Convergence in Metric Spaces

Convergence as an Eventually Local Condition

Neighborhoods describe which points lie sufficiently close to a specified point. Convergence uses this local idea to describe the long-term behavior of a sequence: its terms may begin anywhere, but eventually they must lie as close to the proposed limit as any prescribed positive distance requires. The metric determines what “close” means, so the same sequence of points can behave differently when considered in different metric spaces.

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 an index $N\in\mathbb{N}$ such that $d(x_n,x)<\varepsilon$ whenever $n\geq N$. We write $x_n\to x$ or $\lim_{n\to\infty}x_n=x$.

The index $N$ may depend on $\varepsilon$. A smaller tolerance can require a later starting index. The condition makes no demand on the finitely many terms before $N$; it requires only that every term from $N$ onward lie in the open ball $B_\varepsilon(x)$. In other words, each such ball contains a tail of the sequence.

Convergence is always relative to a chosen metric and space. In $\mathbb{R}$ with its usual metric, the condition becomes $|x_n-x|<\varepsilon$ eventually. In a different metric, the distances being tested may differ, and in a restricted space the possible limit must belong to that space. These distinctions will matter in the examples below.

Worked Example: A Rational Sequence in the Usual Metric

In $\mathbb{R}$ with the usual metric, consider $x_n=(3n-2)/(n+4)$. To test convergence to $3$, compute the distance exactly: $$ d(x_n,3)=\left|\frac{3n-2}{n+4}-3\right| =\left|\frac{3n-2-3n-12}{n+4}\right| =\frac{14}{n+4}. $$ For any $\varepsilon>0$, choose an integer $N\geq1$ with $N>14/\varepsilon-4$. If $n\geq N$, then $n+4\geq N+4>14/\varepsilon$, and consequently $$ d(x_n,3)=\frac{14}{n+4}<\varepsilon. $$ Thus $x_n\to3$. The calculation identifies exactly how large the index must be to meet a given tolerance.

Convergence and Neighborhoods

The definition can be phrased without mentioning a particular radius in advance. Every neighborhood of the proposed limit must contain all sufficiently late terms. This follows from the fact that a neighborhood contains a ball centered at that point, established by the Metric-Ball Neighborhood Basis Theorem.

Theorem (Neighborhood Characterization of Convergence): Let $(X,d)$ be a metric space, let $(x_n)$ be a sequence in $X$, and let $x\in X$. Then $x_n\to x$ if and only if, for every neighborhood $N$ of $x$, there is an index $N_0$ such that $x_n\in N$ whenever $n\geq N_0$.

Proof. Suppose $x_n\to x$, and let $N$ be a neighborhood of $x$. By the Metric-Ball Neighborhood Basis Theorem, there is an $r>0$ such that $B_r(x)\subseteq N$. Convergence gives an index $N_0$ such that $d(x_n,x)<r$ whenever $n\geq N_0$. Thus $x_n\in B_r(x)\subseteq N$ for every $n\geq N_0$.

Conversely, suppose every neighborhood of $x$ contains all sufficiently late terms. For any $\varepsilon>0$, the ball $B_\varepsilon(x)$ is a neighborhood of $x$. By the assumed condition, there is an $N_0$ such that $x_n\in B_\varepsilon(x)$ whenever $n\geq N_0$. This means $d(x_n,x)<\varepsilon$ for all such $n$, which is the definition of $x_n\to x$. $\square$

This characterization explains why convergence is a tail condition: changing finitely many terms cannot affect whether a sequence converges to a specified point. It also clarifies why merely having infinitely many terms close to $x$ is not enough. Convergence requires that, for each neighborhood, every term after some index stays inside it.

Worked Example: Convergence in a Product Metric

Give $\mathbb{R}^2$ the maximum metric $$ d_\infty\bigl((a,b),(c,e)\bigr)=\max\{|a-c|,|b-e|\}. $$ Consider $z_n=(1/n,(-1)^n/n)$ and $z=(0,0)$. The two coordinate distances are $$ \left|\frac{1}{n}-0\right|=\frac{1}{n}, \qquad \left|\frac{(-1)^n}{n}-0\right|=\frac{1}{n}, $$ because $|(-1)^n|=1$. Therefore $$ d_\infty(z_n,z)=\max\left\{\frac{1}{n},\frac{1}{n}\right\}=\frac{1}{n}. $$ Given $\varepsilon>0$, choose an integer $N>1/\varepsilon$. Then $n\geq N$ implies $d_\infty(z_n,z)=1/n\leq1/N<\varepsilon$, so $z_n\to(0,0)$. The alternating sign in the second coordinate does not prevent convergence because its magnitude tends to zero.

The same calculation illustrates the role of the chosen product metric: for the maximum metric, both coordinate errors must be small, and the larger of the two errors is the distance.

Subsequences Preserve a Limit

A subsequence selects terms at strictly increasing indices. Since its indices eventually exceed every fixed index, it cannot keep selecting terms from an initial portion of the original sequence. As a result, convergence passes to every subsequence.

Theorem (Convergence of Subsequences): If $x_n\to x$ in a metric space and $(x_{n_k})$ is a subsequence, then $x_{n_k}\to x$.

Proof. Let $\varepsilon>0$. Since $x_n\to x$, there is an index $N$ such that $d(x_n,x)<\varepsilon$ whenever $n\geq N$. The subsequence indices satisfy $n_k\geq k$, so for every $k\geq N$, we have $n_k\geq k\geq N$. It follows that $$ d(x_{n_k},x)<\varepsilon \qquad\text{for every }k\geq N. $$ This is precisely the definition of $x_{n_k}\to x$. $\square$

The theorem is useful when a complicated sequence is studied through simpler selections of its terms. It also gives a way to disprove convergence: if one can find a subsequence that does not converge to a proposed limit, then the original sequence cannot converge to that limit. A subsequence does not have to contain consecutive terms; it does have to preserve the increasing order of the indices.

Worked Example: A Uniformly Convergent Function Sequence

Let $X=C_b(\mathbb{R})$, the space of bounded continuous real-valued functions on $\mathbb{R}$, with the metric induced by the supremum norm: $$ d(f,g)=\|f-g\|_\infty=\sup_{t\in\mathbb{R}}|f(t)-g(t)|. $$ For each positive integer $n$, define $f_n(t)=t/(n(1+|t|))$, and let $f(t)=0$. Each $f_n$ is continuous and bounded. For every $t\in\mathbb{R}$, $|t|\leq1+|t|$, so $$ |f_n(t)-f(t)|=\frac{|t|}{n(1+|t|)}\leq\frac{1}{n}. $$ Taking the supremum over $t$ gives $\|f_n-f\|_\infty\leq1/n$. Given $\varepsilon>0$, choose an integer $N>1/\varepsilon$. For $n\geq N$, $$ d(f_n,f)=\|f_n-f\|_\infty\leq\frac{1}{n}\leq\frac{1}{N}<\varepsilon. $$ Therefore $f_n\to f$ in this metric. The estimate is uniform in $t$: one index $N$ works for every point of the domain.

Distances Behave Well Under Convergence

The reverse triangle inequality for a metric controls how much the distance to a fixed point can change when one endpoint moves. It therefore converts convergence of points into an estimate about real-valued distances.

Theorem (Continuity of Distance to a Fixed Point): Let $(X,d)$ be a metric space. If $x_n\to x$ and $y\in X$ is fixed, then $d(x_n,y)\to d(x,y)$.

Proof. By the reverse triangle inequality for a metric, $$ \bigl|d(x_n,y)-d(x,y)\bigr|\leq d(x_n,x). $$ Let $\varepsilon>0$. Since $x_n\to x$, there is an index $N$ such that $d(x_n,x)<\varepsilon$ whenever $n\geq N$. For those indices, $$ \bigl|d(x_n,y)-d(x,y)\bigr|\leq d(x_n,x)<\varepsilon. $$ This is the definition of convergence in $\mathbb{R}$, so $d(x_n,y)\to d(x,y)$. $\square$

The same reasoning works when both points vary, provided both sequences converge. Indeed, the triangle inequality gives $$ \bigl|d(x_n,y_n)-d(x,y)\bigr| \leq d(x_n,x)+d(y_n,y). $$ To see this, first use the reverse triangle inequality to bound the left side by $d(x_n,x)+d(y_n,y)$, applying the triangle inequality in both directions between the pairs. If $x_n\to x$ and $y_n\to y$, then for any $\varepsilon>0$, each term on the right is eventually less than $\varepsilon/2$. Hence $d(x_n,y_n)\to d(x,y)$. This estimate is often more useful than expanding the distance formula, especially in metric spaces without coordinates.

Relative Spaces and Common Pitfalls

When $A\subseteq X$ is equipped with the restricted metric, convergence in $A$ requires the sequence and its limit to belong to $A$. The distances between points of $A$ are inherited from $X$, but a point outside $A$ is not a possible limit in the metric space $A$. Thus, convergence in an ambient space does not by itself establish convergence to a point of a restricted space.

Worked Example: Ambient Convergence Without a Limit in the Restricted Space

Let $A=(0,1)$ with the restricted usual metric, and consider $x_n=1/n$, which belongs to $A$ for every $n\geq2$. In the ambient space $\mathbb{R}$, the sequence converges to $0$, since for every $\varepsilon>0$, choosing an integer $N>1/\varepsilon$ ensures $1/n<\varepsilon$ for $n\geq N$. But $0\notin A$, so this is not convergence to a point of the metric space $A$.

In fact, the sequence has no limit in $A$. Suppose it converged in $A$ to some $a\in(0,1)$. Its distances to $0$, computed in the ambient real line, would satisfy $$ a=\left|a-0\right|\leq\left|a-\frac1n\right|+\frac1n. $$ Convergence to $a$ in $A$ would make $|a-1/n|\to0$, while $1/n\to0$. The right side would therefore tend to $0$, forcing $a=0$, contrary to $a\in(0,1)$. This illustrates why the space containing a sequence is part of the convergence question.

Another common mistake is to confuse “infinitely many terms enter every ball” with convergence. For example, in $\mathbb{R}$, the sequence $x_n=(-1)^n$ has infinitely many terms equal to $1$ and infinitely many equal to $-1$. Balls around $1$ contain infinitely many terms, but not all sufficiently late terms: the terms equal to $-1$ keep appearing. Convergence requires eventual membership in each ball, not merely repeated visits. The Uniqueness of Metric Limits Theorem, established earlier in this course, also ensures that a sequence cannot have two distinct limits in the same metric space.

Takeaway: A sequence converges to $x$ when every ball, or equivalently every neighborhood, of $x$ contains all sufficiently late terms. Subsequences preserve convergence, and metric distance estimates let convergence pass to distances. Always specify the metric space in which the sequence and its limit are being considered.

Check Your Understanding

Use the definitions and results above to answer the following questions.

In the definition of convergence, why may the index $N$ depend on $\varepsilon$, and what must hold for every $n\geq N$?
How does eventual membership in every neighborhood of $x$ imply convergence to $x$?
Why must every subsequence of a convergent sequence converge to the same limit?
If $x_n\to x$, what estimate relates $|d(x_n,y)-d(x,y)|$ to $d(x_n,x)$, and what does it imply?
Why does convergence in an ambient space not always give convergence to a point of a restricted metric space?

What You'll Learn