\[ \newcommand{R}{\mathbb{R}} \newcommand{N}{\mathbb{N}} \newcommand{defarrow}{\quad \stackrel{\text{def}}{\Longleftrightarrow} \quad} \]

Completeness

Cauchy sequences

A sequence \(\{x_n\}\) in a metric space \((X,d)\) is a Cauchy sequence (or simply Cauchy) if the distance between its elements tends to zero: \[ \{x_n\}_{n \in \N} \text{ Cauchy } \defarrow d(x_n,x_m) \to 0 \quad\text{ as } \quad m,n \to \infty. \] Equivalently, \[ \forall\: \varepsilon > 0 \quad \exists\: n_\varepsilon; \qquad d(x_m,x_n) < \varepsilon \quad\text{ whenever}\quad m,n \geq n_\varepsilon. \]

Ex.

The sequence \(\{x_n\}_{n \in \N}\) of rational numbers \( x_n = \sum_{k=0}^n \frac{1}{k!} \) is a Cauchy sequence with respect to the distance \(d(x,y) = |x-y|\). For each \(m \geq n \geq 1\), \[ \begin{align*} |x_m-x_n| &= \sum_{k=n+1}^m \frac{1}{k!} \leq \frac{1}{(n+1)!} \sum_{k=0}^{m-(n+1)} \frac{1}{(n+1)^k}\\ &= \frac{1}{(n+1)!} \frac{1 - (\frac{1}{n+1})^{(m-n)}}{1 - \frac{1}{n+1}} \stackrel{n \geq 1}{\leq} \frac{2}{(n + 1)!} \to 0 \quad \text{ as }\quad m \geq n \to \infty.\end{align*}\] Note, however, that \(\lim_{n \to \infty} x_n = e \not\in \mathbb Q\).

The sequence \(\{x_n\}_{n \in \N}\) of functions \(x_n: t \mapsto \sum_{k=0}^n \frac{t^k}{k!}\) is Cauchy in \(BC([0,1],\mathbb R)\). For each \( m \geq n \geq 1\), \[ \|x_m-x_n\|_{BC([0,1],\R)} = \sup_{t \in [0,1]} \Big|\sum_{k=n+1}^m \frac{t^k}{k!} \Big| \leq \sum_{k=n+1}^m \frac{1}{k!} \to 0 \quad\text{ as }\quad m \geq n \to \infty,\] in view of the above calculation. In this case, \(\lim_{n\to \infty} x_n = \exp \in BC([0,1],\R).\)¹⁾

℘ Convergent sequences are Cauchy sequences

In any metric space, \[ x_n \to x \:\text{ as }\: n \to \infty \qquad\text{ implies } \qquad d(x_m,x_n) \to 0 \:\text{ as }\: m,n \to \infty. \] N.b. The opposite is in general not true.

Proof

Let \(\varepsilon > 0\). Since \(x_n \to x\) there exists \(n_\varepsilon\) such that \[ d(x,x_n) < \varepsilon/2 \quad\text{ for }\quad n \geq n_\varepsilon. \] Hence \[ d(x_m,x_n) \leq d(x_m,x) + d(x,x_n) < \varepsilon/2 + \varepsilon/2 = \varepsilon, \] for \(m,n \geq n_\varepsilon\).

℘ Cauchy sequences are bounded

If \((X,\|\cdot\|)\) is a normed space, and \(\{x_n\}_{n \in \N} \subset X\) is Cauchy, then \[ \sup_{n \geq 1}\|x_n\| < B \quad\text{ for some } \quad B \in \R. \]

Proof

According to the definition of a Cauchy sequence, there exists \(n_1\) such that \[ \|x_m - x_n\| < 1 \quad\text{ for } \quad m,n \geq n_1. \] If \(n \leq n_1\), \[\|x_n\| \leq \max_{1 \leq n \leq n_1} \|x_n\| =: \tilde B,\] and, if \( n \geq n_1 \), \[ \|x_n\| \leq \|x_n - x_{n_1}\| + \|x_{n_1}\| < 1 + \|x_{n_1}\|. \] Hence, \( \|x_n\| \leq B := \max\{\tilde B, 1 + \|x_{n_1}\| \} \) for all \(n \in \mathbb N.\)

Complete metric spaces and Banach spaces

A metric space in which every Cauchy sequence converges is called complete. A complete normed space is called a Banach space.

Ex.

\(\R\) is complete with respect to the metric \(d(x,y) = |x-y|\). Hence, \((\R,|\cdot|)\) is a Banach space.
Both \(\R^n\) and \(\mathbb C^n\) are Banach spaces with respect to the norm \(\|x\|_{l_2} = ( \sum_{j=1}^n |x_j|^2 )^{1/2}\).
The space of square-summable sequences \[ l_2 = \Big\{ \{x_j\}_{j\geq 1} \colon \sum_{j=1}^\infty |x_j|^2 < \infty \Big\} \] is complete with respect to the norm \[\|x\|_{l_2} = \big( \sum_{j=1}^\infty |x_j|^2 \big)^{1/2}. \] So is \(l_\infty\), and \(l_p\), for any \(p \geq 1\).
For any interval \(I \subset \R\), \(BC(I,\R)\) and \(BC(I,\mathbb C)\), with norms given by \( \|x\| = \sup_{t \in I}|x(t)|\), are Banach spaces.

℘ Subspaces of complete metric spaces are complete if and only if they are closed

Let \(M \subset X\) be a subspace of a complete metric space \((X,d)\), i.e. \((M,d) \subset (X,d)\). Then \[ M \text{ complete} \quad\Longleftrightarrow\quad M \text{ closed}. \]

Proof

Assume that \(M\) is complete. Closedness means \[ M \ni x_n \to x \in X \quad\Longrightarrow\quad x \in M. \] So assume that \(\{x_n\}_{n \in \N} \subset M\) converges towards \(x \in X\). Since any convergent sequence is Cauchy, \(\{x_n\}_{n \in \N} \subset M\) is Cauchy. But since \(M\) is complete, \(\{x_n\}_{n \in \N}\) converges to an element \(y \in M\). By uniqueness of limits, \(y = x\). Thus \(M\) is closed.

Contrariwise, assume that \(M\) is closed and let \(\{x_n\}_{n \in \N} \subset M\) be a Cauchy sequence. Recall that \((M,d) \subset (X,d)\) carry the induced metric \(d\). Thus \[ \begin{align*} \{x_n\}_{n \in \N} \text{ Cauchy in } M \quad &\Longrightarrow\quad \{x_n\}_{n \in \N} \text{ Cauchy in } X\\ \quad &\stackrel{X \text{ complete}}{\Longrightarrow}\quad M \ni x_n \to x \in X\\ \quad &\stackrel{M \text{ closed}}{\Longrightarrow}\quad x \in M. \end{align*} \] Thus \(M\) is complete.

Ex.

Euclidean space \(\R^n\) can be identified with a subspace of \(l_2(\R)\): \[ \R^n = \{ (x_1, \ldots, x_n, 0 , 0 \ldots) \in l_2\}. \] In this respect, \(\R^n\) is both a complete and a closed subspace of \(l_2\) (any limit of points in \(\R^n\) remains in \(\R^n\)).

℘ BC is complete

Let \(I \subset \R\) be a non-empty interval. Then \(BC(I,\R)\) and \(BC(I,\mathbb C)\) are Banach spaces.

Proof

The proof is identical for \(\R\) and \(\mathbb C\). Also, we already know that \(BC(I,\R)\) is a normed and linear space, so we only need to show that it is complete.

Let \(\{x_n\}_n\) be a Cauchy sequence in \(BC(I,\R)\). We want to prove that it converges to a limit function \(x_0 \in BC(I,\R)\).

Pointwise convergence: For any \(t \in I\), \[ |x_n(t) - x_m(t)| \leq \sup_{t \in I}|x_n(t) - x_m(t)| = \| x_n - x_m \|_{BC(I,\R)}\] Thus, \[ \{x_n\}_n \text{ Cauchy in } BC(I,\R) \quad\Longrightarrow\quad \{x_n(t)\}_n \text{ Cauchy in } \R.\] \(\R\) being complete, there exists a limit in \(\R\), \[ \forall\: t \in I \quad \exists\: x_0(t) := \lim_{n \to \infty} x_n(t) \quad \text{ in } \R,\] and we define the limit function \(x_0\) by \(x_0 := [t \mapsto x_0(t)].\)

Boundedness: For each \(t \in I\) there exists \(n_t \in \mathbb N\) such that \(|x_0(t) - x_{n_t}(t)| < \varepsilon\). Thus \[ |x_0(t)| \leq |x_0(t) - x_{n_t}(t)| + |x_{n_t}(t)| < \varepsilon + \|x_{n_t}\|_{BC(I,\R)} \leq \varepsilon + \sup_{n \in \mathbb N} \|x_n\|_{BC(I,\R)} < c, \] since Cauchy sequences are bounded. Taking the supremum over all \(t \in I\) yields that \[ \|x_0\|_{BC(I,\R)} < c.\]

Convergence in norm: A similar argument shows that \(x_n \to x_0\) in \(BC(I,\R)\). Namely, let \(\varepsilon > 0\). For each \(t \in I\) there exists \(m_t \in \mathbb N\) such that \[ |x_0(t) - x_{m}(t)| < \frac{\varepsilon}{2} \quad\text{ for }\quad m \geq m_t. \] Also, there exists \(n_\varepsilon \in \mathbb N\) such that \[\|x_n-x_m\|_{BC(I,\R)} < \frac{\varepsilon}{2} \quad\text{ for }\quad m,n \geq n_\varepsilon. \] Choose \(m \geq \max\{m_t, n_\varepsilon\}\). Then \[\begin{aligned}|x_n(t) - x_0(t)| &\leq |x_n(t) - x_m(t)| + |x_m(t) - x_0(t)|\\ &\leq \|x_n-x_m\|_{BC(I,\R)} + |x_m(t) - x_0(t)|\\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \quad\text{ for }\quad n \geq n_\varepsilon.\end{aligned}\] Taking the supremum over \(t \in I\) then yields that \[ \|x_n -x_0\|_{BC(I,\R)} < \varepsilon \quad\text{ for }\quad n \geq n_\varepsilon.\] Thus \(x_n \to x_0\) in \(BC(I,\R)\).

Continuity: To prove that \(x_0\) is continuous, pick \(t \in I\) and let \(\varepsilon > 0\). Since \(x_n \to x_0\) in \(BC(I,\R)\) there exists \(n_\varepsilon \in \mathbb N\) such that \[ \|x_0 - x_n\|_{BC(I,\R)} < \frac{\varepsilon}{3} \quad\text{ for }\quad n\geq n_\varepsilon.\] Fix such an \(n\). Since \(x_n\) is continuous, there exists \(\delta := \delta(n,\varepsilon) > 0\) with \[ |x_n(s) - x_n(t)| < \frac{\varepsilon}{3} \quad\text{ for }\quad |s-t| < \delta. \] All taken together, \[\begin{align*} |x_0(s) - x_0(t)| &< |x_0(s) - x_n(s)| + |x_n(s) - x_n(t)| + |x_n(t) - x_0(t)| \\ &\leq \|x_0 - x_n\|_{BC(I,\R)} + |x_n(s) - x_n(t)| + \|x_n - x_0\|_{BC(I,\R)} \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon \quad\text{ for }\quad |s-t| < \delta.\end{align*}\] Hence, \(x_0\) is continuous.

This shows that every Cauchy sequence \(\{x_n\}_n \subset BC(I,\R)\) converges (in \(BC(I,\R)\)) towards an element \(x_0 \in BC(I,\R)\). Thus \(BC(I,\R)\) is complete. ²⁾

› The space of square-summable sequences is complete

The space of square-summable (complex or real) sequences is a Banach space with respect to the norm \( \|x\|_{l_2} = ( \sum_{j\geq 1} |x_j|^2)^{1/2}\).

Proof

This time we pursue the proof for complex-valued sequences (it is no different from the proof for real-valued sequences).

Pointwise convergence: Take \(\{x_n\}_n\) Cauchy in \(l_2\), with \(x_n = (x_n(1),x_n(2), \ldots)\). Then \[ |x_n(j) - x_m(j)| \leq \Big( \sum_{j=1}^\infty |x_n(j) -x_m(j)|^2 \Big)^{1/2} = \|x_n-x_m\|_{l_2} \to 0 \quad\text{ as }\quad m,n \to \infty.\] Thus, for each \(j \in \mathbb N\), \(\{x_n(j)\}_n\) is Cauchy in \(\mathbb C\), and thus convergent: \[ \forall\: j \in \mathbb N \quad \exists \: x_0(j) = \lim_{n \to \infty} x_n(j) \in \mathbb C. \] Let \(x_0 := (x_0(1), x_0(2),\ldots)\).

Boundedness: By the above, for any \(j\in \mathbb N\) there exists \(n_j \in \mathbb N\) such that \[|x_0(j)-x_n(j)|^2 < \frac{1}{2^j} \quad\text{ for } n \geq n_j. \] For any finite \(N \in \mathbb N\), choose \(n \geq \max_{1 \leq j \leq N} n_j\). Then \[ \begin{align*} \Big(\sum_{j=1}^N |x_0(j)|^2\Big)^{1/2} &\leq \Big(\sum_{j=1}^N |x_0(j)-x_n(j)|^2\Big)^{1/2} + \Big(\sum_{j=1}^N |x_n(j)|^2\Big)^{1/2}\\ &< \Big( \sum_{j=1}^N \frac{1}{2^j} \Big)^{1/2} + \|x_n\|_{l_2} \leq 1 + \sup_{n \in \mathbb N}\|x_n\|_{l_2} < c, \end{align*}\] since Cauchy sequences are bounded. The right-hand side is independent of \(N\), so we may now let \(N \to \infty\) on the left-hand side, yielding that \[\|x_0\|_{l_2} < c, \quad\text{ whence }\quad x_0 \in l_2. \]

Convergence in norm: Let \(\varepsilon > 0\). As above, we can find \(n_j\) such that \[|x_0(j)-x_m(j)|^2 < \frac{\varepsilon/2}{2^j} \quad\text{ for } m \geq n_j, \] and \[ \sum_{j=1}^N |x_0(j)-x_m(j)|^2 < \sum_{j=1}^N \frac{\varepsilon/2}{2^j} \leq \varepsilon/2 \quad\text{ for }\quad m \geq \max_{1 \leq j \leq N} n_j.\] Using this, \[ \begin{align*} \Big( \sum_{j=1}^N |x_0(j) - x_n(j)|^2 \Big)^{1/2} &\leq \Big(\sum_{j=1}^N |x_0(j)-x_m(j)|^2\Big)^{1/2} + \Big(\sum_{j=1}^N |x_m(j) - x_n(j)|^2\Big)^{1/2}\leq \varepsilon/2 + \|x_n - x_m \|_{l_2}. \end{align*}\] Since \(\{x_n\}_n\) is Cauchy in \(l_2\) there exists \(n_\varepsilon \in \mathbb N\) such that \[\|x_n - x_m \|_{l_2} < \varepsilon/2 \quad\text{ for }\quad m,n \geq n_\varepsilon. \] Select \(m\) such that this holds (for example, \(m \geq n_\varepsilon + \max_{1 \leq j \leq N} n_j\)). Then \[ \left(\sum_{j=1}^N |x_0(j) - x_n(j)|^2\right)^{1/2} < \varepsilon \quad\text{ for }\quad n \geq n_\varepsilon.\] Since \(n_\varepsilon\) does not depend on \(N\), we may let \(N \to \infty\), to obtain that \[ \|x_0-x_n\|_{l_2} < \varepsilon \quad\text{ for }\quad n \geq n_\varepsilon.\] Hence, \(x_n \stackrel{\text{in } l_2}{\to} x_0\), and \(l_2\) is complete.

¹⁾

The exponential function \(t\mapsto e^t\) is often expressed as \(\mathrm{exp}\) to separate it from the real number \(e = \exp(1)\).

²⁾

The same proof can be used to show that the space of bounded and uniformly continuous functions \(BUC(I,\R)\) is complete. Since on compact sets continuous functions are uniformly continuous, \(BUC([a,b],\R) = BC([a,b],\R)\) for any closed and bounded interval \([a,b] \subset \R\).