20. Mon, Oct. 13 What we have done so far corresponds roughly to Chapters 2 & 3 of Lee. Now we turn to Chapter 4. The first idea is connectedness.

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1 20. Mon, Oct. 13 What we have done so far corresponds roughly to Chapters 2 & 3 of Lee. Now we turn to Chapter 4. The frst dea s connectedness. Essentally, we want to say that a space cannot be decomposed nto two dsjont peces. Defnton A dsconnecton (or separaton) of a space X s a par of dsjont, nonempty open subsets U, V X wth X = U [ V. We say that X s connected f t has no dsconnecton. Example (1) If X s a dscrete space (wth at least two ponts), then any par of dsjont nonempty subsets gves a dsconnecton of X. (2) Let X be the subspace (0, 1) [ (2, 3) of R. ThenX s dsconnected. (3) More generally, f X = A ` B for nonempty spaces A and B, thenx s dsconnected. (4) Another example of a dsconnected subspace of R s the subspace Q. A dsconnecton of Q s gven by ( 1, ) \ Q and (, 1) \ Q. (5) Any set wth the trval topology s connected, snce there s only one nonempty open set. (6) Of the 29 topologes on X = {1, 2, 3}, 19 are connected, and the other 10 are dsconnected. For example, the topology {;, {1},X} s connected, but {;, {1}, {2, 3},X} s not. (7) If X s a space wth the generc pont (or ncluded pont) topology, n whch the nonempty open sets are precsely the ones contanng a specal pont x 0,thenX s connected. (8) If X s a space wth the excluded pont topology, n whch the open proper subsets are the ones mssng a specal pont x 0,thenX s connected. (9) The lower lmt topology R`` s dsconnected, as the bass elements [a, b) are both open and closed (clopen!), whch means that ther complements are open. Proposton Let X be a space. The followng are equvalent: (1) X s dsconnected (2) X = A ` B for nonempty spaces A and B (3) There exsts a nonempty, clopen, proper subset U X (4) There exsts a contnuous surjecton X {0, 1}, where {0, 1} has the dscrete topology. Now let s look at an nterestng example of a connected space. Proposton The only (nonempty) connected subspaces of R are sngletons and ntervals. Proof. It s clear that sngletons are connected. Note that, by an nterval, we mean smply a convex subset of R. It s clear that any connected subset must be an nterval snce f A s connected and a<b<cwth a, c 2 A, thenetherb 2 A or ( 1,b) \ A and (b, 1) \ A gve a separaton of A. So t remans to show that ntervals are connected. Let I R be an nterval wth at least two ponts, and let U I be nonempty and clopen. We wsh to show that U = I. Let a 2 U. Wewll show that U \ [a, 1) =I \ [a, 1). A smlar argument wll show that U \ ( 1,a]=I \ ( 1,a]. Consder the set R a = {b 2 I [a, b] U}. Note that a 2 R a, so that R a s nonempty. If R a s not bounded above, then [a, 1) U I, and we have our concluson. Otherwse, the set R a has a supremum s =supr a n R. Snce we can express s as a lmt of a U-sequence and snce U s closed n I, t follows that f s 2 I then s must also le n U. Note that f s/2 I, thensncei s an nterval we have U \ [a, 1) =[a, s) =I \ [a, 1). On the other hand, as we just sad, f s 2 I then s 2 U. ButU s open, so some -neghborhood of s (n I) lesnu. But no pont n (s, s + /2) can le n U (or I), snce any such pont would then 35

2 also le n R a. Agan, snce I s an nterval we have U \ [a, 1) =[a, s] =I \ [a, 1). One of the most useful results about connected spaces s the followng. Proposton Let f : X! Y be contnuous. If X s connected, then so s f(x) Y. Proof. Ths s a one-lner. Suppose that U f(x) s closed and open. Then f 1 (U) mustbe closed and open, so t must be ether ; or X. Ths forces U = ; or U = f(x). Snce the exponental map exp : [0, 1] connected. More generally, we have Proposton Let q : X! S 1 s a contnuous surjecton, t follows that S 1 s! Y be a quotent map wth X connected. Then Y s connected. Whch of the other constructons we have seen preserve connectedness? All of them! (Well, except that subspaces of connected spaces need not be connected, as we have already seen. Proposton Let A X be connected for each, andassumethatx 0 2 T A 6= ;. Then S A s connected. Proof. Assume each A s connected, and let U S A be nonempty and clopen. Then x 2 U for some x 2 S A. Suppose x 2 A 0.ThenU\ A 0 s nonempty and clopen n A 0,soU\ A 0 = A 0. In other words, A 0 U. Snce x 0 2 A 0, t follows that x 0 2 U. But now for any other A j,we have that x 0 2 A j \ U, so that A j \ U s nonempty and clopen n A j. It follows that A j U. 21. Wed, Oct. 15 Last tme, we ntroduced the dea of connectedness and showed (1) that the connected subsets of R are precsely the ntervals and (2) the mage of a connected space under a contnuous map s connected. Ths mples. Theorem 21.1 (Intermedate Value Theorem). Let f :[a, b] ever ntermedate value between f(a) and f(b). Proof. Ths follows from the fact that the mage s an nterval.! R be contnuous. Then f attans We also showed that an overlappng unon of connected subspaces s connected. As an applcaton, we get that products nteract well wth connectedness. ny Proposton Assume X 6= ; for all 2{1,...,n}. Then X s connected f and only f each X s connected. =1 Proof. ()) Ths follows from Prop 20.5, as p : Y X! X s surjectve (ths uses that all X j are nonempty). (() Suppose each X s connected. By nducton, t su ces to show that X 1 X 2 s connected. Pck any z 2 X 2. We then have the embeddng X 1,! X 1 X 2 gven by x 7! (x, z). Snce X 1 s connected, so s ts mage C n the product. Now for each x 1 2 X 1, we have an embeddng x1 : X 2,! X 1 X 2 gven by y 7! (x 1,y). Let D x1 = x1 (X 2 ) [ C. Note that each D s connected, beng the overlappng unon of two connected subsets. But we can wrte X 1 X 2 as the overlappng unon of all of the D x1, so by the prevous result the product s connected. The followng result s easy but useful. 36

3 Proposton Let A B A and suppose that A s connected. Then so s B. Proof. Exercse Theorem Assume X 6= ; for all 2 I, where s I s arbtrary. Then Y X s connected f and only f each X s connected. Proof. As n the fnte product case, t s mmedate that f the product s connected, then so s each factor. We sketch the other mplcaton. We have already showed that each fnte product s connected. Now let (z ) 2 Y X. For each j 2 I, wrted j = pj 1 (z j ) Y X. For each fnte collecton j 1,...,j k 2 I, let F j1,...,j k = \ Y D j X. j6=j 1,...,j k Then F j1,...,j k = Xj1 X jk, so t follows that F j1,...,j k s connected. Now (z ) 2 F j1,...,j k for every such tuple, so t follows that F = [ F j1,...,j k s connected. It remans to show that F s dense n Y product). Let X (n other words, the closure of F s the whole U = pj 1 1 (U j1 ) \ \pj 1 k (U jk ) be a nonempty bass element. Then U meets f j1,...,j k,sou meets F. Snce U was arbtrary, t follows that F must be dense. Note that the above proof would not have worked wth the box topology. We can show drectly that R N, equpped wth the box topology, s not connected. Consder the subset B R N consstng of bounded sequences. If (z ) 2B,then Y (z 1,z + 1) s a neghborhood of (z )nb. On the other hand, f (z ) /2 B, the same formula gves a neghborhood consstng entrely of unbounded sequences. We conclude that B s a nontrval clopen set n the box topology. Ok, so we have looked at examples and studed ths noton of beng connected, but f you asked your calculus students to descrbe what t should mean for a subset of R to be connected, they probably wouldn t come up wth the no nontrval clopen subsets dea. Instead, they would probably say somethng about beng able to connect-the-dots. In other words, you should be able to draw a lne from one pont to another whle stayng n the subset. Ths leads to the followng dea. Defnton We say that A X s path-connected f for every par a, b of ponts n A, there s a contnuous functon (a path) : I! A wth (0) = a and (1) = b. Ths s not unrelated to the earler noton. Proposton If A X s path-connected, then t s also connected. Proof. Pck a pont a 0 2 A. For any other b 2 A, we have a path b n A from a 0 to b. Then the mage b (I) s a connected subset of A contanng both a 0 and b. It follows that A = [ b2a 37 b(i)

4 s connected, as t s the overlappng unon of connected sets. For subsets A R, wehave A s path-connected ) A s connected, A s an nterval ) A s path-connected. So the two notons concde for subsets of R. But the same s not true n R 2! Example 21.7 (Topologst s sne curve). Let be the graph of sn(1/x) for x 2 (0, ].Then s homeomorphc to (0, ] and s therefore path-connected and connected. Let C be the closure of n R 2. Then C s connected, as t s the closure of a connected subset. However, t s not path-connected (HW VI), as there s no path n C connectng the orgn to the rght end-pont (, 0). 22. Fr, Oct. 17 Path-connectedness has much the same behavor as connectedness. Proposton (1) Images of path-connected spaces are path-connected (2) Overlappng unons of path-connected spaces are path-connected (3) Fnte products of path-connected spaces are path-connected However, the topologst s sne curve shows that closures of path-connected subsets need not be path-connected. Our proof of connectvty of Y X last tme used ths closure property for connected sets, so the earler argument does not adapt easly to path-connectedness. But t turns out to be easer to prove. Theorem Assume X 6= ; for all 2 I, where s I s arbtrary. Then Y X s path-connected f and only f each X s path-connected. Proof. The nterestng drecton s ((). Thus assume that each X s path-connected. Let (x ) and (y ) be ponts n the product Y X. Then for each 2Ithere s a path n X wth (0) = x and (1) = y. By the unversal property of the product, we get a contnuous path =( ):[0, 1]! Y X wth (0) = (x ) and (1) = (y ). The overlappng unon property for (path-)connectedness allows us to make the followng defnton. Defnton Let x 2 X. Wedefnetheconnected component (or smply component) of x n X to be C x = [ C. x2c connected Smlarly, the path-component of X s defned to be PC x = [ x2p connected 38 P.

5 The overlappng unon property guarantees that C x s connected and that PC x s path-connected. Snce path-connected sets are connected, t follows that for any x,wehavepc x C x. An mmedate consequence of the above defnton(s) s that any (path-)connected subset of X s contaned n some (path-)component. Example Consder Q, equpped wth the subspace topology from R. Then the only connected subsets are the sngletons, so C x = {x}. Such a space s sad to be totally dsconnected. Note that for any space X, each component C x s closed as C x s a connected subset contanng x, whch mples C x C x. If X has fntely many components, then each component s the complement of the fnte unon of the remanng components, so each component s also open, and X decomposes as a dsjont unon X = C 1 q C 2 q qc n of ts components. But ths does not happen n general, as the prevous example shows. The stuaton s worse for path-components: they need not be open or closed, as the topologst s sne curve shows. Defnton Let X be a space. We say that X s locally connected f any neghborhood U of any pont x contans a connected neghborhood x 2 V U. Smlarly X s locally pathconnected f any neghborhood U of any pont x contans a path-connected neghborhood x 2 V U. Ths may seem lke a strange defnton, but t has the followng nce consequence. Proposton Let X be a space. The followng are equvalent. (1) X s locally connected (2) X has a bass consstng of connected open sets (3) for every open set U X, the components of U are open n X Proof. We show (1), (3). Suppose that X s locally connected and let U X be open. Take C U to be a component. Let x 2 C. We can then fnd a connected neghborhood x 2 V U. SnceC s the component of x, wemusthavev C, whch shows that C s open. Suppose, on the other hand, that (3) holds. Let U be a neghborhood of x. Then the component C x of x n U s the desred neghborhood V. In partcular, ths says that the components are open f X s locally connected. The locally path-connected property s even better. Proposton Let X be a space. The followng are equvalent. (1) X s locally path-connected (2) X has a bass consstng of path-connected open sets (3) for every open set U X, the path-components of U are open n X (4) for every open set U X, every component of U s path-connected and open n X. Proof. The mplcatons (1), (3) are smlar to the above. We argue for (1), (4). Assume X s locally path-connected, and let C be a component of an open subset U X. Let P C be a nonempty path-component. Then P s open n X. But all of the other path-components of C are also open, so ther unon, whch s the complement of P, must be open. It follows that P s closed. Snce C s connected, we must have P = C. On the other hand, suppose that (4) holds. Let U be a neghborhood of x. Then the component C x of x n U s the desred neghborhood V. 39

6 In partcular, ths says that the components and path-components agree f X s locally pathconnected. Just as path-connected mples connected, locally path-connected mples locally-connected. But, unfortunately, there are no other mplcatons between the four propertes. Example The topologst s sne curve s connected, but not path-connected or locally connected or locally path-connected (see HWVI). Thus t s possble to be connected but not locally so. Example For any space X, thecone on X s defned to be CX = X [0, 1]/X {1}. The cone on any space s always path-connected. In partcular, the cone on the topologst s sne curve s connected and-path connected but not locally connected or locally path-connected. Example A dsjont unon of two topologst s sne curves gves an example that s not connected n any of the four ways. Example Note that f X s locally path-connected, then connectedness s equvalent to path-connectedness. A connected example would be R or a one-pont space. A dsconnected example would be (0, 1) [ (2, 3) or a two pont (dscrete) space. Fnally, we have spaces that are locally connected but not locally path-connected. Example The cocountable topology on R s connected and locally connected but not path-connected or locally path-connected. (See HWVI) Example The cone on the cocountable topology wll gve a connected, path-connected, locally connected space that s not locally path-connected. Example Two copes of R cocountable gve a space that s locally connected but not connected n the other three ways. 40

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