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

# 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

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

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

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

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

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 $[a,b] \subset \R$.