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

Completions

Every metric space (and every normed space) can be made complete. To make this precise, recall that an isomorphism is a bijective (on-to-one and onto) map that preserves the essential structure of something. Likewise, an isometry is a map that preserves distances.1)

Isometries

Two metric spaces $(X,d_X)$ and $(Y,d_Y)$ are called isometric if there exists a bijective isometry between them, i.e., if there exists an invertible function $\varphi \colon X \to Y$ such that $d_X(x_1,x_2) = d_Y(\varphi(x_1),\varphi(x_2)).$ The function $\varphi$ is called an isometry.

Ex.
• The set of sequences with only zeros and ones, $X = \{ (x_1,x_2, \ldots) \in l_\infty \colon \text{ for each } j,\: x_j = 0 \text{ or } x_j = 1\},$ endowed with the $l_\infty$-metric, $d(x,y) = \sup_{j \in \mathbb N}|x_j-y_j|$ is isometric to $X$ endowed with the discrete metric, because $d(x,y) = 1 \quad\text{ unless }\quad x = y.$ The isometry is the identity operator, $\varphi\colon x \mapsto x, (X,\|\cdot\|_{l_\infty}) \to (X,d_\text{discrete})$.
• Let $a < b$. The normed spaces $BC((0,1),\R)$ and $BC((a,b),\R)$ of bounded continuous functions on the intervals $(0,1)$ and $(a,b)$, respectively, are isometric, since $\varphi \colon BC((0,1),\R) \to BC((a,b),\R), \qquad f(\cdot) \stackrel{\varphi}{\mapsto} f\Big(\frac{\cdot - a}{b-a}\Big)$ is an isometry: $d_{BC((0,1),\R)}(f,g) = \sup_{x \in (0,1)} |f(x)-g(x)| = \sup_{x \in (a,b)} \Big|f\big(\frac{x - a}{b-a}\big) - g \big(\frac{x - a}{b-a}\big)\Big| = d_{BC((a,b),\R)}(f,g).$ (Note that $\varphi$ is invertible with inverse $\varphi^{-1} \colon f(\cdot) \mapsto f(a + \cdot (b-a))$. Hence, studying the metric space $BC((a,b),\R)$ is no different from studying $BC((0,1),\R)$.

Isomorphisms

Vector space isomorphisms

A vector space isomorphism is a bijective linear map between two vector spaces, i.e. an invertible function $T \colon X \to Y$ such that $T(x + y) = Tx + Ty \quad\text{ and }\quad T(\lambda x) = \lambda T x \qquad\text{ for all }\quad x,y \in X,\: \lambda \in \mathbb R \: (\text{or } \mathbb C).$ Two vector spaces which allow for such a mapping are called isomorphic, and we write $X \cong Y \defarrow \exists \:\text{ isomorphism }\: T \colon X \to Y.$ N.b. A set which is the image of a vector space isomorphism automatically becomes a vector space (it inherits its linear structure from $X$).

Ex.
• Regarded as a real vector space, the space $\mathbb C^n$ of complex $n$-tuples, $z = (z_1, \ldots, z_n), \qquad z_1, \ldots, z_n \in \mathbb C$ is isomorphic to Euclidean space $\R^{2n}$ via the isomorphism2) $z = (x_1 + iy_1, \ldots, x_n + iy_n) \mapsto (x,y) = (x_1, \ldots, x_n, y_1, \ldots, y_n).$
• The set of polynomials with real coefficients of degree at most $n$, $P_n(\R)$, is a vector space consisting of elements $p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0, \qquad a_0, \ldots, a_n \in \R.$ Why is this a vector space? Because $P_n(\R) \cong \R^{n+1}:$ The mapping $T \colon P_n(\R) \to \R^{n+1}, \qquad a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \mapsto (a_0, a_1, \ldots, a_n)$ is both bijective, $\text{ for any }\: (a_0, \ldots, a_n) \in \R^{n+1} \: \text{ there exists a unique }\: p(x) = \sum_{k=0}^n a_k x^k \in P_n(\R),$ and linear, \begin{align*} T\bigg(\lambda \sum_{k=0}^n a_k x^k + \mu \sum_{k=0}^n b_k x^k\bigg) &= T \bigg( \sum_{k=0}^n (\lambda a_k + \mu b_k) x^k \bigg)\\ &= (\lambda a_0 + \mu b_0, \ldots, \lambda a_n + \mu b_n)\\ &= \lambda (a_0,\ldots, a_n) + \mu (b_0, \ldots, b_n)\\ &= \lambda T\bigg(\sum_{k=0}^n a_k x^k\bigg) + \mu T\bigg(\sum_{k=0}^n b_k x^k\bigg). \end{align*} Hence any linear operation in $P_n(\R)$ corresponds to a linear operation in $\R^{n+1}$, meaning that $P_n(\R)$ and $\R^{n+1}$ are isomorphic as vector spaces.

Isomorphisms on normed spaces

If the vector spaces are normed, they are isomorphic as normed spaces if they are isomorphic (as vector spaces) and isometric (as metric spaces). Sometimes this is called isometrically isomorphic to avoid confusion.

Ex.
• The spaces $BC((0,1),\R)$ and $BC((a,b),\R)$ are isometrically isomorphic, since the isometry $\varphi$ in the example above is also a vector space isomorphism: $\varphi( \lambda f + \mu g)(x) = (\lambda f + \mu g)\Big(\frac{x- a}{b-a}\Big) = \lambda f\Big(\frac{x- a}{b-a}\Big) + \mu g\Big(\frac{x- a}{b-a}\Big) = \lambda \varphi(f)(x) + \mu \varphi(g)(x), \qquad x \in (a,b).$

Embeddings

If a (normed) vector space $X$ is isomorphic to a subspace $M \subset Y$ of another (normed) vector space, we say that it is (continuously) embedded in $Y$, $X \hookrightarrow Y \defarrow \exists \:\text{ isomorphism }\: T \colon X \to M \subset Y.$ To ease terminology we shall use this concept also for isometries between metric spaces.

Ex.
• The vector space of polynomials of degree at most $1$, $P_1(\R)$, is continuously embedded in three-dimensional Euclidean space, $P_1 \hookrightarrow \R^3, \quad\text{ since }\quad P_1 \cong \mathbb R^2 \subset \R^3.$
• The identity operator provides an embedding of the vector space of continuously differentiable functions on an interval $I$ into the vector space of continuous functions on the same interval: 3) $C^1(I,\R) \hookrightarrow C(I,\R).$

Dense sets

A subset $M \subset X$ of a metric space is dense if its closure is the whole space. $M \text { dense in } X \defarrow \overline M = X.$ In this sense, $M$ is 'almost all of $X$'.

Ex.
• $\mathbb Q$ is dense in $\mathbb R$: $\text{ for any } \lambda \in \mathbb R \text{ there is a sequence } \{q_n\}_n \subset \mathbb Q \quad \text{ such that } \quad q_n \to \lambda \quad\text{ in } \R.$
• Stone-Weierstrass4): Let $I = [a,b]$ be a finite and closed interval. The polynomials, $P(\R)$, the infinitely continuosly differentiable functions, $C^\infty(I,\R)$, and the $k$ times continuously differentiable functions, $C^k(I,\R)$, $k \geq 1$, are all dense in the space of bounded and continuous functions $BC(I,\R)$: $\forall\: \varepsilon > 0, \: f \in BC(I,\R)\quad \exists\: f_\text{approx} \in P(\R) \subset C^\infty(I,\R) \subset C^k(I,\R); \quad \sup_{x \in I}|f(x) - f_\text{approx}(x)| < \varepsilon.$

Separability

A metric space is said to be separable if it contains a countable dense set: $X \text{ separable } \defarrow \exists \{x_n\}_{n \in \mathbb N} \subset X; \qquad \overline{\{x_n\}_n} = X.$

Ex.
• Since $\mathbb{Q}$ is countable, and $\overline{\mathbb{Q}} = \R$ (with respect to the distance $d(x,y) = |x-y|$), it follows that $(\R,|\cdot|)$ is separable.
• Using that $\overline{\mathbb{Q}} = \R$ one can show that all the spaces $\R^n, \mathbb C^n, l_p(\R)$ and $l_p(\mathbb C)$ for $1 \leq p < \infty, BC([a,b],\R)$, and $BC([a,b],\mathbb{C})$ are separable (with respect to their standard norms/metrics).
• Neither $l_\infty$ nor $BC((a,b),\R)$ or $BC((a,b),\mathbb{C})$ is separable. (In this respect, these spaces are much 'bigger' than the other spaces considered in this course.)

› Completion theorem

Every metric (normed) space is densely embedded in a complete metric (normed) space.

Ex.
• If we complete $\mathbb Q$ with respect to the metric $d(x,y) = |x-y|$ we get $\R$: $\mathbb Q \stackrel{\text{dense}}{\hookrightarrow} \R.$
• If we complete $C^\infty([0,1],\R)$ with respect to the supremum norm, $\|\cdot\|_\infty$, we get $BC([0,1],\R)$.
• Let $I \subset \R$ be an open interval, and consider (measurable) functions such that the integral $\int_I |f(x)|^2\,dx \quad\text{ exists and is finite.}$ Then $\|f\|_{L_2(I,\R)} := \Big( \int_I |f(x)|^2\,dx\Big)^{1/2}$ defines a norm on these functions, so we have a normed space5). The completion of this space is called the space of square-integrable functions, written $L_2(I,\R).$ The same space can be obtained by completing $C(I,\R)$ with respect to the $L_2$-norm. Hence $(C(I,\R),\|\cdot\|_{L_2}) \stackrel{\text{dense}}{\hookrightarrow} L_2(I,\R).$ A deep result in analysis is that $L_2(I,\R) \cong l_2(\R)$ (isometrically isomorphic): the elements in $l_2(\R)$ may be identified as (generalised) Fourier coefficients of elements in $L_2(I,\R)$.
1)
The word isos is Greek for 'same', 'similar'; morphe is 'shape','form'; and metron is 'measure'.
2)
This is also an isometry, since $|z|^2 = |x|^2 + |y|^2$.
3)
Note that, when $I = \R$, or if $I$ is not closed, neither of these spaces are normed. When $I = [a,b]$ is closed and finite, $BC([a,b],\R) = (C([a,b],\R),\|\cdot\|_\infty)$. Otherwise, $BC(I,\R)$ is strictly smaller than $C(I,\R)$.
4)
There are many versions of this theorem. The classical result states that the set of polynomials are dense in $BC([a,b],\mathbb C)$.
5)
Two functions in this space are equal if they are equal almost everywhere on $I$.