Chapter V Connected Spaces
1. Introduction
In this chapter we introduce the idea of connectedness. Connectedness is a topological property quite different from any property we considered in Chapters 1-4. A connected space \ need not have any of the other topological properties we have discussed so far. Conversely, the only topological properties that imply “\l\lŸ\ is connected” are very extreme such as “ 1” or “ has the trivial topology.”
2. Connectedness
Intuitively, a space is connected if it is all in one piece; equivalently a space is disconnected if it can be written as the union of two nonempty “separated” pieces. To make this precise, we need to decide what “separated” should mean. For example, we think of ‘‘ as connected even though can be written as the union of two disjoint pieces: for example, ‘ œE∪Fß where E œ Ð ∞ß !Ó and F œ Ð!ß ∞Ñ . Evidently, “separated” should mean something more than “disjoint.”
On the other hand, if we remove the point ! to “cut” ‘ , then we probably think of the remaining space \ œ‘ Ö!× as “disconnected.” Here, we can write \ œ E ∪ F , where E œ ( ∞ß !Ñ and FœÐ!ß∞Ñ . E and F are disjoint, nonempty sets and (unlike EF and in the preceding paragraph) they satisfy the following (equivalent) conditions:
i) EF and are open in \ ii) EF and are closed in \ iii) ÐF ∩ cl\\ EÑ ∪ ÐE ∩ cl FÑ œ g that is, each of E and F is disjoint from the closure of the other. (This is true, in fact, even if we use cl‘ instead of cl\ .)
Condition iii) is important enough to deserve a name.
Definition 2.1 Suppose EF and are subspaces of Ð\ßÑEFg . and are called separated if each is disjoint from the closure of the otherÐF∩EÑ∪ÐE∩FÑœg that is, if cl\\ cl .
It follows immediately from the definition that
i) separated sets must be disjoint, and
ii) subsets of separated sets are separated: if Eß F are separated, G © E and H©F , then G and H are also separated.
Example 2.2
1) In‘ , the sets E œ Ð ∞ß !Ó and F œ Ð!ß ∞Ñ are disjoint but not separated . Likewise in ‘##### , the sets EœÖÐBßCÑÀB C Ÿ"× and FœÖÐBßCÑÀÐB 2 Ñ C "× are disjoint but not separatedÞ
213 2) The intervals E œ Ð ∞ß !Ñ and F œ Ð!ß ∞Ñ are separated in ‘ but cl‘‘ E ∩ cl F Á gÞ The same is true for the open balls EœÖÐBßCÑÀB## C "× and F œ ÖÐBß CÑ À ÐB 2 Ñ## C "× in ‘ #. The condition that two sets are separated is stronger than saying they are disjoint, but weaker than saying that the sets have disjoint closures.
Theorem 2.3 In any space Ð\ßg Ñ , the following statements are equivalent:
1) g\ and are the only clopen sets in \ 2) if E©\ and Fr Eœg , then Eœg or Eœ\ 3) \ is not the union of two disjoint nonempty open sets 4) \ is not the union of two disjoint nonempty closed sets 5) \ is not the union of two nonempty separated sets.
Note: Condition 2) is not frequently used. However it is fairly expressive: to say that FrEœg says that no point B in \ can be “approximated arbitrarily closely” from both inside and outside E so, in that sense, E and Fœ\E are pieces of \ that are “separated” from each other.
Proof 1)ÍEEœg 2) This follows because is clopen iff Fr (see Theorem II.4.5.3 ). 1)Ê\œE∪FEF 3) Suppose 3) is false and that where , are disjoint, nonempty and open. Since \EœF is open and nonempty, we have that E is a nonempty proper clopen set in \ , which shows that 1) is false. 3)Í 4) This is clear. 4)Ê\œE∪FEßF 5) If 5) is false, then , where are nonempty and separated. Since clF∩Eœgß we conclude that cl F©F , so F is closed. Similarly, E must be closed. Therefore 4) is false. 5)ÊE 1) Suppose 1) is false and that is a nonempty proper clopen subset of \ . Then Fœ\E is nonempty and clopen, so E and F are separated. Since \œE∪F , 5) is false. ñ
Definition 2.4 A space Ð\ßg Ñ is connected if any (therefore all) of the conditions 1) - 5) in Theorem 2.3 hold. If G©\ , we say that G is connected if G is connected in the subspace topology.
According to the definition, a subspace G©\ is disconnected if we can write GœE∪F , where the following (equivalent) statements are true:
1) EF and are disjoint, nonempty and open in G 2) EF and are disjoint, nonempty and closed in G 3) EF and are nonempty and separated in G .
If GEßFG is disconnected, such a pair of sets will be called a disconnection or separation of .
The following technical theorem and its corollary are very useful in working with connectedness in subspaces.
214 Theorem 2.5 Suppose EßF©G©\ÞEF Then and are separated in G iff EF and are separated in \.
Proof clG\FœG∩ cl F (see Theorem III.7.6 ), so E∩ cl G Fœg iff E∩Ð cl \ F∩GÑœg iffÐE∩GÑ∩ cl\\ Fœg iff E∩ cl FœgÞ Similarly, F∩ cl G\ Eœg iff F∩ cl EœgÞ ñ
Caution: According to Theorem 2.5, GGœE∪FEF is disconnected iff where and are nonempty separated set in GGœE∪FEF iff where and are nonempty separated set in \ . Theorem 2.5 is very useful because it means that we don't have to distinguish here between “separated in G\ ” and “separated in ” because these are equivalent. In contrast, when we say that GG is disconnected if is the union of two disjoint, nonempty open (or closed ) sets EßF in G, then phrase “in GEF\ ”cannot be omitted: the sets , might not be open (or closed ) in . "" " For example, suppose \ œ Ò!ß "Ó and G œ Ò!ß## Ñ ∪ Ð ß "Ó . The sets E œ Ò!ß # Ñ and " FœÐ# ß"Ó are open, closed and separated in G . By Theorem 2.5, they are also separated in ‘ but they are neither open nor closed in ‘ .
Example 2.6
1) Clearly, connectedness is a topological property. More generally, suppose 0À\Ä] is continuous and onto. If F]0ÒFÓ is proper nonempty clopen set in , then " is a proper nonempty clopen set in \ . Therefore a continuous image of a connected space is connected.
2) A discrete space \l\lŸ" is connected iff . In particular, ™ and are not connected.
3) is not connected since we can write as the union of two nonempty separated sets: œÖ;− À;## #×∪Ö;− À; #×. Similarly, we can show is not connected. More generally suppose G©‘ and that G is not an interval . Then there are points +D, where +ß,−G but DÂGÞ Then ÖB−GÀBDלÖB−GÀBŸD× is a nonempty proper clopen set in GG . Therefore is not connected . In fact, a subset GG of ‘ is connected iff is an interval. It is not very hard, using the least upper bound property of ‘‘ , to prove that every interval in is connected. (Try it as an exercise! ) We will give a short proof soon (Corollary 2.12) using a different argument.
4) (The Intermediate Value Theorem ) If \0À\Ä is connected and ‘ is continuous, then ranÐ0Ñ is connected (by part 1 ) so ran Ð0Ñ is an interval ( by part 3 ). Therefore if +ß , − \ and 0Ð+ÑD0Ð,Ñ , there must be a point -−\ for which 0Ð-ÑœDÞ
5) The Cantor set G is not connected (since it is not an interval). But much more is true. Suppose Bß C − E © G and that B CÞ Since G is nowhere dense (see IV.10 ), the interval ÐBß CÑ ©Î G, so we can choose DÂG with BDC . Then FœÐ∞ßDÑ∩E œÐ∞ßDÓ∩E is clopen in E , and F contains B but not C so E is not connected. It follows that every connected subset of G contains at most one point. A space Ð\ßg Ñ is called totally disconnected every connected subset E satisfies lEl Ÿ "Þ The spaces ™ ß ß and are other examples of totally disconnected spaces.
6) \ is connected iff every continuous 0 À \ Ä Ö!ß "× is constant: certainly, if 0 is continuous and not constant, then 0" ÒÖ!×Ó is a proper nonempty clopen set in \ so \ is not
215 connected. Conversely, if \E is not connected and is a proper nonempty clopen set, then the characteristic function ;E À \ Ä Ö!ß "× is continuous but not constant.
Theorem 2.7 Suppose 0 À \ Ä ] . Let > œ ÖÐBß CÑ − \ ‚ ] À C œ 0ÐBÑ× œ “the graph of 0Þ ” If 00 is continuous, then graph of is homeomorphic to the domain of 0 ; in particular, the graph of a continuous function is connected iff its domain is connected.
Proof We want to show that \2À\Ä is homeomorphic to >> . Let be defined by 2ÐBÑ œ ÐBß 0ÐBÑÑÞ Clearly 2 is a one-to-one map from \ onto > . Let + − \ and suppose ÐY ‚ Z Ñ ∩> is a basic open set containing 2Ð+Ñ œ Ð+ß 0Ð+ÑÑÞ Since 00Ð+Ñ−ZßS\+ is continuous and there exists an open set in containing and such that 0ÒSÓ © Z Þ Then + − Y ∩ S , and 2ÒY ∩ SÓ © ÐY ‚ Z Ñ ∩> , so 2 is continuous at + . If Y\2ÒYÓœÐY‚]Ñ∩2 is open in , then >> is open in , so is open. Therefore 2 is a homeomorphism. ñ
Note: It is not true that a function 0 with a connected graph must be continuous. See Example 2.22.
The following lemma makes a simple but very useful observation.
Lemma 2.8 Suppose QßR are separated subsets of \ . If G © Q ∪R and G is connected, then G©Q or G©R .
Proof G∩Q and G∩R are separated (since G∩Q ©Q and G∩R©R ) and G œ ÐG ∩ QÑ ∪ ÐG ∩ RÑ. But G is connected so ÐG ∩ QÑ and ÐG ∩ RÑ cannot form a disconnection of G . Therefore either G∩Qœg (so G©RÑ or G∩Rœg (so G©Q ). ñ
The next theorem and its corollaries are simple but powerful tools for proving that certain sets are connected. Roughly, the theorem states that if we have one “central ” connected set G and other connected sets none of which is separated from G, then the union of all the sets is connected.
Theorem 2.9 Suppose GG−M and α (αα ) are connected subsets of \ and that for each , Gα and are not separated. Then is connected. G∪GWœG α Proof Suppose that WœQ∪R where Q and R are separated. By Lemma 2.8, either G©Q orG©R . Without loss of generality, assume G©Q . By the same reasoning we conclude that for each α , either Gαα ©Q or G ©RÞ But if some G α ©R , then G and G α would be separated. Hence every G©Qα . Therefore Rœg and the pair QßR is not a disconnection of Wñ.
Corollary 2.10 Suppose that for each α −M , Cα is a connected subset of \ and that for all , C . Then C is connected. α"Á−Mα" ∩GÁg Ö α À−M× α
216 Proof If Mœgß then Ö C Àαα −Mלg is connected. If MÁg , pick an −M and let G be α ! α! the “central set” G−MG∩GÁgGG in Theorem 2.9. For all α , αα!! , so α and α are not separated. By Theorem 2.9, C is connected. ÖÀ−M×α α ñ
Corollary 2.11 For each 8− , suppose G88" is a connected subset of \ and that G ∩Gn Ág . Then ∞ is connected. 8œ"G8 Proof Let 8 . Corollary 2.10 (and simple induction) shows that the 's are Eœ855œ" G E8 connected. Then g Á E"# © E © ÞÞÞ © E 8 © ÞÞÞ Another application of Corollary 2.10 gives that ∞∞ is connected. 8œ"Eœ88 8œ" G ñ
Corollary 2.12 Let M©‘‘ Þ Then M is connected iff M is an interval. In particular, is connected, so ‘‘ and g are the only clopen sets in .
Proof We have already shown that if MM is not an interval, then is not connected (Example 2.6.3). So suppose MMœgMœÖ<×M is an interval. If or , then is connected.
Suppose M œÒ!ß"Ó and that EßF are nonempty disjoint closed sets in M . Then there are points +!! − E and , − F for which l+ !!!! , l œ .Ð+ ß , Ñ œ .ÐEß FÑÞ
To see this, define 0 À E ‚ F Ä Ò!ß "Ó by 0ÐBß CÑ œ lB ClÞ E ‚ F is a closed bounded set in ‘# soE‚F is compact. Therefore 0 has aminimum value, occurring at some point Ð+!! ß , Ñ − E ‚ F Ðsee Exercise IV.E23. )
+,!! + ! , ! + !! , Let Dœ ###−Ò!ß"Ó. Since lD,lœl!! ,lœl l l+,l !! , we conclude D  E . Similarly, D  F . Therefore Ò!ß "Ó Á E ∪ F , so Ò!ß " ] is connected.
Suppose + ,Þ The interval Ò+ß ,Ó is homeomorphic to Ò!ß "Ó , so each interval Ò+ß ,Ó is connected. Since Ð+ß ,Ñ œ∞ Ò+ "" ß , Ó , Corollary 2.10 implies that Ð+ß ,Ñ is 8œ" 88 connected. Similarly, Corollary 2.10 shows that each of the following unions is connected:
∞ [+ß , " Ó œ Ò+ß ,Ñ 8œ1 8 ∞ [+" ß,ÓœÐ+ß,Ó 8œ1 8 ∞ [ 8œ" +ß + 8Ó œ Ò+ß ∞Ñ ∞ 8œ" Ð+ß + 8Ñ œ Ð+ß ∞Ñ ∞ [ 8œ1 +8ß+ÓœÐ∞ß+Ó ∞ [ 8œ1 +8ß+ÑœÐ∞ß+Ñ ∞ [ 8œ1 8ß8Óœ‘ ñ
Corollary 2.13 For every 8−‘ , 8 is connected Þ
Proof By Corollary 2.12, ‘‘"8 is connected. can be written as a union of straight lines (each homeomorphic to ‘‘ ) through the origin and Corollary 2.10 implies that 8 is connected. ñ
217 Corollary 2.14 Suppose that for all Bß C − \ there exists a connected set CBC © \ with Bß C − GBC. Then \ is connected.
Proof Certainly \œg is connected. If \Ág , choose +−\Þ By hypothesis there is, for each , a connected set containing both and . By Corollary 2.10, C−\ G+C + C \œ ÖG+C ÀC−\× is connected. ñ
Example 2.15 Suppose G8 #ÞG is a countable subset of ‘‘88 , where Then is connected. In particular, ‘88BßCG is connected. To see this, suppose are any two points in ‘8 .
Choose a straight line PBCBC which is perpendicular to the line segment joining and . For each :−Pßlet G:: be the union of the two line segments B:∪ :C . G is the union of two intervals with a point in common, so G: is connected.
w If :Á: , then G::w ∩GœÖBßC× . So if D−G , then D is in at most one G : . Therefore (since ‡8 Gis countable), there is a : −P for which G::‡‡ ∩Gœg . Then BßC−G ©‘ GÞ So 8 Corollary 2.14 (with GœGÑBC :‡ shows that ‘ G is connected.
The definition of connectedness agrees with our intuition in the sense that every set that you think (intuitively) should be connected is actually connected according to Definition 2.4. But according to Definition 2.4, certain strange sets also turn out “unexpectedly” to be connected. ‘88 might fall into that category. So the official definition forces us to try to expand our intuition about what “connected” means. Question: Is ‘88 connected?
This situation is analogous to what happens with the “%$ - definition” of continuity. Using that definition it turns out that every function that you expect (intuitively) should be continuous actually is continuous. If you have a “problem” with the official definition of continuity, it would be that it almost seems “too generous” it allows some “unexpected” functions also to be
218 continuous. An example is the well-known function from elementary analysis: 0À‘‘ Ä , where " if Bœ : in lowest terms 0ÐBÑ œ ;; !Bif is irrational
0BB is continuous at iff is irrational.
Definition 2.16 Suppose \\ is connected. If Ö:× is not connected, then : is called a cut point in \Þ
If 0À\Ä] is a homeomorphism, then it is easy to check that : is a cut point in \ iff 0Ð:Ñ is a cut point in ]Þ Thereforehomeomorphic spaces have the same number of cut points .
Example 2.17 ‘‘8 is not homeomorphic to if 8 2 Þ
Proof Every point :−‘‘ is a cut point. But Example 2.15 shows that 8 has no cut points when 8 2 Þ
It is also true but much harder to prove that ‘‘78 and are not homeomorphic whenever 7Á8. One way to prove this is to develop theorems about a topological property called dimension. Then it turns out (thankfully) that dim‘‘78œ7Á8œ dim so these spaces are not homeomorphic. One can also prove this result using homology theory a topic developed in algebraic topology.
Example 2.18 How is W" topologically different from Ò!ß "Ó ? Both are compact connected metric spaces with cardinality - , and there is no topological property from Chapters 1-4 that can distinguish between these spaces. The difference has to do with connectivity. The interval Ò!ß "Ó has cut points :: (if is not an endpoint, then : is a cut point) à but W" has no cut points since W"" Ö:× is homeomorphic to Ð!ß "Ñ for every : − W Þ
Corollary 2.19 Suppose Ð\ßgg Ñ and Ð] ßw Ñ are nonempty topological spaces. Then \ ‚ ] is connected iff\] and are connected. (It follows by induction that the same result holds for any finite product of spaces. When infinite products are defined in Chapter 6, it will turn out that the product of any collection of connected spaces is connected.)
Proof ÊSuppose \‚] is connected. Since \‚] Ág , we have \œ1\ Ò\‚]Ó . Therefore \\] is the continuous image of a connected space, so is connected. Similarly, is connected. É\ Let and ] be nonempty connected spaces, and consider any two points Ð+ß,Ñ and Ð-ß.Ñ in \‚]Þ Then \‚Ö,× and Ö-ׂ] are homeomorphic to \ and ] , so these “slices” of the product are connected and both contain the point Ð-ß ,Ñ . By Corollary 2.10, G œ Ð\ ‚ Ö,×Ñ ∪ ÐÖ-× ‚ ] Ñ is a connected set that contains both Ð+ß ,Ñ and Ð-ß .Ñ . By Corollary 2.14, we conclude that \‚] is connected. ñ
(Corollary 2.19 gives an another reason why ‘8 is connected for 8"Ñ .
219 Corollary 2.20 Suppose G\G©E©GE is a connected subset of . If cl , then is connected. In particular, the closure of a connected set is connected.
Proof For each +−EßÖ+× and G are not separated. By Theorem 2.9, is connected. E œ G ∪ ÖÖ+× À + − E× ñ
Example 2.21 By Corollary 2.20, the completion of a connected pseudometric space Ð\ß .Ñ must be connected.
sin 1 !BŸ" Example 2.22 Let 0ÐBÑ œB . Then 0Ð" Ñ œ ! for every 8 − and the graph !Bœ! 8 oscillates more and more rapidly between " and " as BÄ! Þ Part of the graph is pictured below. Of course, 0Bœ!Þ is not continuous at Let > be the graph of the restricted function 1 œ 0lÐ!ß "ÓÞSince 1 is continuous, Theorem 2.7 shows that >> is homeomorphic to Ð!ß "Ó so is connected.
> is sometimes called the “topologist's sine curve.”
Because cl>>œ ∪ÐÖ!ׂÒ"ß"ÓÑ , Corollary 2.20 gives that > ∪E is connected for any set E © Ö!× ‚ Ò "ß "ÓÞ In particular, >>0 œ ∪ ÖÐ!ß !Ñ× Ð the graph of 0Ñ is connected.
Therefore, a function 0 with a connected graph need not be continuous . However, it is true that if the graph of a function 0À‘‘ Ä is a closed connected subset of ‘# , then 0 is continuous. (The proof is easy enough to read: see C.E. Burgess, Continuous Functions and Connected Graphs , The American Mathematical Monthly, April 1990, 337-339.)
3. Path Connectedness and Local Path Connectedness
220 In some spaces \\ , every pair of points can be joined by a path in . This seems like a very intuitive way to describe “connectedness”. However, this property is actually stronger than our definition for a connected space. .
Definition 3.1 A path in \ is a continuous map 0ÀÒ!ß"ÓÄ\Þ The path starts at its initial point 0Ð!Ñ and ends at its terminal point 0Ð"ÑÞ We say 0 is a path from 0Ð!Ñ to 0Ð"ÑÞ
Sometimes it helps to visualize a path by thinking of a point moving in \ from 0Ð!Ñ to 0Ð"Ñ with 0Ð>Ñ representing its position at “time” > − Ò!ß"Ó . Remember, however, that the path, by definition, is the function 0Ð0Ñ©\ , not the set ran . To illustrate the distinction: suppose 0 is a path from BC to . Then the function 1ÀÒ!ß"ÓÄ\ defined by 1Ð>Ñœ0Ð">Ñ is a different path (running in the “opposite direction,” from C to B ), even though ran Ð0Ñ œ ran Ð1ÑÞ
Definition 3.2 A topological space \ is called path connected if, for every pair of points Bß C−\, there is a path from B to C in \.
Note: \BßC−\ is called arcwise connected if, for every pair of points , there exists a homeomorphism 0 À Ò!ß "Ó Ä \ with 0Ð!Ñ œ B and 0Ð"Ñ œ C . Such a path 0 is called an arc from BC to . If a path 0 in a Hausdorff space \ is not an arc, the reason must be that 0 is not one-to-one (why?). It can be proven that a Hausdorff space is path connected iff \ is arcwise connected. Therefore some books use “arcwise connected” to mean the same thing as “path connected.” Theorem 3.3 A path connected space \ is connected.
Proof g\Ág+−\B−\ is connected, so assume and choose a point . For each , there is a path 0+BGœÐ0ÑGBBBB from to . Let ran . Each is connected and contains + . By Corollary 2.10, is connected. \œ ÖGB ÀB−\× ñ
221 Sometimes path connectedness and connectedness are equivalent. For example, a subset M©‘ is connected iff MM is an interval iff is path connected. But in general, the converse to Theorem 3.,3 is false as the next example shows.
sin 1 !BŸ" Example 3.4 Consider 0ÐBÑ œ B . In Example 2.22, we showed that the !Bœ! graph >>0 is connected. However, we claim that there is no path in 0 from Ð!ß !Ñ to Ð"ß !Ñ and therefore >0 is not path connected.
Suppose, on the contrary, that 2 À Ò!ß "Ó Ä>0 is a path from Ð!ß !Ñ to Ð"ß !ÑÞ For > − Ò!ß "Ó , write 2Ð>Ñ œ Ð2"# Ð>Ñß 2 Ð>ÑÑ −> 0". 2 and 2 # are continuous (why? ). Since Ò!ß "Ó is compact, 2 is uniformly continuous (Theorem IV.9.6 ) so we can choose $" ! for which l? @l $"## Ê .Ð2Ð?Ñß 2Ð@ÑÑ " Ê l2 Ð?Ñ 2 Ð@Ñl "Þ
We have ! − 2" ÐÐ!ß !ÑÑ . Let > ‡ œ sup 2 " Ð!ß !ÑÞ Then ! Ÿ > ‡ " . Since 2 " Ð!ß !Ñ is a closed set, >‡" − 2 Ð!ß !Ñ so 2Ð> ‡ Ñ œ Ð!ß !ÑÞ (We can think of t‡ as the last “time” that the path 2 goes through the origin).
‡‡ ‡ ‡ Choose a positive $$""" so that !Ÿ> > $ " . Since 2Ð>Ñœ! and 2Ð> $ Ñ! , we can choose a positive integer R for which
‡‡## !œ2Ð>ÑŸ""R" R 2Ð> $ ÑÞ
‡‡ # By the Intermediate Value Theorem, there exist points ?ß @ − Ð> ß > $ Ñ where 2" Ð?Ñ œ R" #RÐR "Ñ1 1 and 2Ð@Ñœ"#R## . Then 2Ð?Ñœ sin and 2Ð@Ñœ # sin , so l2Ð?Ñ2Ð@Ñlœ" ## . But this is impossible since l?@l$$ "## and therefore l2Ð?Ñ2 Ð@Ñl"Þ ñ
Note: Let > be the graph of the restriction 1 œ 0lÐ! "Ó . For any set E © Ö!× ‚ Ò "ß "Ó , a similar argument shows that >>∪E is not path connected. In particular, cl œ>>> ∪ÐÖ!ׂÒ"ß"ÓÑ is not path connected. But is homeomorphic to Ð!ß"Ó , so is path connected. So the closure of a path connected space need not be path connected .
Definition 3.5 A space Ð\ßg Ñ is called
a) locally connected if for each point B−\ and for each neighborhood R of B , there is a connected open set YB−Y©R such that .
b) locally path connected if for each point B−\ and for each neighborhood R of BYB−Y©R , there is a path connected open set such that .
Note: to say Y is path connected means that any two points in U can be joined by a path in U . Roughly, “locally path connected” means that “nearby points can be joined by short paths.”
Example 3.6
1) ‘8 is connected, locally connected, path connected and locally path connected.
2) A locally path connected space is locally connected.
222 3) Connectedness and path connected are “global” properties of a space \ : they are statements about \ “as a whole.” Local connectedness and local path connectedness are statement about what happens “locally” (in arbitrarily small neighborhoods of points) in \ In general, global properties do not imply local properties, nor vice-versa.
a) Let \ œ Ð!ß "Ñ ∪ Ð"ß #Ó . \ is not connected (and therefore not path connected) but \ is locally path connected (and therefore locally connected). The same relations hold in a discrete space \ with more than one point.
b) Let \\ be the subset of ‘# pictured below. Note that contains the “topologist's sine curve” as a subspace you need to imagine it continuing to oscillate faster and faster as it approaches the vertical line segment in the picture:
The \\ is path connected (therefore connected, but is not locally connected: if : œ Ð!ß !Ñß there is no open connected set containing : inside the neighborhood RœFÐ:Ñ∩\" . Therefore \ is also not locally path connected. #
Notice that Examples a) and b) also show that neither “(path) connected” nor “locally (path) connected” implies the other.
Lemma 3.7 Suppose that 0\+,1,- is a path in from to and is a path from to . Then there exists a path 2\ in from +- to .
Proof 01 ends where begins, so we feel intuitively that we can “join” the two paths “end-to- end” to get a path 2+- from to . The only technical detail to handle is that, by definition, a path 2 must be a function with domain Ò!ß "Ó . To get 2 we simply “join and reparametrize:”
223
" 0Ð#>Ñ ! Ÿ > Ÿ # Define 2ÀÒ!ß"ÓÄ\ by 2Ð>Ñœ " . (You can imagine a point moving 1Ð#> "Ñ# Ÿ > Ÿ " twice as fast as before: first along the path 01 and then continuing along the path .) The function 2ñ is continuous by the Pasting Lemma (see Exercise III.E22 ).
Theorem 3.8 If \\ is connected and locally path connected, then is path connected.
Proof If \œg , then \ is path connected. So assume \Ág . For +−\ , let GœÖB−\Àthere exists a path in \ from + to B× . Then GÁg since +−G (why? ). We want to show that Gœ\Þ
Suppose B−G . Let 0 be a path in \ from + to B . Choose a path connected open set Y containing BC−Y1YBC . For any point , there is a path in from to . By Lemma 3.7, there is a path 2 in \ from + to C , so C − GÞ Therefore B − Y © Gß so G is open.
Suppose BÂG and choose a path connected open set Y containing B . If C−Y , there is a path 1 in YCB from to . Therefore there cannot exist a path in \+C from to or else, by Lemma 3.7, there would be a path 2+BB from to and would be in G . Therefore CÂGB−Y©\G , so , so G is closed and therefore clopen.
\G is connected and is a nonempty clopen set, so Gœ\\ñ . Therefore is path connected.
Here is another situation (particularly useful in complex analysis) where connectedness and path connected coincide:
Corollary 3.9 An open connected set S in ‘8 is path connected.
Proof Suppose B−S . If R is any neighborhood of B in S , then B− intS RœY ©S . Since S is open in ‘‘%88 , and YSYÞ is open in , is also open in Therefore there is an ! such that 8 FÐBÑ©Y%% ©RÞ Since FÐBÑ is an ordinary ball in ‘ ßFÐBÑ % is path connected Þ (Of course,
224 this might not be true for a ball in an arbitrary metric space.) We conclude that S is locally path connected so, by Theorem 3.8, Sñ is path connected.
4. Components
Informally, the “components” of a space \ are its largest connected subspaces. A connected space \\ has exactly one component itself. In a totally disconnected space, for example , the components are the singletons ÖB×Þ In very simple examples, the components “look like” just what you imagine. In more complicated situations, some mild surprises can occur.
Definition 4.1 A component G\ of a space is a maximal connected subspace. (Here, “maximal connected” means: G is connected and if G©H©\ where H is connected, , then GœH )
For any , let and is connected Then ; since is :−\ G: œ ÖEÀ:−E©\ E ×Þ Ö:×−G:: G a union of connected sets each containing :G , :: is connected (Corollary 2.10 ). If G©H and H is connected, then HE was one of the sets in the collection whose union defines G: so H©G:: and therefore G œHÞ Therefore G : is a component of \ that contains : , so can be written as the union of components: . \œ:−\ G:
Of course it can happen that GœG:; when :Á;Àfor example, in a connected space \ , Gœ\: for every :−\Þ But if GÁG :;:; , then G∩GœgÀ if B−G∩G :;:; , then G∪G would be a connected set strictly larger than G: .
The preceding paragraphs show that the distinct components of \\À form a partition of a pairwise disjoint collection whose union is \:µ;:; . If we define to mean that and are in the same component of \µ , then it is easy to see that “ ” is an equivalence relation on \ and that G:: is the equivalence class of .
Theorem 4.2 \\ is the union of its components. Distinct components of are disjoint and each component is a closed connected set.
Proof In light of the preceding comments, we only need to show that each component G: is closed. But this is clear: G©:: cl G and cl G : is connected (Corollary 2.20 ). By maximality, we conclude that Gœ cl GÞ: ñ
It should be clear that a homeomorphism maps components to components. Therefore homeomorphic spaces have the same number of components.
Example 4.3
1) Let \ œ Ò"ß #Ó ∪ Ò$ß %Ó ∪ Ò&ß 'Ó ©‘ Þ \ has three components: Ò"ß #Ó, Ò$ß %Ó, and Ò&ß 'Ó. For each ! Ÿ : Ÿ ", we have G: œ Ò!ß "ÓÞ If G is a component in a space \ that has only finitely many components, then \G is the union of the other finitely many (closed) components. Therefore G is clopen.
225 However, a space can have infinitely many components and in general they need not be " open. For example, if \ œ Ö!× ∪ Ö8 À 8 −‘ × © , then the components are the singleton sets ÖB× (why? ) . The component Ö!× is not open in \ .
2) In ‘# , \ œ#$ B ÐÐ8ß !ÑÑ is not homeomorphic to ] œ B ÐÐ8ß !ÑÑ because 8œ""Î% 8œ" "Î% \] has two components but has three.
3) If G©\ and G is nonempty connected and clopen, then G is a component of \ : for if gÁG©H©\ , then G is clopen in H so if H is connected, then GœHÞ
4) The sets \] and in ‘# pictured below are not homeomorphic since \ contains a cut point :\Ö:× for which has three components. ] contains no such cut point.
Example 4.4 The following examples are meant to help “fine-tune” your intuition about components by pointing out some false assumptions that you need to avoid. (Take a look back at Definition 2.4 to be sure you understand what is meant by a “disconnection.”)
" 1) Let \ œÖ!× ∪ Ö8 À 8 − ×Þ One of the components of \ is Ö! }, but Ö!× " is not clopen in \ . Therefore the sets E œ Ö!× and F œ Ö8 À 8 − × do not form a disconnection of \Þ . A component and its complement may not form a disconnection of \
2) If a space \ is the union of disjoint closed connected sets, these sets need not be components. For example, Ò!ß "Ó œ<−Ò!ß"Ó Ö<× .
226 3) If B and C are in different components of \, then there might not exist a disconnection of for which and For example, consider∞ # , E∪F \ B−E C−FÞ \œP∪P∪"#8œ" V© 8‘ where:
PP"# and are the straight lines with equations Cœ#Cœ#Þ and
" For each 8 − ß V8 is the rectangle ÖÐBß CÑ À lBl œ 8 and lCl œ # 8 ×Þ (The top and bottom edges of the VPPÞV8"#8 's approach the lines and ) The first four 's are pictured:
Each V\V\88 is connected and clopen in . Therefore each is a component of . (See Example 4.3.3.)
The other components of \P are "# and P . For example:
PG"" is connected. Let be the component that contains PG . must be disjoint from each component VPÁG8" , so if , then the additional points in G are from P # that is, GœP"# ∪H where H©PÞ But in that case, P " would be a nontrivial clopen set in G and GPœGÞ would not be connected. Therefore "
Suppose that EF and are disjoint clopen sets in \ for which \œE∪FÞEF and are separated so, by Lemma 2.8, PEF# is either a subset of or a subset of : without loss of generality, assume that P©E## and let :−PÞ Since E is open, : has a neighborhood R©EÞ But R intersects infinitely many connected VE8 's, each of which, therefore, must also be a subset of . Since the top edges of these VP8" 's approach , there are points on PEœEÞP " in cl Therefore " intersects E, so P""# ©EÞ So P ∪P ©EÞ
In particular: Ð!ß#Ñ−P"# and Ð!ß#Ñ−P are in different components of \ , but both are in the same piece E of a disconnection.
Conversely, however: suppose E∪F is a disconnection of some space ] , with B−E and C−F. Then B and C must be in different components of ] . (Why? )
227 4) Suppose \@G\Ö@× is a connected space with a cut point . Let be a component of . ()Draw a few simple pictures before reading on.
It can happen that @Â cl\ G (Would you have guessed that @G must be in cl\ ? )
For example, consider the following subspace\œG∪F of ‘# where
" GÒß"ÓB is the interval # on the -axis and
∞ is a “broom” made up of disjoint “straws” F œ ÖÐ!ß !Ñ× ∪8œ" G88 G (each a copy of Ð!ß "Ó ) extending out from the origin and arranged so that slopeÐG8 Ñ Ä !Þ
The broom FFœ\ is connected (because it's path connected ), so cl\ is connected.
Let @œÐ!ß!Ñ . Each straw G8 is connected and clopen in \Ö@× . Therefore each G8 is a component of \ Ö@×ß and the remaining connected subset, G is the remaining component of \ Ö@×Þ
So @\@ÂG is a cut point of and cl\ .
Note: for this example, @Â cl\ G , but @ is in the closure of the each of the other components G\Ö@×Þ8 of
There is a much more complicated example, due to Knaster and Kuratowski (Fundamenta Mathematicae, v. 2, 1921 ). There, \ is a connected set in ‘# with a cut point @\Ö@× such that is totally disconnected . Intuitively, all the singleton sets ÖB× are “tied together” at the point @\@\ to create the connected space ; removing causes to “explode” into “one-element fragments.” In contrast to the “broom space”, all components in \Ö@× are singletons, so @ is not in the closure (in \ ) of any of them.
228 Here is a description of the Knaster-Kuratowski space \ (sometimes called “Cantor's teepee”). The proof that is has the properties mentioned is omitted. (You can find it on p. 145 of the book Counterexamples in Topology (Steen & Seebach) . Define
# " G œ the Cantor set ( © Ò!ß "Ó Ñ on the B -axis in ‘ , and let @ œ Ð# ß "Ñ .
Hœ { :−GÀ : is the endpoint of one of the “deleted middle thirds” in the construction of G× œ Ö: − G À : œ !Þ+"#$ + + ÞÞÞ+ 8 ÞÞÞbase three , where the + 8 's are eventually equal to !#ÞH or eventually equal to } Of course, is countable.
IœGH ( œ “the points in G that are not isolated on either side in G ”)
Then G œH∪IßH∩Iœg and both H and I are dense in G .
For each :−G , let @:œ the line segment from @ to : and define a subset of @: by
ÖÐBßCÑ−@:ÀC is rational × if :−H Gœ : ÖÐBßCÑ−@:ÀC is irrational × if :−IÞ Cantor's teepee is the space . One can show that is connected and that \œ:−G G: \ \Ö@× is totally disconnected.
5. Sierpinski's Theorem
Let gg be the cofinite topology on . Clearly, Ðß Ñ is connected. But is it path connected? (Try to prove or disprove it.) This innocent sounding question turns out to be harder than you might expect.
If 0ÀÒ!ß"ÓÄÐßÑg is a path from (say) "#ÐßÑ to in g . Then ran Ð0Ñ is a connected set containing at least two points. But every finite subspace of Ðßg Ñ is discrete, so ran Ð0Ñ must be infinite. Therefore ∞ " , where infinitely many of the sets " are nonempty. Ò!ß"Óœ8œ" 0Ð8Ñ 0Ð8Ñ Is this possible? The question is not particularly easy. In fact, the question of whether Ðßg Ñ is path connected is equivalent to the question of whether Ò!ß "Ó can be written as a countable union of pairwise disjoint nonempty closed sets.
The answer lies in a famous old theorem of Sierpinski which states that a compact connected Hausdorff space \ cannot be written as a countable union of two or more nonempty pairwise disjoint closed sets. (Of course, “countable” includes “finite.” But the“finite union” case is trivial: \8 is not a union of nonempty disjoint closed sets Ð8 #Ñ since each set would be clopen\ an impossibility since is connected.)
We will prove Sierpinski's result after a series of several lemmas. The line of argument used is due to R. Engelking. (It is possible to prove Sierpinski's theorem just for the special case \œÒ!ß"Ó. That proof is a little easier but still nontrivial.)
229 Lemma 5.1 If EF and are disjoint closed sets in a compact Hausdorff space \ , then there exist disjoint open sets YZ and with E©YF©ZÞ and
Proof Consider first the case where EœÖB× , a singleton set. For each C−F , choose disjoint open sets YZCC and with B−YC−Z C and C . The open sets Z C cover the compact set F so a finite number of them cover F , say F © ZCC"8 ∪ ÞÞÞ ∪ Z œ Z . Let Y œ Y C " ∩ ÞÞÞ ∩ Y C 8 . Then E©YßF©Z and Y , Z are disjoint open sets. Suppose now that EF and are any pair of disjoint closed sets in \ . For each B−Eß pick disjoint open sets YZBB and such that ÖBשYF©Z B and B . The open Y B 's cover the compact set E , so a finite number of them cover E , say E © YBB"8 ∪ ÞÞÞ ∪ Y œ Y Þ Let
Z œ ZBB"8 ∩ ÞÞÞ ∩ Z. Then E © Y ß F © Z and Y ß Z are disjoint open sets. ñ
Note: If EF and were both finite, an argument analogous to the proof given above would work in any Hausdorff space \ . The proof of Lemma 5.1 illustrates the rule of thumb that “compact sets act like finite sets.”
Lemma 5.2 Suppose SÐ\ßÑœÖJÀ−M× is an open set in the compact space gY . If α α is a family of closed sets in \ for which Yαα © S , then there exist ß ÞÞÞß − M such that " 8 J∩J∩ÞÞÞ∩J©SÞαα"# α 8
Proof For each C−\S , there is an αα such that CÂJαα . Therefore Ö\J À −M× is an open cover of the compact set \ S . There exist αα" ß ÞÞÞ8 − M such that
Ð\Jαα"8 Ñ∪Ð\J2 Ñ∪ÞÞÞ∪Ð\J α Ѫ\S. Taking complements gives that
J∩J∩ÞÞÞ∩J©SÞñαα"# α 8
Definition 5.3 Suppose . The set and is clopen in is :−\ U: œ ÖG©\À :−G G \× called the quasicomponent of \: containing .
U\: is always a closed set in . The next two lemmas give some relationships between the component G::: containing and the quasicomponent UÞ
Lemma 5.4 If :−\ , then G:: ©UÞ
Proof If G:G©GG∩G is any clopen set containing , then :: for otherwise and disconnect . Therefore and is clopen . G∩Ð\GÑ::: G G©ÖG À : − G G × œ U: ñ
230 # Example 5.5 An example where GÁUÞ:: In ‘ , let P 8 be the horizontal line segment Ò !ß "Ó ‚ Ö" × and define \ œ∞ P ∪ ÖÐ!ß !Ñß Ð"ß !Ñ× . 8 8œ" 8
The components of \ are the sets P8 and the singleton sets ÖÐ!ß !Ñ× and ÖÐ"ß !Ñ×Þ
If G\Ð!ß!ÑGP is any clopen set in containing , then intersects infinitely many 8 's so (since the PGP88's are connected) contains those 's. Hence the closed set G contains points arbitrarily close to Ð"ß !Ñ so Ð"ß !Ñ is also in G . Therefore Ð!ß !Ñ and Ð"ß !Ñ are both in UÐ!ß!Ñ , so GÁUÐ!ß!Ñ Ð!ß!Ñ. (In fact, it is easy to check that UÐ!ß!Ñ œ ÖÐ!ß !Ñß Ð"ß !Ñ×Þ)
Lemma 5.6 If \:−\GœU is a compact Hausdorff space and , then :: .
Proof GG©UGœUU:::::: is a maximal connected set and ; therefore if we can prove that is connected. Suppose UœE∪F:: , where EßF are disjoint closed sets in U . We can assume that :−E . We will show that Fœg in other words, that there is no disconnection of U: . U\EF\: is closed in , so and are also closed in . By Lemma 5.1, we can choose disjoint open sets Y and Z in \ with :−E©Y and F©Z . Then U: œE∪F©Y∪ZÞ Since \U is compact and : is an intersection of clopen sets in \ , Lemma 5.2 lets us pick finitely many clopen sets G"8 ß ÞÞÞß G such that U :" © G ∩ ÞÞÞ ∩ G 8 © Y ∪ Z . Let G œ G " ∩ ÞÞÞ ∩ G 8 . G is clopen in \ and U: ©G©Y∪Z . Y∩G is open in \ and, in fact, Y∩G is also closed in \ : since cl Y©\Zß we have that Y∩Gœ cl Y∩Gœ cl Y∩ cl Gª cl ÐY∩GÑÞ Therefore Y∩G is one of the clopen sets containing :UU©Y∩G©YU∩Fœgß whose intersection defines :: , so . Therefore : so FœgÞ ñ
Definition 5.7 A continuum is a compact connected Hausdorff space.
231 Lemma 5.8 SupposeE\gÁEÁ\G is a closed subspace of a continuum and that . If is a component of EG∩EÁg , then Fr .
Proof Let GE:−GÞG©EœE be a component of and let Since cl , we have that G∩Fr EœG∩ cl E∩ cl Ð\EÑÑœG∩ cl Ð\EÑ , so we need to show that G∩cl Ð\EÑÁgÞ We do this by contraposition: assuming G∩ cl Ð\EÑœg , we will prove Eœ\.
is compact so Lemma 5.6 gives and is clopen in E GœG:: œU œ ÖGααα ©EÀ :−G G E×Þ By assumption, G © \ cl Ð\ EÑ , so by Lemma 5.2 there exist indices αα"# , ,..., α 8 for which G©GœG∩GÞÞÞ∩G©\Ð\EÑααα!"# α 8 cl . Since G∩Ð\EÑœg α ! cl , we have
G∩α! Fr Eœg .
GEG©EEœEÞGEαα!!is clopen in and Fr int Since α ! is open in int which is open in \ ,
G\Gαα!! is open in . But is also closed in the closed set EG\\ , so α! is closed in . Since is connected, we conclude that Eœ\ . ñ
Lemma 5.9 Suppose is a continuum and that where the 's are \\œÖJÀ8−×J 88 pairwise disjoint closed sets and JÁg8 for at least two values of 8 . Then for each 8 there exists a continuum G©\88883 such that G∩Jœg and G∩JÁg for at least two 3−Þ
Before proving Lemma 5.9, consider the formal statement of Sierpinski's Theorem..
Theorem 5.10 (Sierpinski) Let be a continuum. If where the 's are \\œÖJÀ8−×J 88 pairwise disjoint closed sets, then at most one J8 is nonempty.
()Of course, the statement of the theorem includes the easy “finite union” case. In proving Sierpinski's theorem we will assume that where the 's are pairwise \œ ÖJ88 À8− × J disjoint closed sets and JÁg8 for at least two values of 8 . Then we will apply Lemma 5.9 to arrive at a contradiction. When all the smoke clears we see that, in fact, there are no continua which satisfy the hypotheses of Lemma 5.9. Lemma 5.9 is really the first part of the proof (by contradiction) of Sierpinski's theorem set off as a preliminary lemma to break the argument into more manageable pieces.
Proof of Lemma 5.9
If Jœg88, let Gœ\Þ
Assume JÁg877 . Choose 7Á8 with J Ág and pick a point :−JÞ By Lemma 5.1, we can choose disjoint open sets YßZ in \ with J878 © Y and : − J © Z . Let G be the component of :ZÞ in cl Certainly G88 is a continuum, and we prove that this choice of G works. We have that G∩Jœg88 (since G© 8 cl Z©\Y ) and that G∩JÁg 87 (since :−G∩J 87 ). Therefore, to complete proof, we need only show that for some 3Á7ß8ßG83 ∩J ÁgÞ
Since :− cl Z©\Y , we have that cl ZÁg ; and cl ZÁ\ because
232 gÁJ88 ©Y. Therefore, by Lemma 5.8, there is a point ;−G ∩ Fr Ð cl ZÑ . Since ;− Fr Ð cl ZÑ and J©Z©ÐZÑß77 int cl we have ;ÂJ . And since ;−ÐZÑ©Z©\Y Fr cl cl , we have that ;ÂJ83383 . But \ is covered by the J 's, so ;−J for some 3Á7ß8Þ Therefore G ∩J ÁgÞ ñ
Proof of Theorem 5.10 We want to show that if ∞ where the 's are disjoint closed \œ8œ" J88 J sets, then at most one JÁg8 . Looking for a contradiction, we suppose at least two J8 's are nonemptyÞ
By Lemma 5.9, there is a continuum G\"""" in with G∩Jœg and such that G has nonempty intersection with a least two 's. We can write ∞ J8"""8 G œG ∩\œG ∩8œ" J ∞∞, where at least two of the sets are nonempty. œ8œ" ÐG∩JÑœ"8 8œ# ÐG∩JÑ "8 G∩J "8
Applying Lemma 5.9 again (to the continuum GG©GG∩Ð"#"# ) we find a continuum such that G"# ∩JÑœG ## ∩J œg and G # intersects at least two of the sets G "8 ∩J Þ Then ∞∞∞ , where at G# œ G #" ∩ G œ8œ# ÐG # ∩ ÐG "8 ∩ J ÑÑ œ 8œ# ÐG #8 ∩ J Ñ œ 8œ$ ÐG #8 ∩ J Ñ least two of the setsG∩J#8 are nonempty.
We continue this process inductively, repeatedly applying Lemma 5.9, to and generate a decreasing sequence of nonempty continua G"# ª G ª ÞÞÞ ª G 8 ª ÞÞÞ such that for each 8ß . This gives ∞∞ ∞ ∞ But this is G∩Jœg88 gœG∩JœG∩\œGÞ8œ" 8 8œ" 8 8œ" 8 8œ" 8 impossible: the 's have the finite intersection property and is compact, so ∞ . G\GÁgñ8 8œ" 8
Example 5.11 By Theorem 5.10, we know that Ò!ß "Ó cannot be written as the union of 7 pairwise disjoint nonempty closed sets if " 7 Ÿ i! . And, of course, Ò!ß "Ó can easily be written as the union of such sets: for example, . What if 7 œ - Ò!ß "Ó œB−Ò!ß"Ó ÖB× i7-! ?
There are other related questions you could ask yourself. For example, can Ò!ß "Ó be written as the union of - disjoint closed sets each of which is uncountable? The answer is “yes.” For example, take a continuous onto function 0 À Ò!ß "Ó Ä Ò!ß "Ó# (a space-filling curve, whose existence you should have seen in an advanced calculus course ). For each B−Ò!ß"Ó , let #" PB œ ÖB× ‚ Ò!ß "Ó œ“the vertical line segment at B in Ò!ß "Ó . Then the sets 0 ÒPB Ó do the job.
We could also ask: is it possible to write Ò!ß "Ó as the union of uncountably many pairwise disjoint closed sets each of which is countably infinite? (See Exercise VIII.E27).
233 Exercises
E1. Suppose \œE∪F where EF and FE are separated.
a) Prove that for any G © \ , cl\F G œ clE ÐE∩GÑ∪ cl ÐF ∩GÑ . b) Conclude that G is closed if G∩E is closed in E and G∩F is closed in F . c) Conclude that G is open if G∩E is open in E and G∩F is open in F . d) Suppose \œE∪F where EF and FE are separated. Prove that if 0À\Ä] and both 0lE and 0lF are continuous, then 0 is continuous.
E2. Prove that Ò!ß "Ó is connected directly from the definition of connected. (Use the least upper bound property of ‘ . Ñ
E3. Suppose both Eß F are closed subsets of Ð\ßg Ñ . Prove that E F is separated from F E . Do the same assuming instead that EF and are both open.
E4. Suppose WÐ\ßÑÞW∩IÁgW∩Ð\IÑÁg is a connected subset of g Prove that if and , then W∩Fr IÁg .
E5. Let Ð\ß .Ñ and Ð] ß .w Ñ be two connected metric spaces. Suppose 5 ! and that Ð+ß,Ñ−\‚]. Let OœÖÐBßCÑ−\‚] À.ÐBß+ÑŸ5 and .w ÐCß,ÑŸ5× .
a) Give an example to show that the complement of O\‚] in might not be connected. b) Prove that the complement of O in \‚] is connected if Ð\ß.Ñ and ]ß.w Ñ are unbounded spaces.
E6. Prove that there does not exist a continuous function 0À‘‘ Ä such that 0ÒÓ© and 0Ò Ó © . Hint: One method: What do you know about ran0 ? What else do you know? ) " Another method: if such an 0 exists, let 1 œ"l0l and let 2 œ 1lÒ!ß "Ó . What do you know about 2?
E7. a) Find the cardinality of the collection of all compact connected subsets of ‘# . b) Find the cardinality of the collection of all connected subsets of ‘# .
E8. Suppose Ð\ß .Ñ is a connected metric space with l\l " . Prove that l\l - .
E9. Suppose each point in a metric space Ð\ß .Ñ has a neighborhood base consisting of clopen sets (such a metric space is sometimes called zero-dimensional ). Prove that Ð\ß .Ñ is totally disconnected.
234 E10. A metric space Ð\ß .Ñ satisfies the %-chain condition if for all % ! and all B , C − \ , there exists a finite set of points BB"# , , ... , B 8"8 , B where BœBßBœC " 8 , and .ÐBßBÑ 33" % for all 3œ1, ..., 8".
a) Give an example of a metric space which satisfies the % -chain condition but which is not connected.
b) Prove that if Ð\ß .Ñ is connected, then Ð\ß .Ñ satisfies the % -chain condition.
c) Prove that if Ð\ß .Ñ is compact and satisfies the % -chain condition, then \ is connected.
" # d) Prove that ÐÖ!Ñ‚Ò"ß"ÓÑ∪ÖÐBß sinB ÑÀ!BŸ"ש‘ is connected. (Use c) to give a different proof than the one given in Example 3.4. )
e) In any space Ð\ßg Ñ , a simple chain from + to , is a finite collection of sets ÖE"8 ß ÞÞÞß E × such that: +−E"3 and +ÂE if 3Á" ,−E83 and ,ÂE if 3Á8 for all 3 œ "ß ÞÞÞß 8 "ß E33" ∩ E Á g E∩Eœg34 if l43lÁ"
Prove \+ß,−\ß is connected iff for every open cover h and every pair of points there is a simple chain from +, to consisting of sets taken from h .
E11. a) Prove that \S is locally connected iff the components of every open set are also open in \. b) The path components of a space are its maximal path connected subsets. Show that \ is locally path connected iff the path components of every open set S\ are also open in .
E12. Let gg be the cofinite topology on . We know that Ðß Ñ is not path connected (because of Sierpinski's Theorem applied to the closed interval Ò!ß "Ó ). Prove that the statement “ Ðg ß Ñ is not path connected” is equivalent to “Sierpinski's Theorem for the case \ œ Ò!ß "ÓÞ ”
E13. Let 8 " and suppose 0 À Ò!ß "Ó Ä‘8 is a homeomorphism (into); then ran Ð0Ñ is called an arc in ‘‘8Þ Use a connectedness argument to prove that an arc is nowhere dense in 8 . Is the same true if Ò!ß "Ó is replaced by the circle W" ?
235 E14. a) Prove that for any space\8 #ß and
if \ has 8 components, then there are nonempty pairwise separated sets L"8 ß ÞÞÞß L for which \ œ L"8 ∪ ÞÞÞ ∪ L (**)
Hints. For a given 8L , do not start with the components and try to group them to form the 8 's. Start with the fact that \\ is not connected. Use induction. When has infinitely many components, then \ has n components for every 8 .
b) Recall that a disconnection of \EßF means a pair of nonempty separated sets for which \œE∪FÞ Remember also that if G is a component of \ , G is not necessarily “one piece in a disconnection of \ ” (see Example 4.4 ). Prove that \8Ð8 #Ñ\ has only finitely many components iff has only finitely many disconnections.
E15. A metric space Ð\ß .Ñ is called locally separable if, for each B − \ , there is an open set Y containing BÐYß.Ñ such that is separable. Prove that a connected, locally separable metric space is separable.
E16. In Ð\ßg Ñ , define BµC if there does not exist a disconnection \œE∪F with B−E and C−Fßi.e., if “ \ can't be split between B and C .” Prove that µ is an equivalence relation and that the equivalence class of a point :UÞ is the quasicomponent : (It follows that \ is the disjoint union of its quasicomponents. )
E17. For the following alphabet (capital Arial font), decide which letters are homeomorphic to each other:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
E18. Suppose 0À‘‘ Ä\œÖÐBßCÑ−# À Bœ! or Cœ!× is continuous and onto. Prove that 0" ÒÖÐ!ß !Ñ×Ó contains at least 3 points.
E19. Show how to write ‘# œE∪F where E and F are nonempty, disjoint, connected, dense and congruent by translation (i.e., bÐ?ß@Ñ− ‘# such that FœÖÐB?ßC@ÑÀÐBßCÑ−E×Ñ .
E20. Suppose \l\l #Þ\E∪FE is connected and Show that can be written as where and F\ are connected proper subsets of .
E21. Prove or disprove: a nonempty product \‚] is totally disconnected iff both \ and ] are totally disconnected.
236 Chapter V Review
Explain why each statement is true, or provide a counterexample.
1. There exists a continuous function 0À‘‘ Ä which is onto and for which 0Ò Ó © Þ
2. Let g‘‘g be the right ray topology on . Then (ßÑ is path connected.
3. There exists a countably infinite compact connected metric space.
4. The letter T is homeomorphic to the letter F.
5. If EF and are connected and not separated, then E∪F is connected.
6. If EF and are nonempty and E∪F is connected, then cl E∩FÁgÞ cl
7. If the graph of a function 0À‘‘ Ä is connected, then 0 is continuous.
8. , with the cofinite topology, is connected.
9. In a space Ð\ßg ), the component containing the point : is a subset of the intersection of all clopen sets containing a point : .
10. If Eß F © Ð\ßg Ñ and E is clopen in E ∪ F , then \ is not connected.
11. If 0 ÀÐ\ßgg ÑÄÐ]ßw Ñ is continuous and onto and Ð\ß g Ñ is path connected, then Ð]ß gw Ñ is path connected.
12. Suppose E©‘ , ±E± " . If E is nowhere dense, then E is not connected.
13. Let G\\‚G be the Cantor set. There exists a nonempty space for which is connected.
14. Let GG be the Cantor set. Then ‘## is connected.
15. If GÐ\ß.ÑÐGß.Ñ is a component of the complete metric space , then is complete.
16. Let SS"#" denote the unit circle in ‘ . is homeomorphic to a subspace of the Cantor set.
17. If \ is the union of an uncountable collection of disjoint nonempty, connected closed sets GGαα, then the 's are components of \ .
18. If \ is the union of a finite collection of disjoint nonempty connected closed sets Gα , then the G\α 's are components of .
19. If \G is the union of a countable collection of disjoint connected closed sets α , then the G\α's are components of .
237 µ µ 20. If Ð\ß .Ñ is connected, then its completion ( \ , . ) is connected.
21. If EFE∩FÁgE©FÞ is connected and is clopen and , then
22. Suppose 0 À‘‘88 Ä is continuous, that 0ÐÐ!ß !ß ÞÞÞß !ÑÑ œ Ð!ß !ß ÞÞÞß !Ñ and 0ÐÐ"ß !ß ÞÞÞß !ÑÑ œ Ð#ß !ß ÞÞÞß !Ñ. Let E denote the set of all fixed points of 0 . It is possible that E is open in ‘8 ( Note: we are not assuming f is a contraction, so f may have more than one fixed point.)
23. Suppose Ð\ß .Ñ is a connected separable metric space with ± \ ± " . Then ± \ ± = - .
24. If a subset EE of ‘ contains an open interval around each of its points, then must be connected.
25. There exists a connected metric space Ð\ß .Ñ with l\l œ i! .
# 26. If ‘ ªE"# ªAA ª ... ª 8 ª ... is a nested sequence of connected sets in the plane, then ∞ A is connected. 8œ" 8 27. ЂÑ∪Ђё ‘ is a connected set in ‘# .
" # 28. Let E œ ÖÐBß sinB Ñ À B !× ©‘ and suppose 0 À cl ÐEÑ Ä Ö!ß "× is continuous. Then 0 must be constant.
29. Suppose \ÁgÞ\ is connected if and only if there are exactly two functions 0−GÐ\Ñ such that 0œ0# .
30. If \0−GÐ\Ñ is nonempty, countable and connected, then every is constant.
31. Every path connected set in ‘# is locally path connected.
32ÞF If is a dense connected set in ‘‘ , then FœÞ
33. In a metric space Ð\ß .Ñ the sets E and F are separated iff .ÐEß FÑ !Þ
34. A nonempty clopen subset of a space \\ must be a component of .
35. Suppose gg and ww are topologies on \©ÞÐ\ßÑ and that gg If g is connected, then Ð\ßg w Ñ is connected.
36. The letter X is homeomorphic to the letter Y.
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