MA651 Topology. Lecture 11. Metric Spaces 2.This text is based on the following books:• ”Topology” by James Dugundgji• ”Fundamental concepts of topology” by Peter O’Neil• ”Linear Algebra and Analysis” by Marc ZamanskyI have intentionally made several mistakes in this text. The first homework assignment is to findthem.63 Complete metric spaces63.1 Cauchy sequencesThe importance of complete metric spaces lies in the fact that in such spaces we can decide whethera sequence is convergent, without necessarily knowing its limit.Definition 63.1. Let (E, d) be a metric space. A sequence (xn) in E is called a Cauchy sequence(d-Cauchy sequence) iflimp,q→∞d(xp, xq) = 0We recall thatlimp,q→∞d(xp, xq) = 0means:∀ ε > 0 ∃N(ε) ∀p, q > N : d(xp, xq) < εThe notion of a d-Cauchy sequence is depended on the particular metric used: The same sequencecan be Cauchy for one metric, but not Cauchy for an equivalent metric.1Example 63.1. In E1, the Euclidean metric de(x, y) = |x − y| is equivalent to the metricdϕ(x, y) = |x1 + |x|−y1 + |y||since the latter is derived from the homeomorphism x → x/(1 + |x|) of E1and ] − 1, +1[. Thesequence {n | n = 1, 2, . . .} in E1is not de-Cauchy sequence, whereas it is dϕ-Cauchy sequence.Properties of Cauchy sequencesProperties of1. Every convergent sequence is Cauchy sequence. For (xn) is a convergent sequence in a metricspace E then there exists x ∈ E such that d(xn, x) tends to zero. The property then followsfromd(xp, xq) 6 d(xp, x) + d(x, xq)This property shows that the concept of a Cauchy sequence is more general than that of aconvergent sequence, so the converse is false. In fact it is strictly more general, for example,there are metric spaces in which a Cauchy sequence may fail to converge, for example Q.Example 63.2. In the space Y =]0, 1] with the E uclidean metric de, the sequence {1/n} isde-Cauchy, yet it does not converge to any y0∈ Y .2. If (xn) is a Cauchy sequence and if for the sequence (yn) we have lim d(xn, yn) = 0, Thenynis a Cauchy sequence. This is so sinced(yp, yq) 6 d(yp, xp) + d(xp, xq) + d(xq, yq)3. If (xn) is a Cauchy sequence so is every one of its subsequences This is evident.4. If (xn) is a Cauchy sequence containing a convergent subsequence (xnk) then (xn) is conver-gent.If, for a subsequence (xnk), there exists x ∈ E such thatlimk→∞d(xnk, x) = 0then we have d(xnk, x) < ε for k > k0. But d(xp, xq) < ε for p, q > N, so that thend(xn, x) 6 d(xn, xnk) + d(xnk, x)and for n > max(N, n0) (and nk> max(N, nk0)) we haved(xn, x) 6 ε + ε263.2 Complete metrics and complete spacesDefinition 63.2. Let Y be a metrizable space. A metric d for Y (that is, one that metrics thegiven topology of Y ) is called complete if every d-Cauchy sequence in Y converges.It must be emphasized that completeness is a property of metrics: One metric for Y may becomplete, whereas another metric may not.A given metric space Y may not be have any complete metric; to denote those that do, we have:Definition 63.3. A metric space Y is called topologically complete (or briefly, complete ) if acomplete metric for Y exists. To indicate that d is a complete metric for Y , we say that Y isd-complete.Less rigorously we can say: A metric space is complete if every Cauchy sequence in it is convergent.We can also say that a metric space is complete if the Cauchy sequences and convergent sequencesare the same, or a sequence is convergent if and only if it is a Cauchy sequence.The importance of this concept lies in the fact that if we have somehow or other verified that aspace is complete it is no longer necessary to find the limit of a sequence in order to know that itis convergent.In the other words, if a metric space is complete and if we have s hown for a sequence (xn), thatlim d(xp, xq) = 0, then we can assert that there is a point X of the space (and the only one) suchthat lim d(xn, x) = 0.Theorem 63.1. Let (Y, d) be a metric space, and assume that d has the property: ∃ ε > 0 ∀y ∈Y : Bd(y, ε) is compact. Then d is complete.Proof. Let ϕ be a d-Cauchy sequence in Y , and choose n so large that δ[ϕ(Tn)] < ε/2; thenϕ(Tn) ⊂ Bd[ϕ(n), ε] and therefore ϕ has an accumulation point y0, thus ϕ → y0.Corollary 63.1. Every local compact met ric space Y is topologically complete. Furthermore, ifY is compact, then every metric d for Y is complete.Proof is left as a homework.Theorem 63.2. A subset A of a complete metric space is a complete subspace if and only if it isclosed.Proof. Let (Y, d) be a complete metric space and A a closed subset in Y . Let (xn) be a Cauchysequence in A for the induced distance. It converges in Y to a point x, but since A is closed,x ∈ A.Conversely, let (Y, d) be a metric space and A a complete subspace. Let x be adherent to A sothat it is the limit of a sequence (xn) of elements of A. Since (xn) converges to x in Y it is a3Cauchy sequence in Y and so also in A. Since A is complete the sequence converges to a point ofA. Since the limit is unique it must be a point x. thus every adherent point belongs to A, and soA is closed.Remarks:1. In the proof of the converse of this result we have shown that every complete subspace of a(not necessarily complete) metric space is closed.2. This proposition shows that in a complete metric space the closed sets and the completesubspaces coincide (and the closed sets and the compact subspaces coincide in a compactspace).64 The Baire property of complete metric spacesTheorem (63.1) gives a sufficient condition for topological completeness; the following theorem isa necessary condition. Because completeness is more prevalent than local compactness, this resultis one of the most important and useful in topology, and has extensive applications in analysis.Theorem 64.1. In a complete metric space E a countable union of closed sets without interiorpoints also has no interior points. (or, a countable intersection of open sets dense in E is alsodense in E.)Definition 64.1. A topological space X is a Baire space if the intersection of each countablyfamily of open sets in Y is dense.Then the Theorem can be written as: Any topologically complete space is a Baire space.Proof. Let (On) be a countable family of dense open sets and let A =TOn. To show thatTOnis everywhere dense it suffices to prove that for every non-empty open ball B in the space E wehaveB ∩\On6= ØLet (rn) be a sequence of strictly positive numbers converging to 0.If B is a non-empty open ball and O1an