\varepsilon$ from $a$.

Take $y=b$, so $\delta(a,b)=1$. The theorem says that the sequence eventually avoids $B_{1/2}(b)$. Indeed, by the definition of the discrete metric, $$ B_{1/2}(b)=\{z\in X:\delta(z,b)<1/2\}=\{b\}. $$ Every sufficiently late term equals $a$, and $a\notin\{b\}$. The ball makes the separation visible even though the space has no coordinates or usual notion of numerical distance.

When Two Sequences Become Close

A useful extension of the same triangle-inequality method concerns two sequences rather than one. Their terms need not agree. It is enough that the distance between corresponding terms tends to zero. If both sequences have limits, the limiting points must then agree. This result is often useful when one sequence is easier to analyze and another is a small perturbation of it.

Theorem (Limits of Asymptotically Close Sequences): Let $(X,d)$ be a metric space. Suppose $u_n\to u$, $v_n\to v$, and $d(u_n,v_n)\to0$. Then $u=v$.

Proof. Let $\varepsilon>0$. From $u_n\to u$, choose $N_1$ such that $d(u,u_n)<\varepsilon/3$ for $n\geq N_1$. From $d(u_n,v_n)\to0$, choose $N_2$ such that $d(u_n,v_n)<\varepsilon/3$ for $n\geq N_2$. From $v_n\to v$, choose $N_3$ such that $d(v_n,v)<\varepsilon/3$ for $n\geq N_3$. For $n\geq\max\{N_1,N_2,N_3\}$, the triangle inequality gives $$ d(u,v)\leq d(u,u_n)+d(u_n,v_n)+d(v_n,v)<\varepsilon. $$ This holds for every $\varepsilon>0$. If $d(u,v)>0$, choosing $\varepsilon=d(u,v)$ would contradict the inequality. Therefore $d(u,v)=0$, and the defining property of a metric implies $u=v$. $\square$

The three terms in the estimate represent three possible sources of separation: from $u$ to $u_n$, between the two sequence terms, and from $v_n$ to $v$. Making each smaller than one third of the desired tolerance ensures that their sum is smaller than the tolerance. This is a general error-budgeting technique: a triangle inequality with several terms can be controlled by assigning each term a portion of the total error.

Worked Example: A Perturbed Sequence Has the Same Limit

In $\mathbb{R}$ with its usual metric, define $$ u_n=1+\frac1n,\qquad v_n=1+\frac1n+\frac{(-1)^n}{n+1}. $$ First, $u_n\to1$, since $|u_n-1|=1/n\to0$. The two sequences are asymptotically close because $$ |u_n-v_n| =\left|\frac{(-1)^n}{n+1}\right| =\frac{1}{n+1}\longrightarrow0. $$ By the theorem, if $v_n$ has a limit, that limit must be $1$. In fact, its convergence can also be checked directly: $$ |v_n-1| =\left|\frac1n+\frac{(-1)^n}{n+1}\right| \leq\frac1n+\frac1{n+1} \leq\frac{2}{n}. $$ For any $\varepsilon>0$, choose $N>2/\varepsilon$. Then $n\geq N$ gives $|v_n-1|\leq2/n\leq2/N<\varepsilon$, so $v_n\to1$. The asymptotic-closeness theorem identifies the limit from the simpler sequence $u_n$; the final estimate verifies convergence directly.

Uniqueness in Function Metrics

The same proof applies when the points of the metric space are functions. No evaluation at individual inputs is needed if the metric is the supremum metric: the metric axioms already provide the triangle inequality. This can be especially helpful when a sequence of functions has a uniform limit, because the candidate limit is a single point in the function space.

Worked Example: A Function Sequence Cannot Have Two Uniform Limits

Let $X=C_b(\mathbb{R})$, with metric $d(f,g)=\|f-g\|_\infty$. Define $f_n(t)=\cos(t)/(n+1)$, and let $f(t)=0$. Since $|\cos(t)|\leq1$ for every $t$, $$ d(f_n,f)=\sup_{t\in\mathbb{R}}\frac{|\cos(t)|}{n+1}=\frac{1}{n+1}. $$ The equality holds because $|\cos(0)|=1$, and the upper bound holds everywhere. Thus $f_n\to f$ in the supremum metric.

Suppose there were another uniform limit $g\in C_b(\mathbb{R})$. The Uniqueness of Metric Limits Theorem, applied in this function metric space, gives $g=f$. The quantitative exclusion argument is also visible here: if $g\ne f$, then $D=\|f-g\|_\infty>0$, and convergence to $f$ forces $f_n$ eventually outside the supremum-norm ball of radius $D/2$ centered at $g$. But convergence to $g$ would require eventual membership in that same ball, which is impossible.

Common Proof Errors and Why the Argument Matters

A frequent error is to say that both distances $d(x_n,x)$ and $d(x_n,y)$ become small without specifying a single index for which they are both small. Convergence supplies separate starting indices, so the proof must take an index at least as large as both. Another error is to use a tolerance that does not force a contradiction. Choosing $D/3$ for each of the two distances works because their sum is $2D/3<D$. Choosing $D/2$ with strict inequalities also works, but using $D$ would not: the triangle inequality would only give $D<2D$, which is no contradiction.

It is also important to use one metric throughout the argument. The triangle inequality compares distances in a particular metric space. If a sequence is considered in a restricted space, both proposed limits must belong to that space, and the distances are those of its restricted metric. The proof does not require the space to be a subset of the real numbers, complete, or compact; it requires only the metric axioms and convergence as defined in this course.

The central technique is therefore simple but widely reusable: measure the separation between proposed limits, connect them through a common sequence term, and make each part of the resulting triangle inequality smaller than a fixed fraction of that separation. With two nearby sequences, insert the distance between their corresponding terms as another part of the same chain.

Takeaway: The triangle inequality prevents a sequence from approaching two distinct points. More generally, if two sequences approach limits while their corresponding terms become arbitrarily close, those limits must coincide.

Check Your Understanding

Use the uniqueness proof and its extensions to answer the following questions.

If two proposed limits are separated by $D>0$, why is $D/3$ a useful tolerance for each convergence estimate?
Why must the index in the uniqueness proof be at least as large as both starting indices supplied by convergence?
What does the eventual exclusion theorem say about a sequence converging to $x$ and a point $y\ne x$?
In the asymptotic-closeness theorem, what three distances are added in the estimate for $d(u,v)$?
Why does the uniqueness argument apply to functions equipped with the supremum metric, even though the points are functions rather than real numbers?