Types of componentsDefinition 6.13. Let X be a space. Define a relation ∼con X by a ∼cb ifthere is some conn ec t ed S ⊂ X with a, b ∈ S. The equivalence classes of Xunder ∼care the connected components (or just components) of X.Define a re l at ion ∼qon X by a ∼qb i f a and b are on the same side ofevery separation: for every separation X = U ∪ V with a ∈ U , we have balso in U. The equivalence classes of X under ∼qare the quasicomponentsof X.The ex i st e nc e of paths from a to b defines an equivalence relation; thepath components of X are the equivalence classes under this relation.There’s a problem with defining “arc components”, specifically, the ex-istence of an arc between two points is not in general a transitive relation.Example 6.14. Let Y be the two-point space 0, 1 with the indiscrete topol-ogy. Let X0= Y × [0, ∞), and let X = {(1, 0)} ∪ {0} × [0, ∞). Then Xis [0, ∞) with the standard topology, but with 0 d oub l ed . There is an arcfrom (0, 1) to any point in X, but no arc from (1, 0) to (0, 0).Then X is path con n ect e d but not arc connected, and the notion of “arccomponent” doesn’t make sense for X.Proposition 6.15. Let X be a space and x ∈ X. Then the path compo-nent of x is a subset of the connected component of x, and the connectedcomponent of x is a subset of the quasicomponent of x.Proof. If there is a path f rom x to y, then the image of this path is aconnected subspace of X containing x and y. This shows the first in cl u si on .If S is a conn ec t ed subspace of X containing x an d y, then any separationX = U ∪ V will have to have S on one side, say S ⊂ U. Otherwise theseparation will i ndu ce a separation of S. So x and y are both in U.Proposition 6.16. If X has only one quasicomponent, then X is connected.The proof is easy and is omi t t ed .Proposition 6.17. The quasicomponent of x is the intersection of all setscontaining x that are both closed and open.Proof. Suppose y is in the quasicomponent of x. Then for each separationX = U ∪ V with x ∈ U, we have y ∈ U. If U is closed and open with x ∈ U ,then either U = X or X = U ∪ (X\U) is a separation. In either case, y ∈ U,so y is in the intersection of all such sets.31Suppose y is not in the quasicomponent of x. Then there is some sepa-ration X = U ∪ V with x ∈ U and y ∈ V . Then U is closed and open andy /∈ U , so y is not in the intersection of all such sets.Example 6.18. For n ∈ N, n > 0, let Rnbe the rectangle in R2withcorners (1 − 1/n, n), (1 − 1/n, −n), (−1 + 1/n, n) and (− 1 + 1 /n, −n), letL1= {−1} ×R and L2= {1} × R. Let S = L1∪ L2∪SnRn. Let S = U ∪ Vbe a separation of S. Suppose L1⊂ U. Then for large n, Rn⊂ U, andtherefore L2⊂ U . However, any given Rnis can be separated from L1∪ L2.So L1∪ L2is a quasicomponent of S. But clearly, L1and L2are differentconnected components of S.Proposition 6.19. Quasicomponents are closed and connected componentsare closed.The proofs are easy and are om it t e d.Example 6.20. The path components of the topologist’s sine curve¯S aboveare S and {0} × [−1, 1], but¯S is connected. Note that S is not closed; sogenerally, path components are not ne ces sar i l y closed.Local connectedness properties and disconnectednessDefinition 6.21. A space X is locally connected if for every poi nt x ∈ X andevery open neighborhood U of x, there is a connected open neighb or hoodV of x with V ⊂ U .A space X is locally path connected if for every point x ∈ X and everyopen ne i ghb orhood U of x, there i s a path-connected open neighborhood Vof x with V ⊂ U.The following needs no proof.Proposition 6.22. A locally path connected space is locally connected.Example 6.23. The topologist’s s i ne curve¯S is connected but not locallypath connected: every sufficiently small neighborhood of (0, 0) has infinitelymany path components.You should be able to find an example of a locally conn ec t ed space thatis not connected.Lemma 6.24. In a space X, suppose S and T are connected sets withnonempty intersection. Then S ∪ T is connected. Similarly, if S and T arepath connected sets with nonempty intersection then S ∪T is path connected.32Proof is easy and is omitted.Proposition 6.25. If X is locally connected, then each connected compo-nent of X is open; therefore the quasicomponents of X coincide with theconnected components.Proof. If S is a connected component of X and S is not open, then there isa point x ∈ S that is a limit point of X\S. If X is locally connected, thenX has a connected open neighborhood U. Since S and U are connected andS ∩ U is nonem pty, S ∪ U is connected, and therefore U ⊂ S. but then x isnot a limit point of X\S. So S is open.Then S is open and closed, but sinc e S is connected no proper subset ofS is both open and closed. So S is a quasicomponent.Proposition 6.26. If X is locally path connected, then the quasicomponentsof X coincide with the path components.Proof. Suppose S is a path component of X. Suppose x is in the closure ofS. Since X is locally path connected, x has a path connected nei ghb or h oodU. Then S ∩ U is nonempty, and therefore S ∪ U is path connected, andU ⊂ S. So¯S ⊂ S and S is closed. Further, such an x cannot be a l i mi tpoint of X\S, and therefore S is open. Since S is connected and open andclosed, S is a quasicom ponent of X.Definition 6.27. A space is totally disconnected if its connected compo-nents are all points.Example 6.28. A space is both totally disconnected and locally connect edif and only if it is discrete.The subs pac es Q and R\Q of R with the standard topology are tot al lydisconnected.Any subspace of a totally disconnected space is totally
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