Topology/Completeness

A sequence {x_n} is a Cauchy sequence when for any ε>0, there is an N such that for any a>N and for any b>N, d(x_a, x_b)<ε.

Theorem: All convergent sequences are Cauchy sequences
A convergent sequence {x_n} will converge to a limit x, implying that there exists an N such that for any a>N, that d(x_a,x)<ε/2. Thus, for any a>N, and for any b>N, d(x_a,x_b)≤d(x_a,x)+d(x_b,x)<ε.

A metric space is complete when all Cauchy sequences converge to a limit.

A subset A of a metric space X is dense in an open set O when O⊆Cl(A). A subset A of a metric space X is everywhere dense when it is dense in X. A subset A of a metric space X is nowhere dense when it is dense in no open set in X.

Theorem

A closed subset of a complete space is itself complete.

Proof

Consider a complete space S and a closed subset C. Consider any Cauchy sequence within C, which is within S, so it has a limit. This limit is a point of contact of this sequence, and consequently, is a point of contact of C, and so is also within C. Thus, C is complete.

Theorem (Cantor's intersection theorem)

The intersection of a sequence of compact sets A_n such that A_n⊆A_n-1 is non-empty if and only if the metric space is complete.

Proof

Select a sequence {x_n} where x_n⊆A_n. There exists a limit point x of that sequence, and hence is a limit point of each A_n. Because each A_n is closed, x is within each A_n, and so

$x\in \bigcap _{i=1}^{\mathrm{\infty }}{A}_n$.

Theorem (Nested balls theorem)

A theorem very similar to the Cantor's intersection theorem is the Nested balls theorem, which states that the intersection of a sequence of the closures of balls A_n such that A_n⊆A_n-1 and such that their sequence of radii r_n approaches 0 is non-empty if and only if the metric space is complete.

Proof

Consider such a sequence of balls A_n, and consider their centers. Since their radii r_n approaches 0, there exists an N such that when n>N, that r_n < ε/2. Consider the centers of the spheres, to be denoted c_n. c_a and c_b must be no farther apart than by ε when both a and b are greater than N, so c_n is a Cauchy sequence, and so has a limit c. The closures of the open spheres, A_n, are closed, and therefore are also complete. The sequence c_n is entirely within A_k for n≥k, so consider the sequence c_n starting at k is entirely within A_k, and since A_k is complete, it contains the limit. Therefore, c is within all the sets, and is thus within their intersection, proving that their intersection is non-empty.

Theorem (Baire's theorem)

A complete metric space is not a countable union of nowhere dense subsets.

Proof

Let A=$\bigcup _{i=1}^{\mathrm{\infty }}{K}_i$ be a complete metric space where each K_i is nowhere dense. Let S₁ be an open ball of radius $\frac{1}{2}$. Let S_n, where n is a natural number greater than 1 to be be an open ball of radius $\frac{1}{2^n}$ contained in S_n-1 which does not meet A_n, which is possible because if it always met A_n, then A₁ would be dense in S_n-1. The centers c_n of the spheres S_n form a Cauchy sequence because when n₁>N and n₂>N, then d(x_n1,x_n2)<$\frac{1}{2^N}$. Therefore, because the space A is complete, it converges to a limit c within A. However, it is not within any K_n, and so it is not within A, a contradiction.

Examples

1. It can be proven that the Euclidean space Rⁿ is complete.
2. The Hilbert space is complete.