So every Cauchy sequence is eventually constant and therefore converges. ... an
isometry also takes Cauchy sequences to Cauchy sequences and non-Cauchy ...
Chapter IV Completeness and Compactness 1. Introduction In Chapter III we introduced topological spaces a generalization of pseudometric spaces. This allowed us to see how certain ideas continuity, for example can be extended into a situation where there is no “distance” between points. Continuity does not really depend on the pseudometric . but only on the topology. Looking at topological spaces also highlighted the particular properties of pseudometric spaces that were really important for certain purposes. For example, it turned out that first countability is the crucial ingredient for proving that sequences are sufficient to describe a topology, and that Hausdorff property, not the metric . , is what matters to prove that limits of sequences are unique. We also looked at some properties that are equivalent in pseudometric spaces for example, second countability and separability but are not equivalent in topological spaces. Most of the earlier definitions and theorems are just basic “tools” for our work. Now we look at some deeper properties of pseudometric spaces and some significant theorems related to them.
2. Complete Pseudometric Spaces Definition 2.1 A sequence ÐB8 Ñ in a pseudometric space Ð\ß .Ñ is called a Cauchy sequence if a% ! b Mß N − such that if 7ß 8 R , then .ÐB7 ß B8 Ñ %. Informally, a sequence ÐB8 Ñ is Cauchy if its terms “get closer and closer to each other.” It should be intuitively clear that this happens if the sequence converges, and the next theorem confirms this. Theorem 2.2 If ÐB8 Ñ Ä B in Ð\ß .Ñ, then ÐB8 Ñ is Cauchy. Proof Let % !. Because ÐB8 Ñ Ä B, there is an R − such that .ÐB8 ß BÑ #% for 8 R . So if 7ß 8 R , we get .ÐB7 ß B8 Ñ Ÿ .ÐB7 ß BÑ .ÐBß B8 Ñ #% #% œ %. Therefore ÐB8 Ñ is Cauchy. ñ A Cauchy sequence does not always converge. For example, look at the space Ðß .Ñ where . is the usual metric. Consider a sequence in that converges to # in ‘à for example, we could used Ð;8 Ñ œ Ð"ß "Þ%ß "Þ%"ß "Þ%"%ß ÞÞÞ Ñ. Since Ð;8 Ñ is convergent in ‘ß Ð;8 Ñ is a Cauchy sequence in (“Cauchy” depends only on . and the numbers ;8 , not whether we are thinking of these ;8 's as elements of or ‘..) But Ð;8 Ñ has no limit in . (Why? To say that “#  ” is part of the answer.)
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Definition 2.3 A pseudometric space Ð\ß .Ñ is called complete if every Cauchy sequence in Ð\ß .Ñ has a limit in \ .
Example 2.4
Throughout this example, . denotes the usual metric on subsets of ‘.
1) Ð‘ß .Ñ is complete. (For the moment, we take this as a simple fact from analysis; however we will prove it soon.) But Ðß .Ñ and Ðß .Ñ are not complete À completeness is not hereditary! 2) Let ÐB8 Ñ be a Cauchy sequence in Ðß .Ñ. Then there is some R such that lB7 B8 l 1 for 7ß 8 R . But the B8 's are integers, so this means that B7 œ B8 for 7ß 8 R . So every Cauchy sequence is eventually constant and therefore converges. Ðß .Ñ is complete. 3) Consider with the metric ." Ð7ß 8Ñ œ l 7" 8" l. Then the sequence ÐB8 Ñ œ Ð8Ñ is Cauchy. If 5 − , then ." ÐB8 ß 5Ñ œ l 8" 5" l Ä 5" Á ! as 8 Ä ∞ , so ÐB8 Ñ does not converge to 5 . The space Ðß ." Ñ is not complete. However Ðß ." Ñ has the discrete topology, so the identity function 0 Ð8Ñ œ 8 is a homeomorphism between Ðß ." Ñ and the complete space (ß .ÑÞ So completeness is not a topological property. A homeomorphism takes convergent sequences to convergent sequences and non convergent sequences to nonconvergent sequences. For example, Ð8Ñ does not converge in Ðß ." Ñ and Ð0 Ð8ÑÑ does not converge in Ðß .Ñ. These two homeomorphic metric spaces have exactly the same convergent sequences but they do not have the same Cauchy sequences. Changing a metric . to a topologically equivalent metric . w does not change the open sets and does not change which sequences converge. But the change may create or destroy Cauchy sequences since this property depends on distance measurements. Another similar example: we know that any open interval Ð+ß ,Ñ is homeomorphic to ‘ . But Ð+ß ,Ñ is not complete and ‘ is complete (where both space have the metric . ). 4) In light of the comments in 3) we can ask: if Ð\ß .Ñ is not complete, might there be a different metric . w µ . for which Ð\ß . w Ñ is complete? To be specific: could we find a metric .w on so . µ . w but where Ðß . w Ñ is complete? We could also ask this question about Ðß .ÑÞ Later in this chapter we will see that the answer is “no” in one case but “yes” in the other. 5) Ðß ." Ñ and ÐÖ 8" À 8 − ×ß .Ñ are both countable discrete spaces, so they are homeomorphic. In fact, the function 0 Ð8Ñ œ 8" is an isometry between these spaces: ." Ð7ß 8Ñ œ | 7" 8" l œ .Ð0 Ð7Ñß 0 Ð8ÑÑÞ An isometry is a homeomorphism, but since it preserves distances, an isometry also takes Cauchy sequences to Cauchy sequences and non-Cauchy sequences to non-Cauchy sequences. Therefore if there is an isometry between two pseudometric spaces, then one space is complete iff the other space is complete. For example, the sequence ÐB8 Ñ œ 8 is Cauchy is Ðß ." Ñ and the isometry 0 carries ÐB8 Ñ to the Cauchy sequence Ð0 ÐB 8 ÑÑ œ Ð 8" Ñ in ÐÖ 8" À 8 − ×ß .Ñ. It is clear that Ð8Ñ has no limit in the domain and Ð 8" Ñ has no limit in the range. Ðß ." Ñ and ÐÖ 8" À 8 − ×ß .Ñ “look exactly alike” not just topologically but as metric spaces. We can think of 0 as simply renaming the points in a distance-preserving way. Theorem 2.5 If B is a cluster point of a Cauchy sequence ÐB8 Ñ in Ð\ß .Ñ, then ÐB8 Ñ Ä B.
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Proof Let % !. Pick R so that .ÐB7 ß B8 Ñ #% when 7ß 8 R Þ Since ÐB8 Ñ clusters at B, we can pick a O R so that BO − F #% ÐBÑ. Then for this O and for 8 R ß .ÐB8 ß BÑ Ÿ .ÐB8 ß BO Ñ .ÐBO ß BÑ #% #% œ % so ÐB8 Ñ Ä B. ñ Corollary 2.6 A Cauchy sequence ÐB8 Ñ in Ð\ß .Ñ converges iff it has a cluster point. Corollary 2.7 In a metric space Ð\ß .Ñ, a Cauchy sequence can have at most one cluster point.
Theorem 2.8 A Cauchy sequence ÐB8 Ñ in Ð\ß .Ñ is bounded that is, the set ÖB" ß B# ß ÞÞÞß B8 ß ÞÞÞ× has finite diameter. Proof Pick R so that .ÐB7 , B8 Ñ " when 7ß 8 R . Then B8 − F" ÐBR Ñ for all 8 R Þ Let < œ max Ö"ß .ÐB" ß BR Ñß .ÐB# ß BR Ñß ÞÞÞß .ÐBR " ß BR Ñ×Þ Therefore, for all 7ß 8 we have .ÐB7 ß B8 Ñ Ÿ .ÐB8 ß BR Ñ .ÐBR ß B7 Ñ Ÿ ## Ñ .> œ ! B
#
># #
> .> œ
>$ '
># #
B$ '
B# #
> .> œ
B# #
B% #%
B$ '
B# #
and, in general, 18 ÐBÑ œ X Ð18" ÐBÑÑ œ ÞÞÞ œ
B# #x
B$ $x
ÞÞÞ
B8" Ð8"Ñx
The functions 18 converge (uniformly) to the solution 1. In this particular problem, we are lucky enough to recognize that the functions 18 ÐBÑ are just the B8 B8 B partial sums of the series ∞ œ " B ∞ Therefore 8œ# 8x 8œ! 8x œ " B / Þ 1ÐBÑ œ " B /B is a solution of our initial value problem, and we know it is valid for all B − M œ Ò "# ß "# Ó. (You can check by substitution that the solution is correct and that it actually is a solution that works for all B − ‘.) Even if we couldn't recognize a neat formula for the limit 1ÐBÑ, we could still make some useful approximations. From the proof of the Contraction Mapping Theorem, we know that 3(18 ß 1Ñ Ÿ
α8 3Ð1! ß1" Ñ "α .
In this example α œ +Q œ "# , so that 3(18 ß 1Ñ Ÿ #
" " œ #8" sup Öl! Ð B# Ñl À B − M× œ #8" † #"$ œ " distance #8# of 1ÐBÑ on the interval Ò "# ß "# Ó.
" #8# .
" #8 3Ð1! ß1" Ñ " "#
Therefore 18 ÐBÑ is uniformly within
Finally, recall that our initial choice of 1! − F was arbitrary. Since lsin Bl Ÿ ", we know that sin − F , and we could just as well have chosen 1! ÐBÑ œ sin B. Then the functions 18 ÐBÑ computed as X Ð1! Ñ œ 1" ß X Ð1# Ñ œ 1# ß ÞÞÞ would be quite different Ðtry computing 1" and 1# Ñ but it would still be true that 18 ÐBÑ Ä 1ÐBÑ œ " B /B uniformly on M À the same limit 1 because the solution 1 is unique. ñ
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The Contraction Mapping Theorem can be used to prove other results for example, the Implicit Function Theorem. (You can see details, for example, in Topology, by James Dugundji.)
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Exercises
E8. +Ñ Suppose 0 À ‘ Ä ‘ is differentiable and that there is a constant O 1 such that |0 w ÐBÑ | Ÿ O for all BÞ Prove that 0 is a contraction (and therefore has a unique fixed point.) b) Give an example of a continuous function 0 À ‘ Ä ‘ such that | 0 ÐBÑ 0 ÐCÑ l lB C l for all B Á C − ‘ but such that 0 has no fixed point. Note: The function 0 is not a contraction mapping. If we allow α œ " in the definition of contraction, then the Contraction Mapping Theorem is not be true not even if we “compensate” by using “ ” instead of “ Ÿ ” in the definition. E9. Let 0 À Ð\ß .Ñ Ä Ð\ß .Ñ, where Ð\ß .Ñ is a nonempty complete metric space. Let 0 5 denote the “ k th iterate of 0 ” that is, 0 composed with itself 5 times. a) Suppose that b5 − for which 0 5 is a contraction. Then, by the Contraction Mapping Theorem, 0 5 has a unique fixed point :. Prove that : is also the unique fixed point for 0 . b) Prove that the function cos À ‘ Ä ‘ is not a contraction. c) Prove that cos5 is a contraction for some 5 − . ÐH3nt: the Mean Value Theorem may be helpful. ) d) Let 5 − be such that 1 œ cos5 is a contraction and let : be the unique fixed point of 1. By a), : is also the unique solution of the equation cos B œ B. Start with 0 as a “first approximation” for : and use the technique in the proof of the Contraction Mapping Theorem to find an 8 − so that | 18 Ð!Ñ : ± < 0.00001. e) For this 8, use a computer or calculator to evaluate 18 Ð!Ñ. (This “solves” the equation cos B œ B with l Error l 0.00001.) E10. Consider the differential equation C w œ B C with the initial condition CÐ!Ñ œ ". Choose a suitable rectangle H and suitable constants Oß Q and + as in the proof of Picard's Theorem. Use the technique in the proof of the contraction mapping theorem to find a solution for the initial value problem. Identify the interval M in the proof. Is the solution you found actually valid on an interval larger than M ?
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5. Completions The set of rationals (with the usual metric . ) is not complete. However, the rationals are a dense subspace of the complete space (‘ß .Ñ. This is a model for the definition of a “completion” for a metric space. µ µ µ Definition 5.1 Ð\ , . Ñ is called a completion of Ð\ß .Ñ if: . |Ð\ ‚ \Ñ œ ., \ is a dense µ µ µ subspace of \ , and Ð\ ,. Ñ is complete. Loosely speaking, a completion of Ð\ß .Ñ adds the additional points needed Ðand no others) to provide limits for the Cauchy sequences that fail to converge in Ð\ß .Ñ . In the case of Ðß .Ñ, we can think of the “additional points” as the irrational numbers, and the resulting completion is Ð‘ß .ÑÞ If a metric space Ð\ß .Ñ is complete, what would a completion look like? Since Ð\ß .Ñ is a µ µ µ complete subspace of Ð\ , . Ñ, \ must be closed in \ . But \ must also be dense. So µ \ œ \ that is, a complete metric space is its own completion. (If Ð\ß .Ñ is a complete µ µ pseudometric space, then \ might not be closed in \ ; but each C − \ \ is at distance ! from some B − \ß so if ÐB8 Ñ Ä C − ] , then ÐB8 Ñ already converges to B − \ and “adding the point C” was unnecessary. Of course different spaces may have the same completion for example, Ð‘ß .Ñ is a completion for both Ðß .Ñ and Ðß .ÑÞ Notice that a completion of Ð\ß .Ñ depends on . , not just the topology g. Þ For example, Ðß .Ñ and ÐÖ 8" À 8 − ×, .Ñ are homeomorphic topological spaces (both have the discrete topology). But the completion of Ðß .Ñ Ðitself!) is not homeomorphic to ÐÖ!× ∪ Ö 8" À 8 − ×ß .Ñ, which is a completion of ÐÖ 8" À 8 − ×ß .ÑÞ Recall that if there is an (onto) isometry 0 between two metric spaces Ð\ß .Ñ and Ð] ß =Ñ, then Ð\ß .Ñ and Ð] ß =Ñ can be regarded as “the same metric space”: we can think of 0 as just “assigning new names” to the points of \ . If 0 À Ð\ß .Ñ Ä Ð] ß =Ñ is an isometry from \ into ] , then we can identify Ð\ß .Ñ with Ð0 Ò\Óß =Ñ an “exact metric copy” of Ð\ß .Ñ inside Ð] ß =ÑÞ So if Ð] ß =Ñ happens to be complete, then Ð0 Ò\Óß =Ñ is dense in the complete space Ðcl] 0 Ò\Óß =Ñ. If we identify Ð0 Ò\Óß =Ñ with Ð\ß .Ñ, then we can call Ðcl] 0 Ò\Óß =Ñ a completion of Ð\ß .Ñ even though \ is not literally a subset of cl] 0 Ò\Ó. Therefore, to find a completion of Ð\ß .Ñ, it is sufficient to find an isometry 0 from Ð\ß .Ñ into some complete space Ð] ß . w Ñ . Example 5.2 Consider with the metric . w Ð8ß 7Ñ œ l 8" 7" l. Ðß . w Ñ is isometric to (Ö 8" À 8 − ×ß .Ñ where . is the usual metric on (0 Ð8Ñ œ 8" is an isometry). These two spaces are “metrically identical.”. Therefore ÐÖ!× ∪ Ö 8" À 8 − ×ß .Ñ contains a dense isometric copy of Ðß . w Ñ. The closure of that “copy” is Ö!× ∪ Ö 8" À 8 − ×, and we can view ÐÖ!× ∪ Ö 8" À 8 − ×ß .Ñ as a completion of Ðß . w Ñ even though is not literally a subspace of Ö!× ∪ Ö 8" À 8 − ×Þ
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The next theorem tells us every metric space has a completion and, as we will see later, it is “essentially” unique.
Theorem 5.3 Every metric space Ð\ß .Ñ has a completion. Proof Our previous comments show that it is sufficient to produce an isometry of Ð\ß .Ñ into some complete metric space: the complete space we will use is ÐG ‡ Ð\Ñß 3Ñ where, as usual, 3 is the metric of uniform convergence. The theorem is trivial if \ œ gß so we assume that we can pick a point : − \ . For any point + − \ , define 9+ À \ Ä ‘ by 9+ ÐBÑ œ .ÐBß +Ñ .ÐBß :ÑÞ The map 9+ is a difference of continuous functions so 9+ is continuous. For each B − \, ‡ l9+ ÐBÑl œ l.ÐBß +Ñ .ÐBß :Ñl Ÿ .Ð+ß :Ñ, so 9+ is bounded. Therefore 9+ − G Ð\ÑÞ Define F À Ð\ß .Ñ Ä ÐG ‡ Ð\Ñß 3Ñ by FÐ+Ñ œ 9+ Þ We complete the proof by showing that F is an isometry. This just involves computing some distances: for any +ß , − \ , 3Ð9+ ß 9, Ñ œ sup Öl9+ ÐBÑ 9, ÐBÑl À B − \× œ sup ÖlÐ.ÐBß +Ñ .ÐBß :ÑÑ Ð.ÐBß ,Ñ .ÐBß :ÑÑl À B − \× œ sup Öl.ÐBß +Ñ .ÐBß ,Ñl À B − \×Þ For any B, l.ÐBß +Ñ .ÐBß ,Ñl Ÿ .Ð+ß ,Ñ. And letting B œ , gives l.ÐBß +Ñ .ÐBß ,Ñl œ .Ð+ß ,Ñ. Therefore 3Ð9+ ß 9, Ñ œ sup Öl.ÐBß +Ñ .ÐBß ,Ñl À B − \× œ .Ð+ß ,ÑÞ
ñ
The completion of Ð\ß .Ñ given in this proof is ÐclG ‡ Ð\Ñ FÒ\Ó,3Ñ. The proof is “slick” but it seems to lack any intuitive content. For example, if we applied the proof to Ðß .Ñ, it would not at all clear that the resulting completion is isometric to Ð‘ß .Ñ (as we would expect). In fact, there is another more intuitive way to construct a completion of Ð\ß .Ñ, but verifying all the details is much more tedious. We will simply sketch this alternate construction here. Call two Cauchy sequences ÐB8 Ñ and ÐC8 Ñ in Ð\ß .Ñ equivalent if .ÐB8 ß C8 Ñ Ä !Þ It is easy to check that µ is an equivalence relation among Cauchy sequences in Ð\ß .Ñ. Clearly, if ÐB8 Ñ Ä D − \ß then any equivalent Cauchy sequence also converges to D . If two nonconvergent Cauchy sequences equivalent, then they are “trying to converge to the same point” but the necessary point is “missing” in \ . µ We denote the equivalence class of a Cauchy sequence ÐB8 Ñ by ÒÐB8 ÑÓ and let \ be the µ set of equivalence classes. Define the distance . between two equivalence classes by µ . ÐÒÐB8 ÑÓß ÒÐC8 ÑÓÑ œ lim .ÐB8 ß C8 ÑÞ (Why must this limit exist?) It is easy to check that 8Ä∞ µ . does not depend on the choice of representative sequences from the equivalence µ ~ classes, and that . is a metric on \ . One then checks (this is only “tricky” part) that µ µ µ Ð\ ß . Ñ is complete: any . -Cauchy sequence of equivalence classes Ð ÒÐB8 ÑÓ Ñ must µ µ converge to an equivalence class in Ð\ ß . Ñ. µ For each B − \ , the sequence ÐBß Bß Bß ÞÞÞÑ is Cauchy, and so ÒÐBß Bß Bß ÞÞÞÑÓ − \ .
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µ µ The mapping 0 À Ð\ß .Ñ Ä Ð\ ß . Ñ given by 0 ÐBÑ œ ÒÐBß Bß Bß ÞÞÞÑÓ is an isometry, and it µ µ µ µ is easy to check that 0 Ò\Ó is dense in Ð\ ß . Ñ so that Ð\ ß . Ñ is a completion for Ð\ß .ÑÞ
This method is one of the standard ways to construct the real numbers from the rationals: ‘ is defined as the set of these equivalence classes of Cauchy sequences of rational numbers. The real numbers can also be constructed as a completion for the rationals by using a method called “Dedekind cuts.” However that approach also uses the order structure in and therefore we cannot mimic it in the general setting of metric spaces. The next theorem gives us the good news that, in the end, it doesn't really matter how we construct a completion because the completion of Ð\ß .Ñ is “essentially” unique: all completions are isometric (and, in a “special” way!). We state the theorem for metric spaces. (Are there modifications of the statement and proof to handle the case where . is merely a pseudometric? ) Theorem 5.4 The completion of Ð\ß .Ñ is unique in the following sense: if Ð] ß =Ñ and Ð^ß >Ñ are both complete metric spaces containing \ as a dense subspace, then there is an (onto) isometry 0 À Ð] ß =Ñ Ä Ð^ß >Ñ such that 0 l\ is the identity map on \ . (In other words, not only are two completions of Ð\ß .Ñ isometric, but there is an isometry between them that holds \ fixed. The isometry merely “renames” the new points in the “outgrowth” ] \Þ ) Proof Suppose C − ] Þ Since \ is dense in ] , we can pick a sequence ÐB8 Ñ in \ which converges to C. Since ÐB8 Ñ is convergent, it is Cauchy in Ð\ß .Ñ, and since >l\ ‚ \ œ . , ÐB8 Ñ is also a Cauchy sequence in the complete space Ð^ß >Ñ. Therefore ÐB8 Ñ Ä some point D − ^ . We define 0 ÐCÑ œ D . Then 0 À Ð] ß =Ñ Ä Ð^ß >ÑÞ (It is easy to check that 0 is well-defined that is, we get the same D no matter which of the possibly many sequences ÐB8 Ñ we first choose converging to C.) In particular, if B − \ß we can choose ÐB8 Ñ to be the constant sequence B8 œ B; then 0 ÐBÑ œ B, so that 0 l\ is the identity map on \ . We need to verify that 0 is an isometry, Suppose C w − ] and we choose ÐB8w Ñ Ä C w . We then have that .ÐB8 ß Bw8 Ñ œ =ÐB8 ß B8w Ñ Ä =ÐCß C w Ñ and .ÐB8 ß Bw8 Ñ œ >ÐB8 ß B8w Ñ Ä >ÐDß D w Ñ œ >Ð0 ÐCÑß 0 ÐCw ÑÑ, so =ÐCß C w Ñ œ >Ð0 ÐCÑß 0 ÐC w ÑÑ ñ
According to Theorem 5.4, the space Ð‘ß .Ñ no matter how we construct it is the completion of the space Ðß .Ñ.
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6. Category In mathematics there are many different ways to compare the “size” of sets. These different methods are useful for different purposes. One of the simplest ways is to say that one set is “bigger” if it has “more points” than another set that is, by comparing their cardinal numbers. In a totally different spirit, we might call one subset of ‘# “bigger” than another if it has a larger area. In analysis, there is a generalization of area. A certain collection ` of subsets of ‘# contains sets that are called measurable, and for each set W − ` a nonnegative real number .ÐWÑ is assigned. .ÐWÑ is called the “measure of W ” and we can think of .ÐWÑ as a kind of “generalized area.” A set with a larger measure is “bigger.” In this section, we will look at a third completely different and useful topological idea for comparing the “size” of certain sets. The most interesting results in this section are be about complete metric spaces, but the basic definitions make sense in any topological space Ð\ß g ÑÞ
Definition 6.1 A subset E of the topological space Ð\ß g Ñ is called nowhere dense in \ if int\ (cl\ E) œ gÞ (In some books, a nowhere dense set is called rare.) A set has empty interior iff its complement is dense. Therefore E is nowhere dense in \ iff int\ (cl\ E) œ g iff \ cl\ E is dense. Intuitively, we can think of an open set S as including some “elbow room” around each of its points if B − S, then all sufficiently nearby points are also in S. Then we think of a nowhere dense set as being “skinny” so skinny that not only does it contain no “elbow room” around any of its points, but even its closure contains no “elbow room” around any of its points.
Example 6.2 1) A closed set J is nowhere dense in \ iff int J œ g iff \ J is dense in \ . In particular, if a singleton set Ö:× is a closed set in \ , then Ö:× is nowhere dense unless : is isolated in \Þ 2) Suppose E © ‘Þ E is nowhere dense iff cl E contains no interval of positive length. For example, each singleton Ö88 œ !Þ>" ># ÞÞÞ>5 ÞÞÞthree , with each >8 − Ö!ß #×Þ ∞
8œ"
We can obviously rewrite this as B œ 2$,88 , where each ,8 − Ö!ß "× and therefore we can define ∞
8œ"
a function 1 À G Ä Ò!ß "Ó as follows: for B − Gß 1ÐBÑ œ 1Ð 2$,88 Ñ œ #,88 . ∞
∞
8œ"
8œ"
More informally,: 1Ð!Þ #," #,# #,$ ÞÞÞthree Ñ œ !Þ," ,# ,$ ÞÞÞtwo
This mapping 1 is not one-to-one . For example, 1Ð (* Ñ œ 1Ð!Þ#!####ÞÞÞÞthree ) œ !Þ"!""""ÞÞÞtwo œ 1Ð )* Ñ œ 1Ð!Þ##!!!!ÞÞÞthree ) œ !Þ""!!!ÞÞÞtwo œ
%$$ %
and
In fact, for +ß , − G , we have 1Ð+Ñ œ 1Ð,Ñ iff the interval Ð+ß ,Ñ is one of the “deleted middle thirds.” It should be clear that 1 is continuous and that if + , in G , then 1Ð+Ñ Ÿ 1Ð,Ñ that is, 1 is weakly increasing. (Is 1 onto? )
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The function 1 is not defined on Ò!ß "Ó G . But Ò!ß "Ó G is, by construction, a union of disjoint open intervals. We can therefore extend the definition of 1 to a mapping K À Ò!ß "Ó Ä Ò!ß "Ó in the following simple-minded way: KÐBÑ œ
1ÐBÑ if B − G 1Ð+Ñ if B − Ð+ß ,Ñ, where Ð+ß ,Ñ is a “deleted middle third”
Since we know 1Ð+Ñ œ 1Ð,Ñ, this amounts to extending the graph of 1 over each “deleted middle third” Ð+ß ,Ñ by using a horizontal line segment from Ð+ß 1Ð+ÑÑ to Ð,ß 1Ð,ÑÑÞ The result is the graph of the continuous function K. K is sometimes called the Cantor-Lebesgue function. It satisfies: i) KÐ!Ñ œ !ß KÐ"Ñ œ " ii) K is continuous iii) K is (weakly) increasing iv) at any point B in a “deleted middle third” Ð+ß ,Ñ, K is differentiable and K w ÐBÑ œ !Þ Recall that G has “measure !.” Therefore we could say “K w ÐBÑ œ ! almost everywhere” even though K rises monotonically from ! to 1!
Note: The technique that we used to extend 1 works in a similar way for any closed set J © ‘ and any continuous function 1 À J Ä ‘. Since ‘ J is open in ‘, we know that we can write ‘ J as the union of a countable collection of disjoint open intervals M8 which have for Ð+ß ,Ñ, Ð ∞ß ,Ñ or Ð+ß ∞Ñ We can extend 0 to a continuous function J À ‘ Ä ‘ simply by extending the graph of 0 over linearly over each M8 À if M8 œ Ð+ß ,Ñ,
then let the graph J over M8 be the straight line segment joining Ð+ß 0 Ð+Ñß >9 Ð,ß 0 Ð,ÑÑÞ if M8 œ Ð ∞ß ,Ñ, then let J have the constant value 0 Ð, 98 I8 if M8 œ Ð+ß ∞Ñ, then let J have the constant value 0 Ð+Ñ on M8 ß There is a much more general theorem that implies that whenever J is a closed subset of Ð\ß .Ñ, then each 0 − GÐJ Ñ and be extended to a function J − GÐ\Ñ. In the particular case \ œ ‘, proving this was easy because ‘ is ordered and we completely understand the structure of the open sets in ‘.. Another curious property of G , mentioned without proof, is that its “difference set” ÖB C À Bß C − G× œ Ò "ß "Ó. Although G has measure !, the difference set in this case has measure #!
206
Exercises E21. Suppose \ is compact and that 0 À \ Ä ] is continuous and onto. Prove that ] is compact. (“A continuous image of a compact space is compact.”)
E22. a) Suppose that \ is compact and ] is Hausdorff. Let 0 À \ Ä ] be a continuous bijection. Prove that 0 is a homeomorphism. b) Let g be the usual topology on Ò!ß "Ó and suppose g" and g# are two other topologies on Ò!ß "Ó such that g" § g § g# Prove that ÐÒ!ß "Ó, g" Ñ is not Hausdorff and that ÐÒ!ß "Ó, g# Ñ is not Á Á compact. (Hint: Consider the identity map 3 À Ò!ß "Ó Ä Ò!ß "ÓÞ ) c) Is part b) true if Ò!ß "Ó is replaced by an arbitrary compact Hausdorff space Ð\ß g Ñ?
E23. Prove that a nonempty space Ð\ß g Ñ is pseudocompact iff every continuous 0 À ‘ Ä ‘ achieves both a maximum and a minimum value. E24. Suppose 0 À Ð\ß .Ñ Ä Ð] ß . w ÑÞ Prove that 0 is continuous iff 0 lO is continuous for every compact set O © \ . E25. Suppose that E and F are nonempty disjoint closed sets in Ð\ß .Ñ and that E is compact. Prove that .ÐEß FÑ !Þ
E26.
Let \ and ] be topological spaces. a) Prove that \ is compact iff every open cover by basic open sets has a finite subcover.
b) Suppose \ ‚ ] is compact. Prove that if \ß ] Á g, then \ and ] are compact. (By induction, a similar statement applies to any finite product.) c) Prove that if \ and ] are compact, then \ ‚ ] is compact. (By induction, a similar statement applies to any finite product.) (Hint: for any B − \ß ÖB× ‚ ] is homeomorphic to ] . Part a) is also relevant.) d) Point out explicitly why the proof in c) cannot be altered to prove that a product of two countably compact spaces is countably compact. (An example of a countably compact space \ for which \ ‚ \ is not even pseudocompact is given in Chapter X, Example 6.8.) Note: In fact, an arbitrary product of compact spaces is compact. This is the “Tychonoff Product Theorem” which we will prove later. It is true that the product of a countably compact space and a compact space is countably compact. You might trying proving this fact. A very similar proof shows that a product of a
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compact space and a Lindelof ¨ space is Lindelof. ¨ That proof does not generalize, however, to show that a product of two Lindelof ¨ spaces is Lindelof. ¨ Can you see why? E27. Prove that if Ð\ß .Ñ is a compact metric space and l \ l œ i! , then \ has infinitely many isolated points.
E28. Suppose \ is any topological space and that ] is compact. Prove that the projection map 1\ À \ ‚ ] Ä \ is closed. (A projection “parallel to a compact factor” is closed.) E29. Let \ be an uncountable set with the discrete topology g . Prove that there does not exist a totally bounded metric d on \ such that g. œ g . E30. a) Give an example of a metric space Ð\ß .Ñ which is totally bounded and an isometry from Ð\ß .Ñ into Ð\ß .Ñ which is not onto. Hint: on the circle W " , start with a point : and keep rotating it around the boundary by increments of some angle " . b) Prove that if Ð\ß .Ñ is totally bounded and if 0 is an isometry from Ð\ß .Ñ into Ð\ß .Ñ, then 0 Ò\Ó must be dense in Ð\ß .Ñ. Hint: Given Bß % !ß cover \ with finitely many #% spheres; if the sequence Bß 0 ÐBÑß 0 Ð0 ÐBÑÑ,ÞÞÞ consists of distinct terms, there must be infinitely many of them in one sphere. c) Show that a compact metric space cannot be isometric to a proper subspace of itself. Hint: you might use part b). d) Prove that if each of two compact metric spaces is isometric to a subspace of the other, then the two spaces are isometric to each other. Note: Part d) is an analogue of the Cantor-Schroeder-Bernstein Theorem for compact metric spaces. E31. Suppose Ð\ß .Ñ is a metric space and that Ð\ß . w Ñ is totally bounded for every metric . w µ . . Must Ð\ß .Ñ be compact?
E32. Let h be an open cover of the compact metric space Ð\ß .Ñ. Prove that there exists a constant $ ! such that for all B: F$ ÐBÑ © Y for some Y − h . (The number $ is called a Lebesgue number for h .) E33. Suppose Ð\ß .Ñ is a metric space with no isolated points. Prove that \ is compact iff .ÐEß FÑ ! for every pair of disjoint closed sets Eß F © \Þ
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Chapter IV Review Explain why each statement is true, or provide a counterexample.
1. Let E denote the set of fixed points of a continuous function 0 À \ Ä \ß where \ is a Hausdorff space. E is closed in \ . 2. Let W denote the set of all Cauchy sequences in which converge to a point in ‘. Then lW ± œ -. 3. For a metric space Ð\ß .Ñ, if every continuous function 0 À \ Ä ‘ assumes a minimum value, then every infinite set in \ has a limit point. 4. There exists a continuous function 0 À ‘ Ä ‘ for which the Cantor set is the set of fixed points of 0 . 5. If \ has the cofinite topology, then every subspace is pseudocompact. 6. Let Ò!ß "Ó have the subspace topology from the Sorgenfrey line. Then Ò!ß "Ó is countably compact. 7. Let Æ œ ÖE © ‘ À E is first category in ‘×, and let g be the topology on ‘ for which Æ is a subbase. Then Ð‘ß g Ñ is a Baire space. 8. Every sequence in Ò!ß "Ó# has a Cauchy subsequence. 9. If E is a dense subspace of Ð\ß .Ñ and every Cauchy sequence in E converges to some point in \ , then Ð\ß .Ñ is complete. 10. If g is the cofinite topology on , then (ß g ) has the fixed point property. 11. Every sequence in Ò!ß "Ñ has a Cauchy subsequence. 12. Let ^ œ Ð!ß 13 Ñ ∩ G , where G is the Cantor set. Then ^ is completely metrizable. 13. The Sorgenfrey plane is countably compact. 14. If every sequence in the metric space Ð\ß .Ñ has a convergent subsequence, then every continuous real valued function on \ must have a minimum value. 15. Suppose Ð\ß .Ñ is complete and 0 À \ Ä \ . The set G œ ÖB − ‘ À 0 is continuous at B× is second category in itself. 16. In the space GÐÒ!ß "ÓÑ, with the metric 3 of uniform convergence, the subset of all polynomials is first category. 17. A nonempty open set in Ð\ß g Ñ cannot be nowhere dense.
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18. Every nonempty nowhere dense subset of ‘ contains an isolated point. 19. There are exactly - nowhere dense subsets of ‘. 20. Suppose F is nonempty subset of with no isolated points. F cannot be completely metrizable. 21. Suppose Ð\ß .Ñ is a countable complete metric space. If E © \ , then ÐEß .Ñ may not be complete, but E is completely metrizable. 22. If E is first category in \ and F © E, then F is first category in E. 23. There are - different metrics . on , each equivalent to the usual metric, for which Ðß .Ñ is complete. 24. In , with the cofinite topology, every infinite subset is sequentially compact. 25. Suppose 0 À Ò!ß "Ó# Ä ‘ and E œ Ö: − (!ß ")# : usual metric, is totally bounded.
`0 `B
and
`0 `C
both exist at :}. Then E, with its
26. A space Ð\ß g Ñ is a Baire space iff \ is second category in itself. 27. Let I denote the subset of the Cantor set G consisting of the endpoints of the open intervals deleted from Ò!ß "Ó in the construction of G . Suppose 0 À Ò!ß "Ó Ä ‘ is continuous and that 0 ÒIÓ © Ò!ß !Þ"Ó. Then 0 ÒGÓ © Ò!ß !Þ"Ó . 28. Let . be a metric on the irrationals which is equivalent to the usual metric and such that Ðß .Ñ is complete. Then Ðß .Ñ must be totally bounded. 29. Suppose Ð\ß .Ñ is a nonempty complete metric space and E is a closed subspace such that ÐEß .Ñ is totally bounded. Then every sequence in E has a cluster point. 30. There is a metric d on Ò!ß "Ó, equivalent to the usual metric, such that ÐÒ!ß "Óß .Ñ is not complete. 31. Let ^ œ Ò!ß "Ó# have the usual topology. Every nonempty closed subset of ^ is second category in itself. 32. Suppose 0 À Ò!ß "Ó8 Ä ‘7 and G œ ÖB − Ò!ß "Ó8 À 0 is continuous at B×. There is a metric . equivalent to the usual metric on G for which ÐGß .Ñ is complete. 33. If the continuum hypothesis (CH) is true, then ‘ cannot be written as the union of fewer than c nowhere dense sets. 34. Suppose E is a complete subspace of Ò!ß "ÓÞ Then E is a K$ set in ‘Þ 35. There are 2- different subsets of ‘ none of which contains an interval of positive length. 36. Suppose Ð\ß .Ñ is a nonempty complete metric space and E is a closed subspace such that
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ÐEß .Ñ is totally bounded. Then every sequence in E has a cluster point. 37. Let . w be a metric on the irrationals which is equivalent to the usual metric and for which Ðß . w Ñ is totally bounded. Then Ðß . w Ñ cannot be complete. 38. The closure of a discrete subspace of ‘ may be uncountable. 39. Suppose 0 À Ò!ß "Ó8 Ä ‘7 and G œ ÖB − Ò!ß "Ó8 À 0 is continuous at B×. Then there is a metric d on G , equivalent to the usual metric, such that ÐGß .Ñ is totally bounded. 40. Suppose Ð\ß g Ñ is compact and 0 À \ Ä ‘ is a continuous function such that 0 ÐBÑ ! for all B − \Þ Then there is an % ! such that 0 ÐBÑ % for all B − \Þ 41. If every point of Ð\ß .Ñ is isolated, then Ð\ß .Ñ is complete. 42. If Ð\ß .Ñ has no isolated points, then its completion has no isolated points. 43. Suppose that for each point B − Ð\ß .Ñ, there is an open set YB such that ÐYB ß .Ñ is complete. Then there is a metric d w µ d on \ such that Ð\ß . w Ñ is complete. µ µ µ 44. For a metric space Ð\ß .Ñ with completion (\ ß . Ñ, it can happen that ± \ \ ± œ i! . µ µ 45. Let . be the usual metric on . There is a completion ( ß . Ñ of Ðß .Ñ for which µ l l œ i! Þ 46. A subspace of ‘ is bounded iff it is totally bounded. 47. A countable metric space must have at least one isolated point. 48. The intersection of a sequence of dense open subsets in must be dense in . 49. If G is the Cantor set, then ‘ G is an J5 set in ‘. 50. A subspace E of a metric space Ð\ß .Ñ is compact iff E is closed and bounded. 51. Let U œ Ö Ò+ß ,Ñ ∩ Ò!ß "Ó À +ß , − ‘ß + , × be the base for a topology g on Ò!ß "Ó. The space ÐÒ!ß "Óß g Ñ is compact. 52. Ðj# ß .Ñ, where . is the usual metric on j# , is sequentially compact. 53. A subspace of a pseudocompact space is pseudocompact. 54. Ðj# ß .Ñ, where . is the usual metric on j# , is countably compact. 55. An uncountable closed set in ‘ must contain an interval of positive length. 56. Suppose G is the Cantor set and 0 À G Ä G is continuous. Then the graph of 0 (a subspace of G ‚ G , with its usual metric) is totally bounded.
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57. Let G denote the Cantor set © Ò!ß "Ó, with the metric .ÐBß CÑ œ
±BC± "±BC±
. Then ÐGß .Ñ is
totally bounded. 58. A discrete subspace of ‘ must be closed in ‘ 59. A subspace of ‘ which is discrete in its relative topology must be countable. 60. If E and F are subspaces of Ð\ß g Ñ and each is discrete in its subspace topology, then E ∪ F is discrete in the subspace topology. 61. Let cos8 denote the composition of cos with itself 8 times. Then for each B − Ò "ß "Ó, there exists an 8 (perhaps depending on B) such that cos8 B œ BÞ 62. Suppose . is any metric on equivalent to the usual metric. Then ÐÒ!ß "Óß .Ñ is totally bounded. 63. Let W " denote the unit circle in ‘# . W " is homeomorphic to a subspace of the Cantor set G . 64. There is a metric space Ð\ß .Ñ satisfying the following condition: for every metric d w µ d, Ð\ß . w Ñ is totally bounded. 65. Ò!ß "Ó# , with the subspace topology from the Sorgenfrey plane, is compact.
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