Metric-Space Convergence and Epsilon Balls

Convergence Means Eventually Entering Every Epsilon-Ball

In a metric space, the distance from each term of a sequence to a proposed limit gives a precise way to describe convergence. The key quantifiers are “for every radius” and “from some index onward”: a sequence may begin far from its limit, but its entire tail must fit inside every ball centered at that limit.

Let \((X,d)\) be a metric space, and let \((x_n)\) be a sequence in \(X\). For \(x\in X\) and \(\varepsilon>0\), the open ball of radius \(\varepsilon\) about \(x\) is \(B_d(x,\varepsilon)=\{y\in X:d(y,x)<\varepsilon\}\).

Definition: The sequence \((x_n)\) converges to \(x\in X\) if for every \(\varepsilon>0\) there is an \(N\in\mathbb{N}\) such that, for every \(n\ge N\), \(d(x_n,x)<\varepsilon\). We write \(x_n\to x\). The index \(N\) may depend on \(\varepsilon\). Equivalently, for every \(\varepsilon>0\), all terms from some index onward lie in \(B_d(x,\varepsilon)\).

The radius \(\varepsilon\) is arbitrary, not a radius selected once and for all. For a small radius, the sequence may need to advance farther before all its terms lie in the ball. The inequality is strict: membership in the open ball means distance \(<\varepsilon\), not distance \(\le\varepsilon\).

Proposition (Ball test for convergence): A sequence \((x_n)\) converges to \(x\) if and only if for every open ball \(B_d(x,\varepsilon)\) with \(\varepsilon>0\), there is an \(N\in\mathbb{N}\) such that \(x_n\in B_d(x,\varepsilon)\) for every \(n\ge N\).

Proof. Suppose \(x_n\to x\). Given any \(\varepsilon>0\), the definition supplies \(N\in\mathbb{N}\) such that \(d(x_n,x)<\varepsilon\) whenever \(n\ge N\). By the definition of the ball, this says exactly that \(x_n\in B_d(x,\varepsilon)\) for every \(n\ge N\). Conversely, suppose the stated ball condition holds. Given any \(\varepsilon>0\), apply it to \(B_d(x,\varepsilon)\). The resulting \(N\) satisfies \(d(x_n,x)<\varepsilon\) for every \(n\ge N\), which is the definition of \(x_n\to x\). \(\square\)

A Distance Estimate That Proves Convergence

In computations, one often finds a formula or estimate for \(d(x_n,x)\). The following criterion makes the connection with convergence explicit. For a real sequence \((r_n)\), the statement \(r_n\to0\) means that for every \(\varepsilon>0\), there is an \(N\in\mathbb{N}\) such that \(|r_n|<\varepsilon\) whenever \(n\ge N\).

Theorem (Distance-to-limit criterion): For a metric space \((X,d)\), a sequence \((x_n)\) converges to \(x\in X\) if and only if the real sequence \(\bigl(d(x_n,x)\bigr)\) converges to \(0\).

Proof. Each \(d(x_n,x)\) is nonnegative, so \(|d(x_n,x)|=d(x_n,x)\). If \(x_n\to x\), then for every \(\varepsilon>0\) there is an \(N\) such that \(d(x_n,x)<\varepsilon\) for all \(n\ge N\). This is precisely \(d(x_n,x)\to0\) as a real sequence. Conversely, if \(d(x_n,x)\to0\), then for every \(\varepsilon>0\) there is an \(N\) such that \(|d(x_n,x)|<\varepsilon\) for all \(n\ge N\). Nonnegativity gives \(d(x_n,x)<\varepsilon\), so the definition gives \(x_n\to x\). \(\square\)

A useful consequence is that an upper bound tending to zero is enough; the exact distance need not be simplified all the way.

Lemma: Suppose \(x\in X\) and \(c_n\ge0\) for each \(n\in\mathbb{N}\). If \(d(x_n,x)\le c_n\) for every \(n\), and \(c_n\to0\) in \(\mathbb{R}\), then \(x_n\to x\).

Proof. Let \(\varepsilon>0\). Since \(c_n\to0\), there is \(N\in\mathbb{N}\) such that \(c_n<\varepsilon\) for every \(n\ge N\). For such \(n\), the assumed bound gives \(d(x_n,x)\le c_n<\varepsilon\). This holds for every \(n\ge N\), so \(x_n\to x\). \(\square\)

Worked Example: Convergence in the Real Line

Give \(\mathbb{R}\) its standard metric \(d(s,t)=|s-t|\). For \(n\in\mathbb{N}\), let \(x_n=(3n+1)/(n+2)\). We will prove that \(x_n\to3\) by finding a tail index for each \(\varepsilon>0\).

First calculate the distance exactly:

d(x_n,3)=\left|\frac{3n+1}{n+2}-3\right| =\left|\frac{3n+1-3(n+2)}{n+2}\right| =\frac{5}{n+2}.

Given \(\varepsilon>0\), choose \(N=\lfloor 5/\varepsilon\rfloor+1\), where \(\lfloor t\rfloor\) is the greatest integer less than or equal to \(t\). Then \(N>5/\varepsilon\). For every \(n\ge N\), \(n+2\ge N+2>5/\varepsilon\), and therefore \(d(x_n,3)=5/(n+2)<\varepsilon\). This proves convergence.

Check with a particular radius: if \(\varepsilon=0.1\), then \(5/\varepsilon=50\), so this choice gives \(N=51\). At \(n=51\), \(x_{51}=154/53\), and \(d(x_{51},3)=|154/53-159/53|=5/53\). Since \(5/53<0.1\) (equivalently, \(500<530\)), the bound holds at the first index in this chosen tail. The general calculation proves it for every \(n\ge51\).

Using the Euclidean Metric

In \(\mathbb{R}^2\), equipped with its Euclidean metric \(d(u,v)=\|u-v\|\), the distance to a proposed limit is the Euclidean norm of the difference vector. A direct estimate of that norm gives an epsilon-ball proof just as it does on the real line.

Worked Example: A Sequence in the Plane

In \(\mathbb{R}^2\) with the Euclidean metric, consider \(x_n=(2+1/n,-1+2/n)\) and \(p=(2,-1)\). Subtract the proposed limit coordinate by coordinate:

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

Given \(\varepsilon>0\), choose \(N=\lfloor\sqrt{5}/\varepsilon\rfloor+1\). Then \(N>\sqrt{5}/\varepsilon\), so for every \(n\ge N\), \(n\ge N>\sqrt{5}/\varepsilon\) and \(d(x_n,p)=\sqrt{5}/n<\varepsilon\). Hence \(x_n\to p\) in the Euclidean metric.

Check with \(\varepsilon=0.1\): since \(\sqrt{5}/0.1=10\sqrt{5}\), and \(22<10\sqrt{5}<23\) (squaring the positive quantities gives \(484<500<529\)), the prescribed index is \(N=23\). At that index, \(x_{23}=(47/23,-21/23)\), so \(x_{23}-p=(1/23,2/23)\) and \(d(x_{23},p)=\sqrt{5}/23\). This is less than \(0.1\), because \(5/529<1/100\), or \(500<529\). The distance formula also gives the required inequality for every \(n\ge23\).

What the Discrete Metric Makes Convergence Mean

On a set \(X\), the discrete metric is \(d(x,y)=0\) when \(x=y\) and \(d(x,y)=1\) when \(x\ne y\). Its distances take only two values. As a result, a ball of radius \(1/2\) around \(x\) contains \(x\) and no other point. This turns the epsilon-ball condition into an exact statement about the terms of a sequence.

Theorem (Convergence for the discrete metric): Let \(X\) have the discrete metric. A sequence \((x_n)\) converges to \(x\in X\) if and only if there is an \(N\in\mathbb{N}\) such that \(x_n=x\) for every \(n\ge N\).

Proof. Suppose \(x_n\to x\). Apply the definition with \(\varepsilon=1/2\). For some \(N\), \(d(x_n,x)<1/2\) whenever \(n\ge N\). In the discrete metric, the distance is either \(0\) or \(1\), and only \(0\) is less than \(1/2\). Thus \(x_n=x\) for every \(n\ge N\). Conversely, suppose \(x_n=x\) for every \(n\ge N\). For any \(\varepsilon>0\), the distance \(d(x_n,x)=0<\varepsilon\) for every \(n\ge N\). This is convergence to \(x\). The argument also applies when \(X\) has just one point. \(\square\)

Worked Example: An Eventually Constant Sequence

Let \(X=\{a,b,c\}\) have the discrete metric, and define \(x_1=b\), \(x_2=c\), \(x_3=b\), and \(x_n=a\) for every \(n\ge4\). We verify convergence to \(a\) using balls. For any \(\varepsilon>0\) and every \(n\ge4\), \(x_n=a\), so \(d(x_n,a)=0<\varepsilon\). Thus \(x_n\to a\).

Check at two radii: \(B_d(a,1/2)=\{a\}\), since \(d(a,a)=0<1/2\) and \(d(b,a)=d(c,a)=1\), which is not less than \(1/2\). Also \(B_d(a,2)=X\), since all three distances from \(a\) are either \(0\) or \(1\), both less than \(2\). From index \(4\) onward the terms lie in both balls, as required; the radius \(1/2\) tests the more restrictive condition.

Conversely, the theorem shows why a sequence that keeps taking a value different from \(a\) cannot converge to \(a\) in this metric: at radius \(1/2\), every term in the tail would have to equal \(a\).

A Common Epsilon Mistake

Common pitfall. To prove \(x_n\to0\) for \(x_n=1/n\) in the standard metric on \(\mathbb{R}\), one sometimes writes: “Choose \(N> \varepsilon\); then \(1/n<\varepsilon\) for \(n\ge N\).” The inequality used here has the wrong dependence on \(\varepsilon\). For example, with \(\varepsilon=0.1\), the choice \(N=1\) satisfies \(N>0.1\), but at \(n=1\) the distance is \(1\), which is not less than \(0.1\).

The correction comes from solving the distance inequality for \(n\): \(1/n<\varepsilon\) holds when \(n>1/\varepsilon\). For each \(\varepsilon>0\), choose \(N=\lfloor1/\varepsilon\rfloor+1\). Then \(n\ge N\) implies \(n\ge N>1/\varepsilon\), and consequently \(d(x_n,0)=1/n<\varepsilon\). For the specific radius \(\varepsilon=0.1\), this gives \(N=11\); at \(n=11\), the distance is \(1/11<0.1\), since \(10<11\), and it is no larger for any \(n\ge11\).

The central habit is to begin with the required inequality \(d(x_n,x)<\varepsilon\), then choose \(N\) so that it holds for every \(n\ge N\). A calculation at one index, or a choice of \(N\) that does not depend correctly on \(\varepsilon\), does not establish convergence.

Check Your Understanding

For each question, pay attention to the metric and to whether the required inequality is strict.

In \(\mathbb{R}\) with its standard metric, prove directly from the definition that \(x_n=4-2/(n+1)\) converges to \(4\). Give a valid choice of \(N\) for an arbitrary \(\varepsilon>0\).
In the Euclidean metric on \(\mathbb{R}^2\), compute the distance from \(y_n=(1-3/n,2+4/n)\) to \(q=(1,2)\), and use it to prove convergence to \(q\).
In a discrete metric space, can a sequence converge to \(x\) if \(x_n\ne x\) for infinitely many indices? Justify your answer using the radius \(1/2\).
For \(x_n=1/n\) in \(\mathbb{R}\), what is the distance to \(0\), and what index condition ensures this distance is less than a given \(\varepsilon>0\)?
Explain why showing \(d(x_N,x)<\varepsilon\) for one index \(N\) is not by itself enough to prove convergence.