3.1.  Convergent Sequences

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We now use the metric structure to describe the asymptotic behavior of an ordered family of points. To say that a sequence \((x_n)\) of points in \(M\) converges to \(x\) means that its terms eventually lie as close to \(x\) as we wish.

Definition 3.1.1. Let \(M\) be a metric space. A sequence \((x_n)\) of points in \(M\) is said to converge in \(M\) if there is a point \(x\in M\) with the following property: For any \(\varepsilon>0\), there exists \(n_0\in \mathbb N^{+}\) such that

\[n\ge n_0\ \Longrightarrow\ d(x_n,x)<\varepsilon.\]

In this case, we also say that \((x_n)\) converges to \(x\), or that \(x\) is the limit of \((x_n)\), and we write \(x_n\to x\) or \(\lim\limits_{n\to \infty} x_n = x\). If \((x_n)\) does not converge, it is said to diverge.

As can be seen from the definition, convergence depends only on the tail of a sequence: changing finitely many initial terms does not affect either convergence or its limit.

Definition 3.1.2. Let \(M\) be a metric space. A sequence \((x_n)\) in \(M\) is said to be bounded if the image of \((x_n)\) is bounded.

Note that the boundedness of a subset of a metric space is not relative to a subspace when the subspace just carries the restricted metric. It therefore follows that the boundedness of a sequence is also not relative to a subspace. We now look into some important properties of convergent sequences in metric spaces.

Theorem 3.1.3. Let \((x_n)\) be a sequence in a metric space \(M\).

1. \((x_n)\) converges to \(x\in M\) if and only if every open ball centered at \(x\) contains \(x_n\) for all but finitely many \(n\).
2. For \(x,y\in M\), if \((x_n)\) converges to \(x\) and to \(y\), then \(x=y\).
3. If \((x_n)\) converges, then \((x_n)\) is bounded.
4. If \(x\in M\) is a limit point of \(A\subseteq M\), then there exists a sequence \((x_n)\) in \(A\) that converges to \(x\).

Proof. (1) Suppose \((x_n)\to x\) and let \(B_r(x)\) be an open ball of \(x\). Corresponding to this \(r\), there exists \(n_0\in \mathbb N^+\) such that \(n\ge n_0\) implies \(x_n\in B_r(x)\). Conversely, suppose that every open ball of \(x\) contains \(x_n\) for all but finitely many \(n\). Fix \(r>0\) and let

\[n_0 > \max {\{ n\in \mathbb N^+ : x_n\notin B_r(x)\}}.\]

Then \(n\ge n_0\) implies \(x_n\in B_r(x)\). Thus \(x_n\to x\). (2) Given \(\varepsilon>0\), there exist \(n_{0_x},n_{0_y} \in \mathbb N^+\) such that

\[n\ge n_{0_x} \Rightarrow d(x_n,x)<\frac \varepsilon 2 \quad \text{and} \quad n\ge n_{0_y} \Rightarrow d(x_n,y)<\frac \varepsilon 2.\]

If \(n_0 = \max {\{n_{0_x}, n_{0_y}\}}\), then we have

\[d(x,y) \le d(x_n,x) + d(x_n,y) < \varepsilon.\]

Since \(\varepsilon\) was arbitrary, we conclude that \(d(x,y)=0\); hence \(x=y\). (3) Suppose \(x_n\to x\). There exists \(n_0\in \mathbb N^+\) such that \(n>n_0\) implies \(x_n\in B_1(x)\). Put

\[\varepsilon = \max {\{d(x_1,x),\ldots,d(x_{n_0},x)\}} + 1.\]

Then we have \(d(x_n,x)<\varepsilon\) for all \(n\in \mathbb N^+\); hence \((x_n)\) is bounded. (4) For each \(n\in \mathbb N^+\), there is a point \(x_n\in A\) such that \(d(x_n,x)<1/n\). Given \(\varepsilon>0\), choose \(n_0\) so that \(1/n_0\le \varepsilon\). Then, \(n\ge n_0\) implies \(d(x_n,x)<\varepsilon\); hence \(x_n\to x\).\(\square\)

When the ambient space also carries algebraic operations, we may ask whether those operations are compatible with taking limits. We begin with sequences of scalars.

Theorem 3.1.4. Suppose \((x_n)\) and \((y_n)\) are sequences in \(\mathbb C\). If \(\lim\limits_{n\to \infty} x_n = x\) and \(\lim\limits_{n\to \infty} y_n = y\), then the following hold:

1. \(\lim\limits_{n\to \infty} (x_n + y_n) = x + y\).
2. \(\lim\limits_{n\to \infty} cx_n = cx\) for any \(c\in \mathbb C\).
3. \(\lim\limits_{n\to \infty} x_ny_n = xy\).
4. If \(x_n\ne 0\) and \(x\ne 0\), then \(\lim\limits_{n\to \infty} (1/x_n) = 1/x\).

Proof. (1) Given \(\varepsilon>0\), there exist \(n_{0_x},n_{0_y} \in \mathbb N^+\) such that

\[n\ge n_{0_x} \Rightarrow |x_n-x|<\frac \varepsilon 2 \quad \text{and} \quad n\ge n_{0_y} \Rightarrow |y_n-y|<\frac \varepsilon 2.\]

If \(n_0 = \max {\{n_{0_x},n_{0_y}\}}\), then \(n\ge n_0\) implies

\[| (x_n + y_n) - (x+y)| \le |x_n-x| + |y_n-y| < \varepsilon.\]

Since \(\varepsilon\) was arbitrary, (1) holds. (2) Given \(\varepsilon>0\), there exists \(n_0\in \mathbb N^+\) such that

\[n\ge n_0 \ \Longrightarrow \ |x_n-x|<\frac{\varepsilon}{|c|+1}.\]

Then, \(n\ge n_0\) implies

\[|cx_n - cx| = |c||x_n-x|<\frac{|c|}{|c|+1}\varepsilon <\varepsilon.\]

Since \(\varepsilon\) was arbitrary, (2) holds. (3) Given \(\varepsilon>0\), there exist \(n_{0_x},n_{0_y}\in \mathbb N^+\) such that

\[n\ge n_{0_x} \Rightarrow |x_n-x|<\sqrt \varepsilon\quad \text{and} \quad n\ge n_{0_y} \Rightarrow |y_n-y|<\sqrt \varepsilon.\]

If \(n_0 = \max {\{n_{0_x},n_{0_y}\}}\), then \(n\ge n_0\) implies

\[|(x_n-x)(y_n-y)| = |x_n-x||y_n-y| < \varepsilon.\]

Thus we have \(\lim\limits_{n\to \infty} (x_n-x)(y_n-y) = 0\). It follows that

\[\lim\limits_{n\to \infty} (x_ny_n - xy) = \lim\limits_{n\to \infty} ((x_n-x)(y_n-y) + x(y_n-y) + y(x_n-x))=0.\]

Therefore

\[\lim\limits_{n\to \infty} x_ny_n = \lim\limits_{n\to \infty} ((x_ny_n - xy)+xy) = xy.\]

(4) Since

\[\left|\frac {1}{x_n} - \frac{1}{x}\right| = \left|\frac{x_n-x}{x_nx}\right|,\]

what we have to show is to ensure that the denominators \(x_n\) stay uniformly away from zero. To do that, we first prove the following lemma, so-called the reverse triangle inequality.

Lemma 3.1.5. In a metric space \((M,d)\), the following inequality holds:

\[d(x,y)\ge |d(x,z) -d(y,z)|.\]

Subproof. If \(d(x,z)\ge d(y,z)\), let \(\delta=d(x,z)-d(y,z)\). In order that

\[d(x,y) + d(y,z) \ge d(x,z) = d(y,z) + \delta ,\]

\(d(x,y)\) must be greater than or equal to \(\delta\). That is, \(d(x,y)\ge d(x,z)-d(y,z)\). If \(d(y,z)\ge d(x,z)\), in the same way, we have \(d(x,y)\ge d(y,z)-d(x,z)\). Therefore \(d(x,y)\ge |d(x,z) -d(y,z)|\).\(\blacksquare\)

Since \(x\ne 0\), we have \(d(x,0)>0\). Then, there exists \(m\in \mathbb N^+\) such that \(n\ge m\) implies \(d(x_n,x)<\frac 1 2 d(x,0)\). By Lemma 3.1.5, we obtain

\[d(x_n,0)\ge d(0,x) - d(x_n,x)>\frac 1 2 d(0,x).\]

whenever \(n\ge m\). Equivalently, \(|x_n| >\frac 1 2 |x|\). Given \(\varepsilon>0\), there is \(n_0>m\) such that \(n\ge n_0\) implies

\[|x_n-x|<\frac 1 2 |x|^2 \varepsilon.\]

Hence, if \(n\ge n_0\), then

\[\left|\frac {1}{x_n} - \frac{1}{x}\right| = \left|\frac{x_n-x}{x_nx}\right|<\frac{2}{|x|^2}|x_n-x|<\varepsilon.\tag*{$\square$}\]

Having established the scalar limit laws, we now consider Euclidean space. Because the Euclidean norm is determined by finitely many coordinates, convergence in \(\mathbb R^k\) can be tested one coordinate at a time. The next theorem makes this reduction precise.

Theorem 3.1.6. Suppose \((\mathbf x_n)\) is a sequence in \(\mathbb R^k\) and \(\mathbf x_n = (x_{1,n},\ldots,x_{k,n})\). Then, \((\mathbf x_n)\) converges to \(\mathbf x = (x_1,\ldots,x_k)\) if and only if

\[\lim\limits_{n\to \infty} x_{i,n} = x_i\]

for \(i=1,2,\dots,k\).

Proof. Suppose \(\mathbf x_n\to \mathbf x\). From the definition of the norm in \(\mathbb R^k\), we have

\[|x_{i,n}-x_i|\le \|\mathbf x_n - \mathbf x\|,\]

for each \(i=1,2,\ldots,k\) and \(n\in \mathbb N^+\). Hence, given \(\varepsilon >0\), there exists \(n_0\in \mathbb N^+\) such that \(n\ge n_0\) implies \(|x_{i,n}-x_i|<\varepsilon\). Therefore \(x_{i,n}\to x_i\) for each \(i=1,2,\ldots,k\). Conversely, suppose \(x_{i,n}\to x_i\) for each \(i=1,2,\ldots,k\). Given \(\varepsilon>0\), there exists \(n_{0_i}\in \mathbb N^+\) such that \(n\ge n_{0_i}\) implies

\[|x_{i,n}-x_i| < \frac {\varepsilon}{\sqrt k}.\]

for each \(i=1,2,\ldots,k\). Put \(n_0 = \max {\{n_{0_1},\ldots,n_{0_k}\}}\). Then \(n\ge n_0\) implies

\[\|\mathbf x_n - \mathbf x\| = \left\{\sum_{i=1}^k |x_{i,n}-x_i|^2\right\}^{\!1/2}<\varepsilon.\]

Since \(\varepsilon\) was arbitrary, \(\mathbf x_n\to \mathbf x\).\(\square\)

References

1. Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.

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