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3.2.  Subsequenes

A sequence may fail to converge as a whole while some of its terms still exhibit convergent behavior. To capture such behavior, we introduce the notion of a subsequence.

Definition 3.2.1. Let \(M\) be a metric space. Given a sequence \((x_n)\), consider a sequence \((n_k)\) in \(\mathbb N^+\), such that \(n_1<n_2<\cdots.\) Then the sequence \((x_{n_k})\), which maps \(k\) to \(x_{n_k}\), is called a subsequence of \((x_n)\). If \((x_{n_k})\) converges, its limit is called a subsequential limit of \((x_n)\).

Theorem 3.2.2. Let \(M\) be a metric space. A sequence \((x_n)\) converges to \(x\) if and only if every subsequence of \((x_n)\) converges to \(x\).

Proof. Suppose \(x_n\to x\) and let \((x_{n_k})\) be a subsequence of \((x_n)\). For any \(\varepsilon>0\), there exists \(n_0\in \mathbb N^+\) such that \(n\ge n_0\) implies \(d(x_n,x)<\varepsilon\). Since \(\{n_k\}_{k\in \mathbb N^+}\) is infinite, there exists \(k_0\in \mathbb N^+\) such that \(k_0\ge n_0\). Hence, \(k\ge k_0\) implies \(d(x_{n_k},x)<\varepsilon\); that is, \((x_{n_k})\to x\). Conversely, suppose every subsequence of \((x_n)\) converges to \(x\). Put \(n_k =k\). Then \((x_{n_k}) = (x_k) = (x_n)\). Therefore \((x_n)\to x\).\(\square\)

Theorem 3.2.3. If \((x_n)\) is a sequence of points in a compact metric space \(M\), then some subsequence of \((x_n)\) converges.

Proof. If \(\{x_n\}_{n\in \mathbb N^{+}}\) is finite, then there is \(x\in \{x_n\}_{n\in \mathbb N^{+}}\) and a sequence \((n_k)\) with \(n_1<n_2<\cdots\), such that

\[x_{n_1} = x_{n_2} = \cdots = x.\]

The subsequence \((x_{n_k})\) converges to \(x\). If \(\{x_n\}_{n\in \mathbb N^{+}}\) is infinite, Theorem 2.2.17 shows that \(\{x_n\}_{n\in \mathbb N^{+}}\) has a limit point \(x\in M\). Since every open ball of \(x\) meets \(\{x_n\}_{n\in \mathbb N^{+}}\), so we may obtain a sequence \((n_k)\) with \(n_1<n_2<\cdots\) such that \(d(x_{n_k},x)<1/k\). Then, the subsequence \((x_{n_k})\) converges to \(x\).\(\square\)

Specializing this theorem to Euclidean space yields the sequential form of the Bolzano–Weierstrass theorem. It is the sequence-theoretic counterpart of the set-theoretic version stated in Corollary 2.2.18.

Corollary 3.2.4 (Bolzano–Weierstrass theorem). Every bounded sequence of points in \(\mathbb R^k\) contains a convergent subsequence.

Proof. If \((\mathbf x_n)\) is a bounded sequence in \(\mathbb R^k\), then some \(k\)-cell contains \(\{\mathbf x_n\}_{n\in \mathbb N^{+}}\). Since every \(k\)-cell is compact, Theorem 3.2.3 completes the proof.\(\square\)

Theorem 3.2.5. Let \(M\) be a metric space and \((x_n)\) be a sequence in \(M\). Then, the set of all subsequential limits of \((x_n)\) is closed.

Proof. Let \(A\) be the set of all subsequential limits of \((x_n)\) and let \(y\in M\) be a limit point of \(A\). We show that there exists a subsequence of \((x_n)\) that converges to \(y\). There is \(n\in \mathbb N^{+}\) such that \(x_n \ne y\). Indeed, if \(x_1 = x_2 = \cdots = y\), then \(A = \{y\}\), which contradicts the fact that \(A\) is infinite. Choose \(n_1=n\) and put \(\delta = d(x_{n_1},y)\). Suppose \(n_1,\ldots, n_{k-1}\) are chosen. Since \(y\) is a limit point of \(A\), there is \(z\in A\) such that \(d(z,y)<2^{-k}\delta\). Since \(z\) is a subsequential limit of \((x_n)\), there exists \(n_k>n_{k-1}\) such that \(d(x_{n_k},z)<2^{-k}\delta\). Thus

\[d(x_{n_k},y)\le d(x_{n_k},z) + d(z,y) <2^{1-k}\delta.\]

Therefore, the subsequence \((x_{n_k})\) converges to \(y\); hence \(y\in A\).\(\square\)

References

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