Abstract topology studies the the notions of nearness and proximity of points by axiomatising the concept of an open neighbourhood of a point. Intuitively, an open set is one with the property that if it contains a point x, then it contains all points near x. In the hyperreal context we can make this idea quite explicit by taking "near" to mean "infinitely close"n. As we shall see, this leads to a very natural formulation and treatment of many topological ideas.
If A IR and r E IR, then the following are standard definitions: • r is an interior point of A if (r-e, r +e) A for some real e > 0. The interior of A is the set A0 of interior points of A. • r is a closure point of A if ( r -e, r + e) intersects A for every real c > 0. The (topological) closure of A is the set A of closure points of • r is a limit point of A if for every real c > 0, (r -c, r + e) intersects A in a point other than r. The set A' of limit points of A is the derived set of A. It follows readily from these definitions that Ao A A = A u A'. 114 10. Topology of the Reals Now, r is an interior point of A if all points within some positive real distance of r belong to A. Our discussion of permanence in Section 7.10 suggests that this property should be equivalent to requiring that all points infinitely close tonr should belong to *A. This is confirmed by the first part of the next result. Theorem 10.1.1 If Aos;;;; lR and r E JR, (1) r is interior to A if and only ifr x implies x E *A, i.e., iff hal(r) s;;;; (2) r is a limit point of A if and only if there is an x -=1-r such that · r x E *A, i.e., iff hal(r) n *A contains a point other than r. (3) r is a closure point of A if and only if r is infinitely close to some x E *A, i.e., iff hal(r) n *A is nonempty. Proof (1) Let r E A0• Then (rn-c, r +c) s;;;; A for some real c > 0. Then the sentence ('Vx E lR) (irn-xi < c ----+ x E A) (i) is true. But now if r x in *JR, then ir -xi < c, so by universal transfer of (i), x E *A. This shows that hal(r) s;;;; *A. (An alternative way of putting this is to observe that hal(r) s;;;; *(rn-c, r +c) s;;;; *A.) Conversely, if hal( r) s;;;; *A, then the sentence (3c E *JR+) ('Vx E *JR) (irn-xi < c ----+ x E *A) is seen to be true by interpreting c as any positive infinitesimal. But then by existential transfer there is some real c > 0 for which (i) holds, so (rn-c, r +c) s;;;; A and hence r E A0• (2) If r E A', then the sentence (Vc E JR+) (3x E lR)(xn-=/-r Air-xl < c Ax E A) (ii) is true. Now take c to be a positive infinitesimal. Then by transfer of (ii), there is a hyperreal x -=1-r with irn-xi < c, whence r x, and X E *A. Conversely, suppose there exists x E hal(r) n *A with x -=1-r. Then if c > 0 is real, irn-xl < c, since r "'x, and thus the sentence (3x E *lR) (x -=/-r A lr-xi < c AxnE *A) is true. By transfer then, there exists a real number distinct from r that belongs to ( r -c, r + c) n A. This shows that r is a limit point of A. 10.2 Open and Closed Sets 115 (3) If r E A, then either r E A, in which case r E hal(r) n *A, or else roEnA', in which case hal(r) n *Ao=/= 0, again by part (2). Conversely, if there exists x E hal(r) n *A, then either x = *An.IR =A, or else x =/= r, and sorE A' by (2). Thus r E AUA' =A. 10.2 Open and Closed Sets If A JR, then • A is open if all its points are interior to it, i.e., Ao =A. • A is (topologically) closed if it contains all its closure points, i.e., A = A U A', this is equivalent to requiring that A contain all its limit points, i.e., A' A. In view of Theorem 10.1.1, it follows that • A is open if and only if for all r E A, if x is infinitely close to r, then X E *Ao; • A is closed if and only if for all real r, if r is infinitely close to some x E *A, then r E A. Theorem 10.2.1 (1) A is open in lR if and only if its complement A c = lR -A is closed in JR. (2) The collection of open sets is closed under finite intersections and arbitrary unions. (3) The collection of topologically closed sets is closed under finite unions and arbitrary intersections. Proof (1) Observe that x E *(Ac) iff x rf. *A, by transfer of r, sonrE D Since A = Suppose A c is closed. To show A is open, let x c::= r E A. Then we must show x E *A. Now, if x E *(Ac), then we would have r c::= x E *(Ac), making r a closure point of Ac, so as Ac is closed, r E Ac, contradicting rEA. Thus we must have x rf. *(Ac), implying x E *A, as desired, by the above observation. The converse is similar, and given as an exercise. 116 10. Topology of the Reals (2) Let A1, ... , An be open. If x r E A1 n · · · noAn, then for each i, x rv r E Ai, and sonx E *Ai. Hence X E *Al n .n.o. n *An(= *(Al n ... nUAn)· This shows that A1 n ··•·(nAn is open. Now let {Ai : i E I} be a collection of open sets. If X r E uiEJ Ai, ·( then for some j, Xr E Aj, so X E *Aj *(UiEJ Ai)· Hence uiE;(Ai is open. (3) Exercise. 0 Theorem 10o2o2 For any real number r, hal(r) = n{*A: rEA and A is open}. Proof. We have already observed that if r E A IR and A is open, then hal(r) *A. On the other hand, if x hal(r), then x '/=-r, so there must exist some realn£ > 0 such that lr-xi > £. Put A = (ro-£, r +c) JR. Then rEA and A is open, but x *A= {y E *IR(: lr-Yl < c}. 0 A topology on a set X is defined axiomatically to be a collection of subsets of X that includes 0 and X and is closed under finite intersections and arbitrary unions. The members of this collection are declared to be the open sets, and their complements are called closed. In such a setting it is possible to study any topological idea that can be characterised by properties of open and closed sets, even in the absence of a notion of numerical distance between points. For instance, the halo of a point r E X could be defined by the equation of Theorem 10.2.2, leading to a nonnumerical account of "infinite closeness" of points. Exercise 10o2o3 Show that the proof of 10.2.1(2) does not work for infinite intersections by showing that 1003 Compactness A set B IRis compact if every open cover of B has a finite subcover, i.e., if whenever B UiEI Ai and each Ai is open in IR, then there is a finite 10.3 Compactness 117 This concept does not appear out of thin air. It emerged from studies in the nineteenth century of bounded and closed intervals in the real line, leading to a proof that such intervals are compact in the sense just de.: fined (Beine-Borel theorem). Since the definition refers only to open sets, it becomes the appropriate one to use for an abstract topological space where there is no notion of numerical distance to specify the notion of boundedness. Robinson's Compactness Criterion. B is a compact subset of if and only if every member of * B is infinitely close to some member of B, i.e., iff *B U hal(r). rEB Since the members of B are all real, the only such member that a given x E * B could be infinitely close to is its shadow. Thus another way to state Robinson's criterion is if x E *B, then x is limited and sh(x) E B. This criterion gives an intuitively appealing and useful characterisation of the notion of compactness. Constructions involving open covers are replaced by elementary reasoning about hyperreal points. For instance: • The open interval (0, 1) lR is not compact, because if cis a positive infinitesimal, then c E *(0, 1) as 0 < c < 1, but cis not infinitely close to any member of (0, 1) because its shadow is 0 (0, 1). • Any closed interval [a, b] lR is compact, because if x E *[a, b], then a :::; x b, sonx is limited and its shadow r must also satisfy an:::; r b. Thus x r E [a, b]. • Any finite set is compact, because if B is finite, then *B = B, so each member ofn*B is infinitely close to itself in B. • If B lR is unbounded above, in the sense that (Vx E JR) (3y E B) (x < y), then B cannot be compact: taking any unlimited x E *JR, by transfer there exists y > x with y E *B . Then y is unlimited, so cannot be infinitely close to any member of B. Similarly, B cannot be compact if it is unbounded below. Altogether then, a compact set must be bounded above and below. • If B is not closed, then B cannot be compact: it must have a closure point r that does not belong to B. As a closure point, r is infinitely close to some x E *B. But then x is not infinitely close to any member of B, since sh(x) = r B. 118 10. Topology of the Reals Hence a compact set must be closed. Proof of Robinson's Criterion. We will show that Robinson's criterion fails if and only if the standard definition of compactness fails. If Robinson's criterion fails, there is a hyperreal b E *B that is not infinitely close to any member of B. Then for each r E B, b j:. r, so there must be a real E:r > 0 such that lb-ri 2: E:r. Then {(r -En r + E:r) : r E B} is an open cover of B. But this cover can have no finite subcover: if, say, then by properties of enlargements of sets (3.10), Since bE *B, it then follows that bE *(ri -Eri,ri+e:rJ, and hence jb-ri l < E:ri for some i, contradicting the definition of Er; . Thus compactness fails for B. For the converse, suppose B is not compact, so that there is an open cover C = {Ai : i E J} of B that has no finite subcover. Each r in B belongs to Aj for some j E J, and hence r E (r-e:, r +c) Aj for some c E R+ because Ai is open. But then using the density of the rationals we can find some rational numbers p, q with r E (p, q) Aj. This shows that there is an open cover C' of B by intervals with rational end points, each of which is included in a member of C. Because the rationals are countable, there are only countably many intervals with rational end points, so we can enumerate C' and write C' = {(pn , Qn) : n E N}, where (pn : n EN) and (qn : n EN) are sequences of rational numbers. Now, C' includes no finite subcovering of B, or else this would lead to a finite subcover from C, since each member of C' is included in a member of C. Thus for each k EN, We can express this fact by the following true sentence: (Vk EN) (3x E B) (\fn E N) [n ::::; k ---+ -.(pn < x < Qn)] . Now take an unlimited K E *N. Then by transfer there exists some hyperreal x E * B such that the statement Pn < x < Qn is false for all n E *N with n ::::; K. In particular, Pn < x < Qn is false for all standard n. But now x cannot be infinitely close to any r E B, because such a point r belongs to (Pn , Qn) for some n E N, and so if x r, then Pn < x < Qn· Hence Robinson's criterion fails, and the proof is complete. D 10.4 Compactness and (Uniform) Continuity 119 Robinson's criterion can be established for abstract compact topological spaces (and leads to a beautifully simple proof of Tychonoff's theorem that the product of compact spaces is compact). In that context the reduction to a countable cover via the density of Q in JR is not generally applicable, and instead a special principle of hyperreal analysis, known as enlargement (cf. Chapter 14), is needed to establish the criterion. Theorem 10.3.1 (Heine-Borel) A set B 1R is compact if and only if it is closed and bounded. Proof We have already seen that if B satisfies Robinson's criterion, then it is closed and bounded (above and below). Conversely, if B is closed and bounded, then there is some real b such that (Vx E B) ( JxJ b). Now, to prove Robinson's criterion, suppose x E *B. Then by transfer, Jxl b E JR. Hence x is limited, and so has a shadow r E JR. Then r x E * B, and so r E B because B is closed. Thus we have shown that x is infinitely close to the member r of B, proving that B is compact. D Exercise 10.3.2 Use Robinson's criterion to prove that in JR a closed subset of a compact set is compact. 10.4 Compactness and (Uniform) Continuity Compactness is an inherently topological notion, being preserved by continuous transformations. Here is a simple hyperreal proof of that fact. Theorem 10.4.1 The continuous image of a compact set is compact. Proof Let f be a continuous real function, and B a compact subset of lR included in the domain of f. Now, it is true, by definition of f(B), that (Vy E f(B)) (3x E B) (y = f(x)). Thus by transfer, if y E *(!(B)), then y = f(x) for some x E *B. Since B is compact, x r for some r E B. Then by continuity of J, f(x) f(r), i.e., y is infinitely close to f(r) E f(B). This shows by Robinson's criterion that f(B) is compact. D In Theorem 7.7.2 it was shown that a continuous function on a closed interval [a, b] is uniformly continuous. We can now see that it is the compactness of [a, b] that accounts for this phenomenon: 120 10. Topology of the Reals Theorem 10.4.2 Ifof is continuous on a compact set B JR, then f is uniformly continuous on B. Proof By Theorem 7.7.1 we have to show that for all x,y E *B, x y implies f(x) f(y). But if x,y E *B, then by compactness x rEB andnyo s E B for some r, s. Thus if x y, then r s, and so r = s, as both are real. Hence by continuity ofnf at rEB, f(x) f(r) and f(y) f(r), whence f(x) f(y) as desired. 0 10.5 Topologies on the Hyperreals There is no canonical way to extend the definitions of interior, closure, and limit point-and hence the definitions of open and closed sets-to general subsets of *lR. These definitions depend on the concept of an open neighbourhood (r -c, r +c) of a point r, and one option would be to allow r to be any member of *JR but to continue to require that the radius c be a positive real number. Here (ro-c, r +c) is the hyperreal interval { x E *JR : r -c < x < r + c}, and we call it a real-radius neighbourhood of r when cis a real number. Thus a subset of *lR will be called real-open if it is a union of realradius neighbourhoods. Examples of real-open sets include the sets of limited numbers, unlimited numbers, positive (respectively negative) unlimited numbers, appreciable numbers, positive (respectively negative) appreciable numbers, and any galaxy (Section 5.4). The class of real-open sets is closed under arbitrary unions, but is not a topology on *lR because it is not closed under finite intersections. For instance, if (ro-c, r +c) and (s-8, s + 8) are real-radius neighbourhoods that overlap in such a way that s -8 is infinitely close to r + c, then the intersection (s-8, r +oc) has infinitesimal width and so does not contain any real-radius neighbourhoods, hence is not real-open. This suggests that the overlaps between real-radius neighbourhoods are in some sense "too small"n. One way to remedy this is to modify the neighbourhoods by removing those members that are infinitely close to the end points, retaining those that are well inside the interval (as defined in Section 8.11), thereby forcing any overlaps to be appreciable. To formalise this, put (( r -c, r + c)) {x E *lR: xois well inside (ro-c, r +oc)} -{ x E *lR : hal( x) ( r -c, r + c)}. A set of the form ((r-c, r+c)) with real c will be called an S-neighbourhood, and an S-open set is one that is a union of S-neighbourhoods. The S-open sets form the S-topology on *JR, first introduced by Abraham Robinson. 10.5 Topologies on the Hyperreals 121 (The "S-" prefix here is for "standard" and is typically used when a standard concept, or the nonstandard characterisation of some standard concept, is applied more widely to nonstandard entities. In Theorem 11.14.4, S-openness of a set B will be related to the notion of B being open when it includes the halo of each of its points.) Note that when c E JR+, each point of ((r -c, r +c)) is of appreciable distance from r-c and from r+c. Alternatively, we can define ((r-c, r+c)) as consisting of those points x whose distance from r is appreciably less than c, in the sense that sh!ro-xl < c, i.e., c-jr -xl is appreciable. The intersection of two S-neighbourhoods ((r -c, r +oc)) and ((s -6, s + 6)) is S-open, because if t belongs to this intersection, then ((t -1, t + 1)) ((ro-c, r +c)) n ((s-8, s + 6)), where 1 is any positive real number such that 1U:::; min{c-shjro-tj, 6-shjs-tj}. From this it follows that the intersection of S-open sets is S-open. Exercise 10.5.1 Show that (i) any S-open set is real-open; (ii) each S-open set is a union of halos, but a union of halos need not be S-open; (ii) no real-radius neighbourhood can be S-open. D Another topology on *JR is obtained if neighbourhoods are allowed to have any positive hyperreal (possibly infinitesimal) as radius. Thus any hyperreal open interval (a, b) can be a neighbourhood, and we define a set A *IR to be interval-open if it is a union of intervals (a, b) with a, b E *R The interval-open sets form the interval topology on *JR. It is immediate that all real-open sets are interval-open, but the converse is not true. There are many hyperreal intervals (a, b) that are interval-open but not real-open, for instance any having a b. Also, any halo becomes interval-open but is not real-open. Consider furthermore the construction n{A :rEA and A is open}. If "open" here means real-open (or S-open), then this gives the halo of r. If it means interval-open, then the result is just { r}. Any hyperreal interval (a, b) can be readily constructed as the intersection of two real-radius neighbourhoods (in many ways). Therefore any 122 10. Topology of the Reals topology that includes the real-radius neighbourhoods must include all hyperreal intervals, and hence all interval-open sets. Exercise 10.5.2 Let A be an open subset of JR. (i) Show that *A is interval-open in *JR. (ii) Suppose A is the union of a sequence (An : n E N) of pairwise disjoint open intervals in JR, with the length of An being less than l. Use transfer to show that some element of *A is infinitely close to n something not in *A. Deduce that *A is not S-open. (iii) Show further that *A contains a point that does not belong to any real-radius neighbourhood that is included in *A. Hence deduce the stronger result that *A is not real-open. The relationships between various topologies on *JR will be explored further in Section 11.14.