aa r X i v : . [ m a t h . GN ] J a n A Report on HausdorffCompactifications of R Arnold Tan JunhanMichaelmas 2018 Mini Projects: Analytic Topology
University of Oxford
FoCS Analytic Topology
Contents R
75 Conclusion 10
The goal of this report is to investigate the variety of Hausdorff compactifications of R . The Alexandroffone-point compactification, the two-point compactification r´8 , , and the Stone- ˇ Cech compactificationare all clearly different. The ultimate aim is to show that there are in fact uncountably many. An inter-mediate aim is to exhibit one compactification of R different from all the compactifications already mentioned.We will often just write δX to refer to a compactification x l, δX y of a space X . We will compare two T compactifications of a space X by writing x l , δ X y ď x l , δ X y to mean that there is a continuous function L : δ X Ñ δ X such that L ˝ l “ l . (Such a function will automatically be onto.) It is not hard to see thatif δ X ď δ X and δ X ď δ X then δ X and δ X are homeomorphic as topological spaces.Let us declare two compactifications x l , δ X y and x l , δ X y to be equivalent if δ X ď δ X and δ X ď δ X .Then ď gives us a partial ordering on the set of equivalence classes of compactifications. This will be usefulfor us towards the end of the report, where we shall apply Zorn’s Lemma to this poset of equivalence classes.For that purpose, let us also recall here that an element p P P of a poset p P, ďq is maximal if whenever wehave q P P with p ď q , then p “ q . (When the equivalence class of a compactification is maximal – withrespect to ď , among all compactifications with some given property – we will simply say the compactificationis maximal.) On the other hand p P P is a greatest element if q ď p for all q P P . Writing p ă q to mean p ď q , p ‰ q (and writing p ć q otherwise), we see that p is maximal iff p ć q for all q P P . A greatestelement in a poset is unique and certainly maximal, however we may have several different maximal elements.A chain , or linearly ordered set , is a poset p P, ďq in which we have comparability of elements: for all p, q P P ,either p ď q or q ď p . In a chain, the notions of maximal and greatest element do coincide. The reader is surely familiar with the idea that the essence of the
Stone- ˇ Cech compactification x h, βX y canbe captured via a certain characterizing property. We run through the steps of showing this, and then,borrowing some of these ideas, we will exhibit a compactification of R that turns out to be different from x h, βX y . Definition 2.1.
Let X be a (nonempty) Tychonoff space.Let t f λ : λ P Λ u be a list of all bounded continuous functions from X to R .For each λ , let I λ be the smallest closed interval such that ran p f λ q Ď Iλ . That is, let I λ “ r inf ran p f λ q , sup ran p f λ qs .Let Y “ ś λ P Λ I λ be the Tychonoff product of the I λ .Define h : X Ñ Y such that for each λ P Λ , h p x qp λ q “ f λ p x q . Let βX “ cl Y p h p X qq .Define the Stone- ˇ Cech compactification of X to be x h, βX y . Let us briefly check that this is indeed a T compactification of X : Page 1FoCS Analytic Topology • Y “ ś λ P Λ I λ , which is compact (by Tychonoff’s Theorem) and T , since this is true for each of the I λ .Therefore, since βX is a subspace of Y , it is T ; since it is closed in Y , it is compact. • h is injective. Suppose we have distinct points x, y P X . Since X is Tychonoff (and hence t y u is closed),there is a continuous function f : X Ñ r , s such that f p x q “ , f pt y uq “ t u . f is bounded, so thereis λ P Λ with f “ f λ . Then, h p x qp λ q “ f λ p x q “ ‰ “ f λ p y q “ h p y qp λ q , so h p x q ‰ h p y q ; • h is continuous. A subbasic open set in Y has the form U λ ˆ ś µ ‰ λ I µ , where U λ is open in I λ . Set U “ p U λ ˆ ś µ ‰ λ I µ q X βX , then h ´ p U q “ t x P X : h p x q P U u “ t x P X : h p x qp λ q P U λ u “ t x P X : f λ p x q P U λ u “ f ´ λ p U λ q , and this is open, since f λ is continuous; • h ´ is continuous. It is enough to see that whenever x P U , where U is open in h p X q , there is an open V Q h p x q in h p X q such that h ´ p V q Ď U . Well, since x R X z U , which is closed, and X is Tychonoff,there is some continuous function f : X Ñ r , s such that f p x q “ and f p X z U q “ t u . f is a boundedcontinuous function from X to R , so there is λ P Λ with f “ f λ . Hence f λ p x q “ and f λ p X z U q “ t u .Note that V λ : “ r , q is open in r , s “ I λ , so V : “ p V λ ˆ ś µ ‰ λ I µ q X h p X q is open in h p X q . We have h ´ p V q “ t x P X : h p x q P V u “ t x P X : h p x qp λ q P V λ u “ t x P X : f λ p x q P r , qu “ f ´ λ r , q , but x P f ´ λ r , q Ď U , so x P h ´ p V q Ď U ; • cl βX p h p X qq “ βX holds. βX is the smallest closed set in Y containing h p X q , so it is the smallestclosed set in βX containing h p X q , because βX is closed in Y by construction. Lemma 2.2.
Let X be a Tychonoff space, and I be a closed bounded interval in R . Let f : X Ñ I becontinuous. Then there exists a continuous function βf : βX Ñ I such that βf ˝ h “ f .Proof. f is bounded and continuous, so there is some λ P Λ such that f “ f λ .Define βf : βX Ñ I, y ÞÑ y p λ q . This is a projection, so it is continuous.Furthermore, for all x P X , we have βf ˝ h p x q “ h p x qp λ q “ f λ p x q “ f p x q . Lemma 2.3.
Let X be a Tychonoff space, and Z “ ś µ P M I µ be a product of closed bounded intervals in R .Let f : X Ñ Z be continuous. Then there exists a continuous function βf : βX Ñ Z such that βf ˝ h “ f .Proof. Define f µ : X Ñ I µ , x ÞÑ f p x qp µ q . f µ “ π µ ˝ f , so it is continuous. Apply Lemma 2.2 to see thatthere exists a continuous function βf µ : βX Ñ I µ such that βf µ ˝ h “ f µ .Now define βf : βX Ñ Z such that for all µ , βf p x qp µ q “ βf µ p x q . That is, π µ ˝ βf “ βf µ .It remains to see that βf is continuous.A subbasic open set in Z has the form U “ U µ ˆ ś ν ‰ µ I ν , where U µ is open in I µ . We have βf ´ p U q “ t x P βX : βf p x q P U u “ t x P βX : βf p x qp µ q P U µ u “ t x P βX : βf µ p x q P U µ u “ p βf µ q ´ p U µ q . This is open, since βf µ is continuous. Lemma 2.4.
Any Tychonoff space X can be embedded in a product of closed bounded intervals.Proof. βX is a subset of such a product! Theorem 2.5 (The Stone- ˇ Cech Property) . Let X be a Tychonoff space.Say a compactification p k, γX q of X has the Stone- ˇ Cech property if whenever K is a compact T spaceand f : X Ñ K is continuous, there exists a continuous map γf : γX Ñ K such that γf ˝ k “ f .( γf will automatically be unique, since it is already determined on the dense set k p X q Ď γX .)Then p h, βX q has the Stone- ˇ Cech property.
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Proof.
Since K is compact T , it is Tychonoff. By Lemma 2.4, without loss of generality there is a product Z “ ś µ P M I µ of closed bounded intervals such that K Ď Z . Viewing f as a continuous function X Ñ Z ,Lemma 2.3 gives us a continuous function βf : βX Ñ Z such that βf ˝ h “ f . It only remains to see thatthe image of βf lies in K .K is compact in the Hausdorff space Z , hence K is closed in Z , so p βf q ´ p K q is closed in βX . Also, f p X q Ď K implies h p X q Ď p βf q ´ p K q . Since h p X q is dense in βX , we must have p βf q ´ p K q “ βX . Theorem 2.6. If p k, γX q is a Hausdorff compactification of X that has the Stone- ˇ Cech property, then x k , γ X y ď x k, γX y for any other compactification x k , γ X y .Proof. Take K “ γ X and f “ k in the definition of p k, γX q having the Stone- ˇ Cech property, to see thatthere exists a continuous map γh : γX Ñ γ X such that γk ˝ k “ k . This is precisely the statement that x k , γ X y ď x k, γX y .Since the Stone- ˇ Cech compactification has the Stone- ˇ Cech property, we deduce:
Corollary 2.7. x h, βX y is the largest compactification of X . On the other hand, we could set x k , γ X y in Theorem 2.6 to be the Stone- ˇ Cech compactification, to getanother corollary:
Corollary 2.8. If x k, γX y is a Hausdorff compactification of X that has the Stone- ˇ Cech property, then x h, βX y ď x k, γX y . This says that the Stone- ˇ Cech compactification is the smallest one having the Stone- ˇ Cech extension property.Suppose now that we consider the problem of extending a given family of bounded continuous functions on X , rather than all bounded continuous functions.For example, suppose we are asked to construct a Hausdorff compactification x k, γ R y of R that has the fol-lowing property: whenever f : R Ñ R is of the form f p x q “ cos p nx q for some n P Z , there exists a continuousfunction γf : γ R Ñ R such that γf ˝ k “ f .We give a construction of such a compactification x k, γ R y , by altering that of x h, βX y . We note beforehandthat cos p nx q “ cos p´ nx q for each n P Z , and the constant function cos p q “ extends trivially to anycompactification, so we need only consider n ě . Proposition 2.9.
Consider a set t f n : n P N u of functions from R to r´ , s , where f p x q “ tanh p x q , f n p x q “ cos p x q @ n ě . Let Y “ ś n P N r´ , s .Define k : R Ñ Y such that for each n P N , k p x qp n q “ f n p x q . Let γ R “ cl Y p k p R qq .Then x k, γ R y is a compactification of R .Proof. We check this is a compactification. • Y “ ś n P N r´ , s is compact (by Tychonoff’s Theorem) and T , since this is true for r´ , s . Therefore,since γX is a subspace of Y , it is T ; since it is closed in Y , it is compact. • k is injective. Suppose we have distinct points x, y P R . f p x q “ tanh p x q is strictly monotone, henceinjective. Therefore, k p x qp n q “ f p x q ‰ f p y q “ k p y qp n q , so k p x q ‰ k p y q ; • k is continuous. A subbasic open set in Y has the form U n ˆ ś m ‰ n r´ , s , where U n is open in r´ , s .Set U “ p U n ˆ ś m ‰ n I m q X γ R , then k ´ p U q “ t x P R : k p x q P U u “ t x P R : k p x qp n q P U n u “ t x P R : f n p x q P U n u “ f ´ n p U n q , and this is open, since f n is continuous; Page 3FoCS Analytic Topology • k ´ is continuous. It is enough to see that whenever x P U , where U is open in k p R q , there is an open V Q k p x q in k p R q such that k ´ p V q Ď U . Well, since x R R z U , which is closed, f p x q “ tanh p x q is suchthat f p x q R cl r´ , s p f p R z U qq .Then f p x q P V : “ r´ , sz cl r´ , s p f p R z U qq , which is open in r´ , s .The set V : “ p V ˆ ś n ‰ r´ , sq X k p R q is open in k p R q , and we have k ´ p V q “ t x P R : k p x q P V u “ t x P R : k p x qp q P V u “ t x P R : f p x q P V qu “ f ´ p V q , so x P f ´ V “ k ´ p V q shows that k p x q P V .Finally, note that k ´ p V q Ď U , since f p R z U q Ď r´ , sz V implies f ´ p V q “ R z f ´ pr´ , sz V q Ď U ; • cl γ R p k p R qq “ γ R holds.Next, let us show that each f n does extend continuously onto x k, γ R y . Lemma 2.10.
Let f n : R Ñ R , x ÞÑ cos p nx q , where n ě . Then there exists a continuous function γf n : γ R Ñ r´ , s such that γf n ˝ k “ f n .Proof. Simply define γf n : γ R Ñ r´ , s , y ÞÑ y p n q . This is a projection, so it is continuous.Furthermore, for all x P R , we have γf n ˝ k p x q “ k p x qp n q “ f n p x q .This already gives us the result that whenever f : R Ñ R is of the form f p x q “ cos p nx q for some n P Z ,there exists a continuous function γf : γ R Ñ R such that γf ˝ k “ f .Next, we show that x k, γ R y is the smallest compactification to which f n extends continuously for each n ě . Proposition 2.11.
Suppose x l, δ R y is a Hausdorff compactification of R that has the following property:for each n ě , there exists a continuous function δf n : δ R Ñ r´ , s such that δf n ˝ l “ f n . Then x k, γ R y ď x l, δ R y .Proof. Define F : δ R Ñ γ R as follows: for each y P δ R and n P N , let F p y qp n q “ δf n p y q . Clearly, F ˝ l “ k ,since for all x P R , F ˝ l p x qp n q “ F p l p x qqp n q “ δf n p l p x qq “ f n p x q “ k p x qp n q . It remains to see that F is continuous.Recall that γ R Ď ś n P N r´ , s , and a subbasic open set in ś n P N r´ , s has the form U n ˆ ś m ‰ n I m , where U n is open in r´ , s . Let U “ p U n ˆ ś m ‰ n I m q X γ R , then F ´ p U q “ t y P δ R : F p y q P U u “ t y P δ R : F p y qp n q P U n u “ t y P δ R : δf n p y q P U n u “ p δf n q ´ p U n q . This is open, since δf n is continuous.Note that this proposition does not quite tell us that x k, γ R y is the smallest compactification to which f p x q “ cos p nx q extends continuously for each n P Z , because among the f n is the function f p x q “ tanh p x q ,which we added to the family in order to construct γ R . We did this so that the family would separate pointsand closed sets ; for a more general construction see Folland (1999).Nevertheless, we shall show in the next section that this compactification is genuinely different from the oneswe have seen before. The compactification we have just constructed is genuinely different from any of the one-point, two-point,or Stone- ˇ Cech compactification of R . It cannot be the one-point or two-point compactification, because thefunction R Ñ R , x ÞÑ cos p x q does not extend continuously to either of these: Proposition 3.1.
Let f : R Ñ R , x ÞÑ cos p x q . There is no continuous function extending f to either theAlexandroff one-point compactification or the two point compactification of R . Page 4FoCS Analytic Topology
Proof.
Suppose for a contradiction we did have such an extension ˜ f .One way to write the one-point compactification is as x i , S y where i : R Ñ S , x ÞÑ p x ` x , x ´ ` x q . Notethat lim n Ñ8 i p n q “ p , q . Hence, by the continuity of ˜ f we would have ˜ f p , q “ ˜ f p lim n Ñ8 i p n qq “ lim n Ñ8 p ˜ f ˝ i qp n q “ lim n Ñ8 cos p n q , but this does not exist in R .Similarly, we may write the two-point compactification as x i , r´ , sy where i : R Ñ r´ , s , x ÞÑ tanh p n q .Now lim n Ñ8 i p n q “ , so ˜ f p q “ ˜ f p lim n Ñ8 i p n qq “ lim n Ñ8 p ˜ f ˝ i qp n q “ lim n Ñ8 cos p n q again contradicts that this limit does not exist.To show that γ R is not homeomorphic to β R , we will show that the former is metrisable while the latter is not.The following is a standard result. Lemma 3.2.
A countable product of metric spaces is metrisable.Proof.
The result is easy for finite products. (Alternatively, if you like, it is deducible from the case ofcountably infinite products, by setting all-but-finitely-many of the factors to be singletons.)Let tp X n , d n q : n P N u be a countably infinite family of metric spaces. Claim : We may assume each d n is bounded above by . Proof : To prove the claim, it is enough to see that any metric d on any space X has an equivalentmetric d defined by d p x, y q “ min t , d p x, y qu . This is a metric on X : – it is non-negative, and zero if and only if x “ y ; – it is symmetric in its variables; – min t , d p x, z qu ď min t , d p x, y qu ` min t , d p y, z qu .If d p x, y q , d p y, z q ď , thenmin t , d p x, z qu ď d p x, z q ď d p x, y q ` d p y, z q “ min t , d p x, y qu ` min t , d p y, z qu . Otherwise, without loss d p x, y q ą , thenmin t , d p x, z qu ď “ min t , d p x, y qu ď min t , d p x, y qu ` min t , d p y, z qu .d induces the same topology as d does on X . Indeed, wite B dr p x q and B d r p x q respectively for the openballs of radius r centered at x , with respect to d and d respectively. Since d ď d , we certainly have B dr p x q Ď B d r p x q for all r ą , x P X , so the topology induced by d is finer than that induced by d . Onthe other hand, for all r ą , x P X , we have B d r p x q Ď B dr p x q where r “ min t , r u . Indeed, suppose y P B d r p x q . Then min t , d p x, y qu ă min t , r u , so d p x, y q ă r .By the claim, we may assume each d n is bounded above by , so it makes sense to define, for x, y P ś n P N X n , d p x, y q “ Σ n P N d n p x p n q , y p n qq n , since this series converges to a value no greater than the convergent series Σ n P N n “ . Claim : d defined above is a metric on the product space ś n P N X n . Page 5FoCS Analytic Topology Proof : – it is non-negative, and zero if and only if every term in the series is zero, if and only if x and y agree on every component, if and only if x “ y ; – it is symmetric in its variables; – Σ n P N d n p x p n q ,z p n qq n ď Σ n P N d n p x p n q ,y p n qq n ` Σ n P N d n p y p n q ,z p n qq n follows immediately from the triangleinequalities for the individual d n .It remains to check that this metric induces the usual product topology on ś n P N X n .Given r ą , N P N , and x, y P ś n P N X n , we certainly have d N p x p N q , y p N qq ă r whenever d p x, y q ă r N .Therefore, the projections π N : p ś n P N X n , d q Ñ p X N , d N q are continuous with respect to these metrics.Therefore the topology τ d induced by d on the product space is finer than the Tychonoff topology τ . One wayto see this is via the universal property of the product: the projection maps π N : p ś n P N X n , d q Ñ p X N , d N q give rise to a unique continuous map i : p ś n P N X n , τ d q Ñ p ś n P N X n , τ q such that i ˝ π N “ π N for each N . Ofcourse, setting i to be the identity map satisfies this equation, and therefore we must have that the identityis continuous as a map p ś n P N X n , τ d q Ñ p ś n P N X n , τ q . In particular, taking the preimage of each open setunder the identity map, we see that τ Ď τ d .On the other hand, we show that any open set U in p ś n P N X n , τ d q is also open in the Tychonoff topology.Let x P U . There is some r ą with B dr p x q Ď U . Choose some k large enough so that Σ n “ k ` n “ k ă r .For each n P t , . . . , k u , define U n “ B d n r { p x p n qq . Then, x P k č n “ π ´ n p U n q Ď B dr p x q Ď U. Indeed, whenever y P Ş kn “ π ´ n p U n q , we have d n p x p n q , y p n qq ă r { for each n P t , . . . , k u , so d p x, y q “ Σ kn “ d n p x p n q , y p n qq n ` Σ n “ k ` d n p x p n q , y p n qq n ă r ` r “ r. Since Ş kn “ π ´ n p U n q P τ , we have shown that U is open in the Tychonoff topology, as required! Corollary 3.3. γ R is metrisable.Proof. γ R can be embedded into the product ś n P N r´ , s , which by the lemma above can be given a metricspace structure. Identifying γ R with its image in the product space, it will inherit the subspace metricinduced by the metric on the product space. Lemma 3.4.
A non-compact Tychonoff space has no maximal metrisable T compactifications.Proof. Suppose X is a non-compact metric space, and x m, ηX y is a metrisable T compactification. Weconstruct another compactification that is strictly larger. m p X q is homeomorphic to X , hence non-compact,hence m p X q ‰ ηX . Pick any x P ηX z m p X q . Since cl ηX p m p X qq “ ηX , there is a sequence p x n q of distinctpoints in m p X q converging to x in the metric d on ηX . (The open ball B p x q must meet m p X q at somepoint x ; the open ball B min t ´ i ,d p x,x i qu p x q must meet m p X q at some point x i ` for each i ě . In this waywe construct an infinite sequence of distinct points of m p X q whose distance to x tends to .)Consider the disjoint subsets S “ t x i : i is even u and S “ t x i : i is odd u of m p X q . Each S i is closed in m p X q , since no any sequence in S i has a limit in m p X q . (If the limit of such a sequence existed, it wouldhave to be x , but this is in ηX z m p X q .) Since m p X q is a subset of the metric space ηX , it is metrisableand hence normal, so by Urysohn’s Lemma there exists a continuous function F : m p X q Ñ r , s such that F p S q “ t u , F p S q “ t u . This function does not extend continuously to ηX . For if it did, then we wouldhave F p x q “ F p lim n Ñ8 x n q “ lim n Ñ8 F p x n q “ , Page 6FoCS Analytic Topologyand similarly F p x q “ F p lim n Ñ8 x n ` q “ lim n Ñ8 F p x n ` q “ , which taken together produce an obvious contradiction.Consider the function m : X Ñ ηX ˆr , s , s ÞÑ p m p s q , F p s qq . This is continuous since both of its componentsare continuous. Then x m , ˜ X y where ˜ X “ cl ηX ˆr , s p m p X qq is a T compactification of X : • ˜ X is compact and T , since it is a closed subspace of the compact T space ηX ˆ r , s ; • m is injective since its first component is injective; • m is continuous; • m is continuous as the composition of the first projection π : m p X q Ñ π p m p X qq and the map m ´ : m p X q Ñ X ; • cl ˜ X p m p X qq “ ˜ X holds.This compactification is larger than x m, ηX y , because there exists a continuous function π : ˜ X Ñ ηX suchthat π ˝ m “ m ; this is simply the first projection π : p z, t q ÞÑ z .On the other hand, F extends continuously to ˜ X ; consider ˜ F : ˜ X Ñ r , s , p z, t q ÞÑ t . This is just thesecond projection, so it is continuous, and we have ˜ F ˝ m “ F . Since F did not extend continuously to ηX , we conclude that there is no homeomorphism from ηX to ˜ X (or else we could compose ˜ F with such ahomeomorphism to get an extension of F to ηX ). In particular, ˜ X is a strictly larger compactification of X . Corollary 3.5.
For any non-compact Tychonoff space X , the Stone- ˇ Cech compactification x h, βX y is notmetrisable.Proof. βX is maximal among all compactifications, hence if it were metrisable it would be maximal amongall metrisable compactifications. Corollary 3.6. β R is not metrisable.Proof. R is a non-compact metric space!Now, clearly β R was homeomorphic to γ R , since one is metrisable and the other is not. We therefore obtainour desired result: Corollary 3.7. γ R is not homeomorphic to β R . R Our final task is to show that there are uncountably many different T compactifications of R .For this, we introduce the concept of the inverse limit (which really is a limit, in the categorical sense) of asequence of spaces with maps between them. Definition 4.1.
Suppose that x X n , d n y , for n P N , is a pair such that X n is a topological space, and d n : X n ` Ñ X n is continuous.The inverse limit x X ω , d ω,n y of the sequence xx X n , d n y : n P N y is defined as follows. Let X ω “ t x P ź n P N X n : @ n, x p n q “ d n p x p n ` qqu , and d ω,n “ π n : X ω Ñ X n be the restriction of the n th projection to X ω , so d ω,n p x q “ π n p x q “ x p n q for each x P X ω . Page 7FoCS Analytic TopologyObserve that each d ω,n is continuous, as the restriction of a continuous function. Observe also that for each n , d ω,n “ d n ˝ d ω,n ` , since d n ˝ d ω,n ` p x q “ d n p x p n ` qq “ x p n q “ d ω,n p x q . We now give a property that characterises the inverse limit.
Proposition 4.2.
Suppose x Y, x g n : n P N yy is any pair such that Y is a topological space, each g n : Y Ñ X n is continuous, and for all n , g n “ d n ˝ g n ` . Then there is a continuous function g : Y Ñ X ω such that forall n , g n “ d ω,n ˝ g .Proof. Simply define g : Y Ñ X ω , y ÞÑ x where x p n q “ g n p y q . This is well-defined, because for each n wehave x p n q “ d n p x p n ` qq : x p n q “ g n p y q “ d n ˝ g n ` p y q “ d n p g n ` p y qq “ d n p x p n ` qq . We also have g n “ d ω,n ˝ g , because d ω,n ˝ g p y q “ d ω,n p x q “ x p n q “ g n p y q . It only remains to show that g is continuous. We show that the preimage under g of each subbasic open setis open. Let U “ U j ˆ ś n ‰ j X n , where U j is open in X j . Then, g ´ p U X X ω q “ t y P Y : g p y q P U u “ t y P Y : g j p y q P U j u “ g ´ j p U j q . This is the continuous preimage of an open set, hence it is open.Let us make a few more easy observations.Firstly, if each d n is onto, then each d ω,n is onto. Indeed, given x n P X n , we can recursively find x i P X i foreach i ą n such that d i p x i q “ x i ´ , by surjectivity of the d i . We can also define, for i ă n , x i “ d i ` p x i ` q .Define x P X ω by x p n q “ x n ; then d ω,n p x q “ x p n q “ x n .Also, if all of the spaces X n are compact Hausdorff, then X ω is compact Hausdorff. Indeed, ś n P N X n isHausdorff and compact by Tychonoff’s theorem, so if we know that X ω is a closed subspace, then it isHausdorff and compact. It remains to see that X ω is closed in ś n P N X n . Well, X ω “ č N P N t x P ź n P N X n : x p N q “ d N p x p N ` qqu “ č N P N ψ ´ p ∆ X N ˆ X N q , where ψ : ś n P N X n Ñ X N ˆ X N , x ÞÑ p π N p x q , d N ˝ π N ` p x qq “ p x N , d N p x p N ` qqq is continuous, sinceeach component is continuous in x . Since X N is Hausdorff, the diagonal ∆ X N ˆ X N “ tp x, x q : x P X N u isclosed in X N ˆ X N . Therefore each ψ ´ p ∆ X N ˆ X N q is closed as the continuous preimage of a closed set.Hence X ω is closed, as the intersection of closed sets. Lemma 4.3. If xx g n , δ n R y : n P N y is a sequence of metrisasble T compactifications of R such that for all n , δ n R ď δ n ` R , then there exists a metrisable T compactification δ ω R of R such that for all n , δ n R ď δ ω R .Proof. By assumption, for each n there is an onto function d n : δ n ` R Ñ δ n R such that d n ˝ g n ` “ g n . Letus take the inverse limit of the system xx δ n R , d n y : n P N y . Call it x X ω , d ω,n y . By definition X ω is a subspaceof a countable product of the spaces δ n R , and is therefore metrisable by metrisability of each of the δ n R .We have remarked above that X ω must be compact Hausdorff, since each individual space δ n p R q is. Sincewe have a pair x R , x g n : n P N yy such that R is a topological space, each g n : R Ñ δ n R is continuous, andfor all n , g n “ d n ˝ g n ` , by Proposition 4.2 there is a continuous function g : R Ñ X ω such that for all n , g n “ d ω,n ˝ g . Let δ ω R “ cl X ω p g p R qq . Then x g, δ ω R y is the desired compactification: • δ ω R is compact T and metrisable, since it is a closed subspace of the compact T and metrisable space X ω ; Page 8FoCS Analytic Topology • g is injective since g is injective; • g is continuous by assumption; • Suppose U is open in R . We claim g p U q is open in g p R q . Well, d ω, ˝ g p U q “ g p U q is open in δ R . (Itis open in g p R q , which is in turn open in δ R as R is locally compact). Then, d ´ ω, p g p U qq is open in δ ω R , and g p U q “ d ´ ω, p g p U qq X g p R q shows that g p U q is open in g p R q . Altogether this shows that g ´ is continuous; • cl δ ω R p g p R qq “ δ ω R holds. • for all n , δ n R ď δ ω R . This is witnessed by the continuous functions d ω,n : δ ω R Ñ δ n R . We haveremarked that they are onto because the d n are onto; furthermore, for each n we have g n “ d ω,n ˝ g .We are almost ready to show that R has uncountably many (non-equivalent) T compactifications. For this,let us recall Zorn’s Lemma. Lemma 4.4 (Zorn’s Lemma) . Let A “ p A, ďq be a nonempty poset in which every nonempty chain has anupper bound. Then A has a maximal element. Theorem 4.5. R has uncountably many T compactifications.Proof. Suppose R has only countably many T compactifications. In particular R has only countably many metrisable T compactifications. We may assume without loss that there are countably infinitely many ofthese. (If there are only finitely many metrisable T compactifications, then certainly one of these is maximalamong all the others; this contradicts Lemma 3.4.)Let the set of all metrisable T compactifications of R be A “ tx h n , δ n R y : n P N u . (We write x h n , δ n R y forease of notation, but we really mean its class rx h n , δ n R ys , of course.)This is a poset. We show that it has a maximal element, by checking that it satisfies the conditions of Zorn’sLemma. A is nonempty, since it contains x k, γ R y . Suppose C Ď A is a nonempty chain. We need to exhibitan upper bound for C . We split into two cases: • If C has only finitely many elements, write these as x h , δ R y ď . . . ď x h r , δ r R y . Then δ r R is a greatestelement of the chain, hence certainly an upper bound. • If C has countably infinitely many elements, write C “ tx h n , δ n R y : n P N u . Let us assume without lossthat C has no maximal element. (A maximal element in a chain would also be a greatest element andhence an upper bound for the chain, so we would be done.) Note also that each nonempty finite subset C of C is still a chain, and by the above case, C has a greatest element max t C u . We now construct asequence xx h n i , δ n i R y : i P N y of compactifications in C such that for all i , δ n i R ď δ n i ` R .Let n “ . x h n , δ n R y is not a maximal element of the chain, so there is n ą n with δ n R ă δ n R .For r ą , at the r th stage consider the finite subchain C r “ t δ i R : 0 ď i ď n r u ; max t C r u is not amaximal element of C , so there is n r ` ą n r with max t C r u ă δ n r ` R .We have inductively defined a sequence xx h n i , δ n i R y : i P N y such that for each i , δ n i R ď δ n i ` R .Therefore Lemma 4.3 applied to this sequence xx h n i , δ n i R y : i P N y (now considered as a sequence ofactual compactifications rather than classes of these) tells us that there exists a metrisable T com-pactification δ ω R such that for each i , δ n i R ď δ ω R . δ ω R is an element of A ; let us show that it is an upper bound for C .Well, for each r P N we have n r ě r so δ r R is among δ R , . . . , δ n r R . Therefore δ r R ď max t C r u ă δ n r ` R ď δ ω R , as required. Page 9FoCS Analytic TopologyWe have now shown that A satisfies the conditions of Zorn’s Lemma, and so has a maximal element x h max , δ max R y . That is, x h max , δ max R y is maximal among all metrisable T compactifications of R .This contradicts Lemma 3.4. Therefore R could not have only countably many T compactifications! In Section 2, for the problem of finding a compactification of R to which the family f n p x q “ cos p x q extendedcontinuously, we could have gone a different route by defining x k, γ R y as follows. Take k : R Ñ r´ , s ˆ r´ , s , x ÞÑ p tanh p x q , cos p x qq , and let γ R be the closure of the image of k in r´ , s ˆ r´ , s . Indeed, a bit of thought shows that if wehave found a compactification x k, γ R y onto which f p x q “ cos p x q extends continuously, then for each n P Z , f n p x q “ cos p nx q will also extend continuously.This relies on the fact that each f n p x q “ cos p nx q can be expanded as a polynomial T n in cos p x q : cos p nx q “ T n p cos p x qq , where, in fact, T n is the n th Chebyshev polynomial.Therefore, if we have a compactification x k, γ R y and a continuous function γf : γ R Ñ R such that γ f ˝ k “ f , then this would also yield, for each n P Z , a continuous function γf n : γ R Ñ R such that γf n ˝ k “ f n . Simply take γf n “ T n ˝ γf : γf n ˝ k “ T n ˝ γf ˝ k “ T n ˝ f “ f n . The advantage of this approach is that we can instantly see this space is metrisable, as a subspace of r´ , s ˆ r´ , s . This means we do not need to rely on the result that a countable product of metric spacesis metrisable.Notice also that we did not prove that our choice of x k, γ R y was smallest among all compactifications towhich the family f n p x q “ cos p nx q extends continuously – this was not necessary for us to show that x k, γ R y is distinct from the one-point, two-point, and Stone- ˇ Cech compactification. Page 10FoCS Analytic Topology
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