2.2 SubspacesAny subset of a topological space inherits the structure of a space.Definition 2.6. Let (X, T ) be a space and let S be a subset of X. Thesubspace topology on S is the one whose open sets are exactly the intersec-tions of open subsets of X with S. Precisely, let S = {U ∩ S|U ∈ T }. Then(S, S) is the subspace topology on S.This topology on S is also called the indu ced topology.Proposition 2.7. Let X be a s pace and let S be a subset of X. The su b-space topology on S is a topology on S (that is, it satisfies the axioms for atopological space).The proof is straightfoward.Example 2.8. Some exampl e s of subspaces of R wit h the standard topol-ogy:• The various kinds of intervals [0, 1], [0, 1), (0, 1], (0, 1).• The set {0} ∪ {1/n|n ∈ N}.• The rational numbers Q.• The irrational numbers R\Q.2.3 Subbases and basesIf you want give someone the definition of a particular topological space, youmust specif y the point set and the set of all open sets. One way to specifythe set of all open sets is to let the axioms do the work.Proposition 2.9. Suppose X is a set and C is a set of subsets of S. LetB0be the set of all intersections of finitely many elements of C, and letB = B0∪ {X, ∅}. Let T be the set of all unions of elements of B. Then Tis the set of open sets for a topology on X.Proof. Clearly unions of elements of T are in T . Let U, V ∈ T , with U =Si∈IUiand V =Sj∈JVjfor some Ui, Vj∈ B. ThenU ∩ V = ([i∈IUi) ∩ ([j∈JVj) =[i∈I,j∈J(Ui∩ Vj).Since each Ui∩ Vjis clearly in B, it follows that U ∩ V is in T and T isclosed under finite intersections. Of course, since X and ∅ are in B, theyare in T as well, and T is the set of open sets for a topology on X.5Definition 2.10. Let (X, T ) be a topological space. A set B of open setsin X is a basis for (X, T ) if the set of all unions of elements of B is T . Aset C of open sets in X is a s ubbasis for (X, T ) if the se t of all intersectionsof finitely many elements of X, toget h er with X and ∅, form a basis for(X, T ).So Proposition 2.9 asserts that any set of subsets of a set X is a sub-basis for some topology on X. In parti c ul ar , one can specify a topol ogy byspecifying a subbasis.Proposition 2.11. Let X be a set and s u ppose B is a set of subsets of X.Suppos e for any U, V ∈ B, the intersection U ∩ V can be written as a u ni onof members of B. Further suppose that each point in X is in some U in B.Then B is a basis for some topology on X.Proof. Define T to be the set of all possible unions of sets of members ofB. Clearly T satisfies the axiom about unions. For U, V ∈ T , we writeU =Si∈IUiand V =Sj∈JVjwhere all the Uiand Vjare in B. Then U ∩ VisSi∈I,j∈J(Ui∩Vj). Since each Ui∩Vjcan be written as a union of membersof B, this U ∩ V can be written as such a union and inductively, T satisfiesthe axiom about unions. Since each p ∈ X is in some set in B, the unionof all memb er s of B is T , and therefore X is in T . S i nc e ∅ is the emptyunion, ∅ is i n T . So T satisfies the axioms for a topology, and B is a basi sfor T .Example 2.12. Bases and subbases can be used to desc ri be the spaces wehave already seen:• The set of open intervals is a basis for the standar d topol ogy on R.• The set of singleton subsets is a basis for the discrete top ol ogy on anyset X.• The topology on a set X with the subbasis{X\{p}|p ∈ X}is the finite complement topology.• If S is a sub se t of a space X and B is a basis for X, then {S ∩U|U ∈ B}is a basis for S with the subspace topol ogy.63 Continuity and homeomorphismsOften in mathematics, one is interested not only in studying mathematicalobjects, but also in studying functions between mat h emat i cal objects th atpreserve some of the structure. In topology, we mainly consider continuousfunctions.Definition 3.1. Suppose X and Y are topological spaces. A functionf : X → Y is continuous if for every open subset U of Y , the preimagef−1(U) is an open subset of X.This definition seems a little mysterious at first. In the ǫ–δ definitionfor the continuity of a func t ion f : R → R, we say f is continuous if forevery point p, it pulls back every small open interval around f(p) to a setcontaining a small open interval around p. This easily translates to sayingthat a continuous func t i on f pulls back every neighborhood of f(p) to aneighborhood of p. Since an open set i s a neighborhood of any of the pointsit contains, continuity should mean that a function pu ll s back open sets toopen sets.Continuity means nearby points get sent to ne arby points. However closetogether I want points to end up, if they start out close enough together, thefunction sends them to points that close together. however small an openset I want to hit wi t h my function, the set of points that hit it are still anopen set.Bases make it more convenient to check that a function is continuous.Proposition 3.2. Suppose X and Y are topological spaces and Y has abasis B. Let f : X → Y be a function and suppose f−1(U) is open for eachbasis element U ∈ B. Then f is continuous.Proof. Let V ⊂ Y be open. Write V as a union of elements {Ui}i∈Iof B:V =[i∈IUi.Then:f−1(V ) = f−1([i∈IUi) =[i∈If−1(Ui),which is open.Some near-trivial observations about continuity:7Proposition 3.3. • For any space X, the identity map from X to itselfis continuous.• Compositions of continuous maps are continuous: given continuousmaps between spaces f : X → Y and g : Y → Z, the composition g ◦f : X → Z is continuous.• If S is a subspace of a space X, then the i ncl u si on S → X is continu-ous.• If X has t he discrete topology, then any function from X to any spaceY is continuous; the discrete topology on X is the only topology on Xwith this property.• If X has the indiscrete topology, then any function from any space Yto X is continuous; the indiscrete topology on X is the only topologyon X with this property.Definition 3.4. Suppose X and Y are top ol ogic al spaces. A homeomor-phism from X to Y is an invertible, continuous function with continuousinverse. The spaces X and Y are homoemorphic if a …