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

# 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 isomorphism
^{2)}\[ z = (x_1 + iy_1, \ldots, x_n + iy_n) \mapsto (x,y) = (x_1, \ldots, x_n, y_1, \ldots, y_n).\]

**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-Weierstrass**^{4)}: 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 space
^{5)}. 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)}

*isos*is Greek for 'same', 'similar';

*morphe*is 'shape','form'; and

*metron*is 'measure'.

^{2)}

^{3)}

^{4)}

^{5)}