Prove Uniqueness of a Sequence Limit

From Epsilon Estimates to Uniqueness

In “Prove a Basic Epsilon-N Limit,” the central task was to make the distance between a sequence term and a proposed limit smaller than an arbitrary tolerance. Uniqueness reverses the emphasis: suppose there are two proposed limits, and compare their distance by going through the same sequence term. The triangle inequality makes that distance no larger than the sum of two errors, and both errors can be made as small as needed.

The real-sequence result, Theorem (Uniqueness of Sequence Limits), was established earlier in this course. Here we examine its proof strategy in a setting where sequence terms need not be real numbers. This generalization makes clear which parts of the argument depend on real-number order and which require only a notion of distance.

Definition: Let \((X,d)\) be a metric space, let \((x_n)\) be a sequence in \(X\), and let \(x\in X\). The sequence \((x_n)\) converges to \(x\) if for every \(\varepsilon>0\), there is a positive integer \(N\) such that \(d(x_n,x)<\varepsilon\) for every integer \(n\geq N\). We write \(x_n\to x\).

The notation changes from absolute error \(|a_n-L|\) to metric distance \(d(x_n,x)\). The quantifier structure is otherwise the same: one threshold must control every term in the tail. In a uniqueness proof, two limits may initially provide different thresholds. Taking the larger threshold ensures that both estimates hold for the same terms.

Name both candidates.
Assume \(x_n\to x\) and \(x_n\to y\), and identify the distance \(d(x,y)\) that must be shown to be zero.

Fix an arbitrary tolerance.
Choose \(\varepsilon>0\). Convergence to each candidate provides a threshold for an error smaller than \(\varepsilon/2\).

Use one common tail.
Take the larger of the two thresholds, so that both distance estimates hold for every later term.

Apply the triangle inequality.
Compare \(x\) to \(y\) by passing through \(x_n\), then let the arbitrary tolerance force \(d(x,y)=0\).

Uniqueness in a Metric Space

The key inequality is valid for every index \(n\), whether or not the sequence converges. What convergence supplies is the ability to make both terms on its right-hand side small at once.

Theorem (Uniqueness of Limits in a Metric Space): Let \((X,d)\) be a metric space and let \((x_n)\) be a sequence in \(X\). If \(x_n\to x\) and \(x_n\to y\), then \(x=y\).

Proof. Let \(\varepsilon>0\) be arbitrary. Since \(x_n\to x\), there is a positive integer \(N_x\) such that \(d(x_n,x)<\varepsilon/2\) for every \(n\geq N_x\). Since \(x_n\to y\), there is a positive integer \(N_y\) such that \(d(x_n,y)<\varepsilon/2\) for every \(n\geq N_y\). Define \(N=\max\{N_x,N_y\}\). For every \(n\geq N\), both estimates hold. The triangle inequality gives

d(x,y)\leq d(x,x_n)+d(x_n,y)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.

Thus \(d(x,y)<\varepsilon\) for every \(\varepsilon>0\). This forces \(d(x,y)=0\): if \(d(x,y)>0\), choosing \(\varepsilon=d(x,y)\) would contradict the strict inequality just proved. By the defining property of a metric, \(d(x,y)=0\) implies \(x=y\). Therefore the limit is unique. \(\square\)

The choice \(\varepsilon/2\) is a convenient way to split the available tolerance between the two errors. More generally, one could ask each error to be smaller than any pair of positive bounds whose sum is at most \(\varepsilon\). What matters is that the two estimates apply to the same \(n\), which is why the proof takes the maximum of the thresholds.

Worked Example: A Sequence in the Plane

Consider the Euclidean metric on \(\mathbb{R}^2\), given by \(d((u_1,u_2),(v_1,v_2))=\sqrt{(u_1-v_1)^2+(u_2-v_2)^2}\). Define \(x_n=(3+1/n,-2-2/n)\), and let \(p=(3,-2)\). Direct calculation gives

d(x_n,p) =\sqrt{\left(3+\frac1n-3\right)^2+\left(-2-\frac2n+2\right)^2} =\sqrt{\frac{1}{n^2}+\frac{4}{n^2}} =\frac{\sqrt{5}}{n}.

Given \(\varepsilon>0\), choose a positive integer \(N>\sqrt{5}/\varepsilon\). Such an integer exists, and for \(n\geq N\), \(\sqrt{5}/n\leq\sqrt{5}/N<\varepsilon\). Therefore \(x_n\to p\). The uniqueness theorem now shows that no point other than \(p\) can also be the limit of this sequence.

For instance, \(q=(4,-2)\) is not another limit: \(d(p,q)=1\), so the theorem rules out \(x_n\to q\). This conclusion does not require a separate coordinate-by-coordinate analysis of a hypothetical convergence to \(q\); it follows from the general metric-space argument.

Fixed Separation as a Test for False Limits

The uniqueness theorem has a useful contrapositive form for practice: if two points have positive distance, a sequence cannot converge to both. A related geometric fact explains why. Around two distinct points, sufficiently small open balls do not overlap.

Theorem (Disjoint Balls Around Distinct Points): Let \((X,d)\) be a metric space, and let \(p,q\in X\) with \(p\neq q\). The open balls of radius \(d(p,q)/2\) centered at \(p\) and \(q\) are disjoint.

Proof. Set \(\delta=d(p,q)\). Because \(p\neq q\) and \(d\) is a metric, \(\delta>0\). Suppose, to obtain a contradiction, that some \(z\in X\) belongs to both open balls. Then \(d(p,z)<\delta/2\) and \(d(z,q)<\delta/2\). The triangle inequality would imply

\delta=d(p,q)\leq d(p,z)+d(z,q)<\frac{\delta}{2}+\frac{\delta}{2}=\delta,

which is impossible. Hence the two open balls are disjoint. \(\square\)

If a sequence converged to both \(p\) and \(q\), then sufficiently late terms would have to lie in both of these balls. The theorem shows that this cannot happen. This picture is useful, but the epsilon proof is still important: it records the thresholds and verifies exactly why the two required estimates hold simultaneously.

Worked Example: Rejecting a Proposed Limit on the Real Line

Let \(a_n=1/(n+1)\). We first verify that \(a_n\to0\). Given \(\varepsilon>0\), choose a positive integer \(N>1/\varepsilon\). For every \(n\geq N\), the denominator is positive and \(n+1\geq n\geq N\), so

|a_n-0|=\frac{1}{n+1}\leq\frac{1}{N}<\varepsilon.

Thus \(0\) is a limit. Now consider the proposed alternative \(1\). For every positive integer \(n\), \(0<1/(n+1)\leq1/2\), and hence

|a_n-1|=1-\frac{1}{n+1}\geq\frac12.

Convergence to \(1\) would require, for \(\varepsilon=1/3\), that \(|a_n-1|<1/3\) for every sufficiently large \(n\). The displayed lower bound shows this never happens. So \(1\) is not a limit. This example uses a direct fixed-tolerance test to reject a candidate; the uniqueness theorem gives the broader conclusion that any limit, if one exists, must be \(0\).

What the Proof Uses—and What It Does Not

For real sequences, the earlier Theorem (Uniqueness of Sequence Limits) is the familiar special case with \(X=\mathbb{R}\) and \(d(u,v)=|u-v|\). The proof in a metric space depends only on three ingredients: convergence is defined through distance, the triangle inequality holds, and distance zero occurs only between equal points. It does not use completeness, order, compactness, or any formula for the sequence.

That distinction prevents a common proof error: trying to compare two proposed limits by subtracting them when the sequence values are not numbers. In an arbitrary metric space, subtraction may not even be defined. Instead, compare distances and pass through a sequence term:

d(x,y)\leq d(x,x_n)+d(x_n,y).

Another frequent gap is using two separate convergence thresholds without checking that the same index satisfies both estimates. The maximum \(N=\max\{N_x,N_y\}\) resolves this: every \(n\geq N\) also satisfies \(n\geq N_x\) and \(n\geq N_y\). Finally, the proof must turn “smaller than every positive tolerance” into zero. The contradiction choice \(\varepsilon=d(x,y)\) supplies that last step explicitly.

Worked Example: Convergence in the Discrete Metric

Let \(X\) be any set with the discrete metric \(d(u,v)=0\) when \(u=v\), and \(d(u,v)=1\) when \(u\neq v\). A sequence \((x_n)\) converges to \(x\) exactly when it is eventually equal to \(x\). To prove the forward direction, use convergence with \(\varepsilon=1/2\). There is an \(N\) such that \(d(x_n,x)<1/2\) for every \(n\geq N\). Since the only possible distances are \(0\) and \(1\), this forces \(d(x_n,x)=0\), hence \(x_n=x\), for all \(n\geq N\).

Conversely, suppose \(x_n=x\) for every \(n\geq N_0\). For any \(\varepsilon>0\) and every \(n\geq N_0\), \(d(x_n,x)=0<\varepsilon\), so \(x_n\to x\). If the same sequence converged to \(y\) as well, it would eventually equal \(x\) and eventually equal \(y\). Taking an index beyond both starting points gives \(x_n=x\) and \(x_n=y\) at that index, so \(x=y\). This concrete case illustrates the general theorem: the distance may have a particularly simple form, but uniqueness still comes from the same metric structure.

When asked to prove uniqueness, first name the distance between the proposed limits. Then fix an arbitrary tolerance, obtain one threshold for each convergence statement, and take the larger threshold. The triangle inequality links the two errors to the fixed distance. This proof pattern is short, but each of its steps protects a necessary part of the argument.

Check Your Understanding

Use the metric definition and the triangle inequality to answer each question.

Why does a uniqueness proof take the maximum of the two thresholds supplied by convergence?
In the metric-space proof, why is it useful to request an error smaller than \(\varepsilon/2\) from each proposed limit?
Which defining property of a metric turns \(d(x,y)=0\) into \(x=y\)?
Why are the open balls of radius \(d(p,q)/2\) around distinct points disjoint?
For the discrete metric, which tolerance forces a convergent sequence to be eventually equal to its limit?
Which parts of the metric-space uniqueness proof would not require the sequence terms to be real numbers?

Prove Uniqueness of a Sequence Limit

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