5.4 Metric SpacesDefinition 5.7. A metric space is a set X with a distance function d: X ×X → R satisfying the following axioms:• d(x, y) = d(y, x) for any x, y ∈ X (sym met r y ) ,• d(x, z) ≤ d(x, y) + d(y, z) for any x, y, z ∈ X (triangle inequality), and• d(x, y) ≥ 0 for any x, y ∈ X, with d(x, y) = 0 if and only if x = y.Example 5.8. For any n ≥ 1, the n-dimensional Cartesian power Rnof thereal line is a metric space with the Euclidean (Pyth agor ean ) metric:d((x1, . . . , xn), (y1, . . . , yn)) =vuutnXi=1(xi− yi)2.Definition 5.9. In a metri c space (X, d), the open ball of radius r ∈ Rcentered at x ∈ X, is the setBr(x) = {y ∈ X|d(x, y) < r}.Example 5.10. In Rn, these are the interiors of round spheres of the givenradius, centered at the gi ven point.Proposition 5.11. The set of all open metric balls in (X, d) form a basisfor a topology on X.Proof . Let x, y ∈ X and let r, s > 0. If Br(x)∩Bs(y) = ∅, there is nothing toshow. So consider any z ∈ Br(x) ∩ Bs(y). Of course, this means d(x, z) < rand d(y, z) < s. Select t > 0 such that:t < min{r − d(x, z), s − d(y, z)}.Suppose that w ∈ Bt(z). Thend(w, x) ≤ d(w, z) + d(z, x) < t + d(x, z) < r − d(x, z) + d(x, z) = r,by the triangle inequality and the choice of t. So w is in Br(x). Similarly, anyw ∈ Bt(z) is also in Bs(y), and therefore Bt(z) ⊂ Br(x) ∩ Bs(y). So for anyz ∈ Br(x) ∩ Bs(y), t he re is a t(z) > 0 such that Bt(z)(z) ⊂ Br(x) ∩ Bs(y).In particular, Br(x) ∩ Bs(y) is a neighborhood of each z it contains. SoBr(x) ∩ Bs(y) is open.18Definition 5.12. For any met r ic space (X, d), the topology with basis givenby the set of all open metric balls is the metric topology on (X, d).The metric topology on R is the same as the order topology. Note thatwe have finally defined a reasonable topology on Rnfor n > 1.Example 5.13. Any set X is a metric space with th e distance functiond(x, y) = 1. This metric space topology is the discrete topology.5.5 The topology on a fini te productThe following is the standard way to use the topology on a pair of spaces toput a topology on the Cartesian p roduct of their point sets.Definition 5.14. Let (X, T ) and (Y, S) be spaces. The product topology onX × Y is the one with basis{U × V |U ∈ T , V ∈ U}.Inductively we can define a topology on Rnby Rn= Rn−1× R, where Rhas the standard topology. Luckily, this topology t u rn s out to be the sameas the metric topology. This will probably be a homework exercise.Proposition 5.15. Some observations:• For any y ∈ Y , the inclusion X → X × Y by x 7→ ( x, y) is continuous.Similarly inclusions from Y to X × Y are continuous.• There are canoni cal coordinate projections p1: X×Y → X (by (x, y) 7→y) and p2: X × Y → Y ; these are continuous.• The product topology on X × Y is the coarsest topology in which t hecoordinate projections are continuous.• The product topology is characterized by the following property: Forany space Z and any continuous maps f : Z → X and g : Z → Y , thereis a unique continuous map f ×g : Z → X ×Y such that p1◦( f ×g) = fand p2◦ (f × g) = g.5.6 Countability propertiesAs we saw in the example of an un cou ntable well ordering, there are un-countable spaces that behave very badly. On the other hand, ther e are un-countable spaces, like R with the standard topology, in which we still havea lot of control. The following definition s help distinguish these situations .19Definition 5.16. Let X be a space.• X is separable if X contains a countable den se subset.• X is second countable if X has a countable basis.• X is first countable if for each p ∈ X, there is a c ountable neigh-borhood basis: a set Npof open neighbor h oods of p, such that everyneighborhood of p contains some member of Np.Proposition 5.17. Any second countable space is both first countable andsepara ble.Proof . Let B be a countable basis for X. For each p ∈ X, the set {U ∈B|p ∈ U } is a countable neighborhood basi s . Let S consist of one point fromeach membe r of B. Then S is a countable dense subse t of X.Proposition 5.18. Any subspace of a first countable space is first countable,and any subspace of a second countable space is second countable.Proof i dea: We find our new basis or neighborhood bases by intersecting theold ones with the subset.Proposition 5.19. Finite products of separable spaces are separable.Proof i dea: The product of the countable dense su bs et s is a countable densesubset.Slightly less trivial:Proposition 5.20. Any separable m etr ic space is second countable.Proof . Let X be a metric space with metric d and countable dense subset S.We will show that the set B of open balls wi th rational radii around points i nS form a countable basis f or X. Consider the basic open set Br(x), for somex ∈ X and r > 0. Let y ∈ Br(x). There is an s > 0 such that Bs(y) ⊂ Br(x).Let z ∈ S ∩ Bs3(y) and let t ∈ (s3,2s3) ∩ Q. Then y ∈ Bt(z), and it followsfrom the triangle inequality that Bt(z) ⊂ Bs(y) ⊂ Br(x). So we have shownthat every point in Br(x) is i ns i de a set in B that is contained in Br(x).Therefore Br(x) is a union of members of B, and therefore B is a basis.Since B is in one-to-one correspondance with the product S × ( ( 0, ∞) ∩ Q)of countabl e sets, B is