Convergence in Subspaces

Convergence When the Space Changes

A subspace keeps the distances between its points, but it has fewer points available as possible limits. That distinction matters for convergence. A sequence in a subset $A$ may approach a point of the ambient space $X$ that is not in $A$; the sequence then has an ambient limit, but that point cannot be its limit in the subspace $A$.

The restricted metric makes the basic comparison precise. Whenever the proposed limit belongs to $A$, convergence in $A$ and convergence in $X$ are equivalent: the distances used in the two convergence tests are identical. We will also use the Sequential Characterization of Closed Sets from earlier in this course to see what changes when $A$ is closed.

Definition: Let $(X,d)$ be a metric space, let $A$ be a nonempty subset of $X$, and equip $A$ with the restricted metric $d_A$. A sequence $(a_n)$ in $A$ converges in $A$ to $a\in A$ if, for every $\varepsilon>0$, there is an index $N$ such that $d_A(a_n,a)<\varepsilon$ whenever $n\geq N$. We write $a_n\to a$ in $A$, or specify the space when needed.

The limit in this definition must be a point of $A$. This is not an additional condition imposed on the distances; it is part of what it means to be a limit in the metric space $(A,d_A)$. In contrast, when we say a sequence in $A$ converges in $X$, its proposed limit is allowed to be any point of $X$.

The Same Limit Test at a Point of the Subspace

Suppose $(a_n)$ is a sequence in $A$, and $a\in A$. For every index $n$, both $a_n$ and $a$ are in the subspace. Consequently, the definition of the restricted metric gives $d_A(a_n,a)=d(a_n,a)$. The two convergence tests therefore use the same real number for every $n$.

Theorem (Convergence in a Subspace): Let $(X,d)$ be a metric space, let $A\subseteq X$ be nonempty, and let $(a_n)$ be a sequence in $A$. For every $a\in A$, the sequence $(a_n)$ converges to $a$ in $A$ if and only if it converges to $a$ in $X$.

Proof. Fix $a\in A$. For each $n$, the points $a_n,a\in A$, so $d_A(a_n,a)=d(a_n,a)$. If $a_n\to a$ in $A$, then for every $\varepsilon>0$, some $N$ satisfies $d_A(a_n,a)<\varepsilon$ for all $n\geq N$. The equality of distances gives $d(a_n,a)<\varepsilon$ for all $n\geq N$, so $a_n\to a$ in $X$. Conversely, if $a_n\to a$ in $X$, then for every $\varepsilon>0$, some $N$ satisfies $d(a_n,a)<\varepsilon$ for all $n\geq N$. The same equality gives $d_A(a_n,a)<\varepsilon$ for all $n\geq N$, so $a_n\to a$ in $A$. $\square$

The requirement $a\in A$ is essential. The theorem compares convergence to a point that is available in both spaces. It does not say that every ambient limit of a sequence in $A$ is also a limit in $A$.

Worked Example: Convergence at an Included Endpoint

Let $A=[0,2]$ be a subspace of $\mathbb{R}$, with its usual metric, and define $a_n=2/(n+1)$ for $n\geq1$. Each $a_n$ belongs to $A$, and the proposed limit $0$ also belongs to $A$. In the ambient space, $$ |a_n-0|=\frac{2}{n+1}. $$ Given $\varepsilon>0$, choose an integer $N$ such that $N+1>2/\varepsilon$. Then for every $n\geq N$, $$ d_A(a_n,0)=d(a_n,0)=\frac{2}{n+1}\leq\frac{2}{N+1}<\varepsilon. $$ Thus $a_n\to0$ in $A$, as well as in $\mathbb{R}$. The endpoint is included in $A$, so it is a legitimate subspace limit.

When the Ambient Limit Is Missing

Now suppose $a_n\in A$ and $a_n\to x$ in $X$, but $x\notin A$. The sequence cannot converge in $A$ to $x$, because $x$ is not a point of $A$. Nor can it converge in $A$ to a different point $a\in A$: the Convergence in a Subspace Theorem would then imply $a_n\to a$ in $X$, contradicting uniqueness of metric limits, established earlier in the course.

Corollary: If a sequence in $A$ converges in $X$ to a point $x\notin A$, it has no limit in the subspace $A$.

Proof. Suppose instead that the sequence converges in $A$ to some $a\in A$. By the Convergence in a Subspace Theorem, it converges in $X$ to $a$. It already converges in $X$ to $x$. Uniqueness of metric limits gives $a=x$, which is impossible because $a\in A$ and $x\notin A$. Therefore it has no limit in $A$. $\square$

Worked Example: A Sequence Approaching a Missing Endpoint

Let $A=(0,1)\subseteq\mathbb{R}$, and take $a_n=1/(n+1)$ for $n\geq1$. Every term belongs to $A$. In $\mathbb{R}$, the sequence converges to $0$, since for every $\varepsilon>0$, choosing $N$ with $N+1>1/\varepsilon$ gives $$ |a_n-0|=\frac{1}{n+1}\leq\frac{1}{N+1}<\varepsilon $$ for all $n\geq N$. But $0\notin A$, so this is not convergence in $A$. The corollary shows that the sequence has no limit in $A$ at all. In particular, it cannot converge to some other point of $(0,1)$.

The same distinction appears for more complicated subsets. The formula for the distances may be familiar, but a limit is always taken in a specified space. An ambient limit outside the subset records how the points sit in $X$; it does not add that missing point to $A$.

Worked Example: Points on a Curve with an Omitted Limit

In $\mathbb{R}^2$ with the Euclidean metric, let $$ A=\{(t,t^2):0<t\leq1\}, \qquad a_n=(1/n,1/n^2). $$ Each $a_n$ is in $A$. The point $(0,0)$ is not in $A$, because its first coordinate would require $t=0$, which is excluded. The Euclidean distance from $a_n$ to $(0,0)$ is $$ d(a_n,(0,0))=\sqrt{\frac{1}{n^2}+\frac{1}{n^4}} =\frac{\sqrt{n^2+1}}{n^2}. $$ For $n\geq1$, $n^2+1\leq2n^2$, so $$ d(a_n,(0,0))\leq\frac{\sqrt{2}}{n}. $$ Given $\varepsilon>0$, choose $N>\sqrt{2}/\varepsilon$. Then $n\geq N$ implies $d(a_n,(0,0))<\varepsilon$, and hence $a_n\to(0,0)$ in $\mathbb{R}^2$. Since the limit point is not in $A$, the sequence does not converge in $A$. This example shows that the issue is membership of the limit, not the dimension or shape of the subset.

Closed Subspaces Keep Ambient Limits

Closedness gives a useful guarantee: a sequence of points in a closed subset cannot converge in the ambient space to a point outside that subset. This follows directly from the Sequential Characterization of Closed Sets, which says that a set is closed if and only if it contains the limits of all convergent sequences of its points.

Theorem (Ambient Limits in a Closed Subspace): Let $A$ be a closed subset of a metric space $(X,d)$, and let $(a_n)$ be a sequence in $A$. If $a_n\to x$ in $X$, then $x\in A$ and $a_n\to x$ in $A$.

Proof. Since $A$ is closed and every $a_n$ belongs to $A$, the Sequential Characterization of Closed Sets implies that the ambient limit $x$ belongs to $A$. Now both the terms $a_n$ and the limit $x$ lie in $A$. The Convergence in a Subspace Theorem therefore applies and gives $a_n\to x$ in $A$. $\square$

This theorem is an implication about sequences whose ambient limit is already known to exist. It does not claim that every sequence in a closed subspace converges; a sequence may fail to converge in $X$ in the first place. Closedness controls where existing limits can lie, not whether a sequence has a limit.

Worked Example: A Closed Subset Contains Its Sequence Limit

Let $A=\{x\in\mathbb{R}:x\geq2\}$, a closed subset of $\mathbb{R}$, and define $a_n=2+3/n$. For every $n$, $a_n\geq2$, so $a_n\in A$. Moreover, $$ |a_n-2|=\frac{3}{n}. $$ Given $\varepsilon>0$, choose $N>3/\varepsilon$. For all $n\geq N$, $3/n\leq3/N<\varepsilon$, so $a_n\to2$ in $\mathbb{R}$. The limit $2$ is in $A$, and the theorem gives convergence in $A$ as well. Directly, the restricted distance satisfies $d_A(a_n,2)=d(a_n,2)=3/n$, so the same estimate verifies the subspace convergence.

A Reliable Way to Check the Space

When a sequence is described as convergent, keep track of both the sequence and the proposed limit. For a sequence in $A$, ask first whether the limit is meant in $A$ or in $X$. If the proposed point belongs to $A$, the Convergence in a Subspace Theorem transfers the convergence statement exactly. If it does not belong to $A$, it cannot be a subspace limit; uniqueness rules out a different subspace limit when the sequence already has that ambient limit.

Closedness is a separate check. If $A$ is closed, an ambiently convergent sequence in $A$ must have its limit in $A$. If $A$ is not closed, such a sequence may converge to a missing point, as the examples in $(0,1)$ and on the curve demonstrate. Do not infer that a sequence fails to converge merely because its limit is absent from the subset: it may still converge in the ambient space. The correct conclusion is that it does not converge in the subspace.

Takeaway: For a sequence in $A$ and a proposed limit $a\in A$, convergence in $A$ is equivalent to convergence in $X$, because the restricted and ambient distances agree. An ambient limit outside $A$ is not a subspace limit; when $A$ is closed, such a limit cannot occur for a sequence in $A$.

Check Your Understanding

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

Why are the convergence tests in $A$ and $X$ identical when both the sequence terms and the proposed limit belong to $A$?
If a sequence in $A$ converges in $X$ to $x\notin A$, can it converge in $A$ to a different point? Explain.
What does closedness of $A$ imply about the ambient limit of a convergent sequence in $A$?
Can a sequence in a nonclosed subspace converge in the ambient space without converging in the subspace? Give an example.
Does closedness guarantee that every sequence in $A$ converges? Why or why not?

What You'll Learn