Week 4 3. Complete metric spaces, Hilbert spaces Example 1: Consider the sequence {xn = 1/n} in the space ℝ\{0} with the usual Euclidean metric. Does it have a limit? No, it doesn’t, as x = 0 isn’t included in the space within the framework of which the sequence in defined. Sequences that don’t converge in ‘our’ space, but converge in a space including ‘ours’ as a subset, are called Cauchy sequences. ۞ A sequence {xn} in a metric space with a metric d(x, y) is said to be a Cauchy sequence if

Theorem 1: All convergent sequences are also Cauchy sequences. Proof: Rewrite the definition of a convergent series with ε replaced by ½ ε: It is then clear that the maximum distance between the terms of the sequence with numbers n > N is less than ε, i.e. QED. Comment: Colloquially, a mathematician would say: the notion of a convergent sequence is stronger than that of a Cauchy sequence.

The opposite to Theorem 1 isn’t correct, i. e The opposite to Theorem 1 isn’t correct, i.e. some Cauchy sequences converge, other don’t. There are spaces, however, where all Cauchy sequences converge. ۞ A metric space S is called complete if all Cauchy sequences in S converge to a limit in S. Example 2: ℝ is complete (due to the so-called Cauchy Criterion – see its proof in the Misc. Section of this module’s website). ℝn is also complete for all n (due to the multi-dimensional generalisation of the Cauchy Criterion).

Theorem 2: Let S and S' be metric spaces with the same metric, and S' ⊂ S. Consider a sequence {xn} which belongs to both S and S', and assume that it is a Cauchy sequence in one of the two spaces. Then, {xn} is a Cauchy sequence in the other space as well. Proof: Obvious.

Example 3: The space ℝ\{0} with the usual Euclidean metric is incomplete (see Example 1). Example 4: The space S of all rational numbers with the usual Euclidean metric is incomplete. To prove this, consider the following sequence (defined recursively): Observe that...

All xn are rational numbers  {xn}  S. Also, {xn}  ℝ and converges there to L = ½ (√5 – 1). Since {xn} converges in ℝ, it is a Cauchy sequence (due to Theorem 1). Since {xn} is a Cauchy sequence in ℝ, it’s a Cauchy sequence in S too (due to Theorem 2). Thus, even though {xn} belongs to S, its limit doesn’t (because L is an irrational number). Thus, S is incomplete. It remains to prove that the limit L of {xn} is indeed ½ (√5 – 1).

۞ An inner product space, which is complete as a metric space with is called a Hilbert space. Example 5: ℝn with the usual Euclidean metric (i.e. resulting from the usual Euclidean inner product) is a Hilbert space.

۞ The set of continuous, square-integrable functions defined on an interval (a, b), and the usual inner product for functions, make a Hilbert space called L2(a, b). We’ll also use this definition with a = −∞ and/or b = +∞. Comment: Even if a and b are finite, the condition of square integrability in the definition of L2(a, b) cannot be omitted. To understand why, consider f(x) = 1/x and g(x) = 2/x on the interval (0, 1), in which case d(f, g) = ∞.

Comment: Some textbooks omit the word “continuous” in the definition of L2(a, b). What are the implications?... Without the requirement of continuity, this definition is inconsistent with that of a metric space (strictly speaking). Indeed, consider a continuous function f(x) L2(a, b) and a discontinuous function f1(x) which differs from f(x) at a single point. The distance between f(x) and f1(x) is zero, but they actually differ – hence, they violate Axiom (B) of metric spaces! Still, many omit the requirement of continuity, with the implication that L2(a, b) consists of “classes” of functions, such that the set of points where two functions from the same class differ has zero measure (e.g. these functions differ at several isolated points).