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# Intro Topology & Geom MATH 4510

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This 4 page Class Notes was uploaded by Miss Noel Mertz on Monday October 26, 2015. The Class Notes belongs to MATH 4510 at University of Utah taught by Staff in Fall. Since its upload, it has received 36 views. For similar materials see /class/229951/math-4510-university-of-utah in Mathematics (M) at University of Utah.

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Date Created: 10/26/15

Math 45101 Summer 2003 Introduction to Topology There will be no problems due for this section The nal exam will be held during class hour on Tuesdaay Jul 29 and will consist of shortanswer questions on the material of the second half of the course Elementary Topology Supplementary Notes part IV Connectness De nition Let X be a topological space X is connected if and only if X cannot be written as a disjoint union of two nonempty open sets Here are some equivalent formulations 1 A topological space X is connected if and only if X cannot be written as a disjoint union of two nonempty closed sets 2 Let X be a connected topological space Then whenever we write X U U V with U and V open and nonempty then U V 31 0 De nition Let A be a subset of a topological space X A is disconnected if we can findsets U and V open in X suchthat A C UUV andA Uy Q A Vy Q Ais connected if it is not disconnected Examples 1 Let X be the unit interval 0 1 Then the set of rational points of X is disconnected So is the set of irrational points 2 Any subset of the Cantor set with two points is disconnected 3 Proposition 0 1 is connected In fact any interval is connected Proposition The closure of a connected set is connected Propositon Let A and B be connected subsets of X If A B 31 0 then A U B is connected Propositions QC and 9D state the appropriate generalization to larger unions De nition Let X be a topological space De ne the equivalence relation on X xEy if there exists a connected set which contains both at and y For x E X the connected component of x is the set Cgc y E X yEx The sets Cgc are the connected components of X Proposition Every point in X belongs to precisely one connected component That is X is the disjoint union of its components The connected components are closed subsets of X Proposition A connected subset of R is an interval An open set on the line is a disjoint union of open intervals Proposition Let f X a Y be a continuous map If A is connected in X then fA is connected in Y Proposition Let f be a continuous function on the interval 1 b Then for every y between u and fb there is a c E 1 b such that fc 7 Examples in Rquot l A subset A of R is starlike if there is a point 10 in A such that for every other point a E A the line segment joining 10 to a is in A Starlike sets are connected 927 is an example of a subset of R2 which is surprisingly connected Look also at the graph of sinlac PathConnectedness De nition Let X be a topological space A path in X is the image of a continuous map 5 0 l a X Note that a path is directed it has an initial point and an endpoint If s is a path then 5 1 denotes the path in the reverse direction If 51 and 52 are paths such that the endpoint of 51 is the initial point of 52 we can combine them to form the product path 8182 De nition X is pathconnected if for any two points at y in X there is a path from x to y Note that 927 provides an example of a connected set in R2 which is not path connected We could de ne path components as we did components then this example has 2 path components Proposition An open set in R is connected if and only if it is path connected Hausdor 39 spaces De nition A topological space X is Hausdor 39 if and only if for x 31 y x and y have disjoint neighborhoods More precisely there are open sets UV with U V Q and x E U y E V Examples Any metric space is Hausdorff Let X be the curve in polar coordinates in the plane 7 60 3 6 S 27r as a point set For two rays R1R2 straight lines emanating from the origin let R1R2 be the set of points in X lying in the cone between R1 and R2 going counterclockwise Then the collection of R1 R2 is a base for a non Hausdorff topology on X because the points 6 0 and 6 27139 have the same neighborhoods Compactness De nition Let X 9 be a topological space A a subset of X A is compact if and only if every open covering of A has a finite subcover To be precise an open covering of A is a collection C C Q of open sets such that for x E A there is a U E C for which at E U Then A is compact if this condition is satisfied for any open covering C of A there are U1 U7 in C which form an open covering of A Examples 1 A finite set is compact A space X with the trivial topology is compact The arrow is compact 2 Rquot is not compact in fact any open ball B0r in R is not cornpact 3 An unbounded set in Rquot is not cornpact 3 A closed subset of a compact set is compact Proposition In a Hausdorff space a compact set is closed Proposition A compact set in R is closed and bounded Proposition Let f X a Y be a continuous map If X is compact so is Proposition Let X be a compact space f X a R a continuous function Then there are points a b in X such that M mama x e X my minfltxgt x e X So what are the compact subsets of Rquot Proposition A closed bounded set in R is compact To prove this we will need an important theorem of Lebesgue 3 Lebesgue s Theorem Any covering of any set in R has a countable subcovering Now the proof goes like this 1 A closed bounded interval I in R is compact This is the heart of the proof and is really just the Heine Borel theorem for those of you Who remember that Suppose we have a countable cover U1 U2 of I Which has no finite subcover Then for every 11 there is an 7 6 I not in U1 U U U Let a be the least upper bound of the set of 7 Since I is closed a E I so there is an N such that a 6 UN But alas all but finitely many of the 7 are also in UN since a is the least upper bound of the 7 so for some n gt N 7 6 U1 U U U a contradiction 2 A closed bounded rectangle in R2 is compact This is a slightly compicated unravelling of What it means to be a covering in a product space and is the basis of the induction proof to get to the general R 3 A closed bounded set in R2 is compact because it is a closed subset of a closed bounded rectangle Proposition Let X and Y be compact Hausdorff spaces A continuous bijection f X a Y is a homeomorphism Let g Y a X be the inverse to f We7ll show that for every closed set C in X 940 is closed in Y Suppose that O is closed in X Then 0 is compact so fO is compact in Y so is closed in Y But fO 940

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