Math 5801 General Topology and Knot Theory

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1 Lecture 23-10/17/2012 Math 5801 Ohio State University October 17, 2012 Course Info Reading for Friday, October 19 Chapter 3.26, pgs HW 8 for Monday, October 22 Chapter 2.24: 3, 5a-d, 8a-d, 12a-f (see pg. 66 for required definitions) Chapter 2.25: 1, 2a-c

2 Connectedness How does path connectedness relate to connectedness? Proposition 217 (Path connected implies connected) If X is a path connected space then X is connected. Proof. Fix x0 X for each x X let fx : [0, 1] X be a path from x0 to x. X = x X fx([0, 1]) is a union of connected sets. What about the other direction? Connectedness Proposition 218 (Connected does not imply path connected) There is a space which is connected but not path connected. Example 219 (Ordered square is connected but not path connected) Let I 2 o = I I with dictionary order topology. I 2 o is a linear continuum and hence connected. Suppose f : [0, 1] I 2 o is continuous and satisfies f (0) = (0, 0) and f (1) = (1, 1) For each r [0, 1] the set {r} ( 1 3, 2 3 ) is open so f 1 ({r} ( 1 3, 2 3 )) is open in [0, 1]. Thus it contains some q Q [0, 1]. Thus π1 f is a surjection from countable Q [0, 1] to uncountable [0, 1]. Contradiction.

3 Connectedness Example 220 (Topologist s sine curve) Let X R 2 be the following subspace U = {(x, sin( 1 x )) x > 0} V = {(0, y) 0 y 1} X = U V U is path connected and hence connected. U X U so X is connected. No path can connect (0, y) to (1, sin 1). Suppose f : [0, 1] X is a path from (0, y) to (1, sin 1) with f (t) = (x(t), y(t)). Assume x(0) = 0 and x(t) > 0 for t > 0 otherwise restrict f to smaller domain. Choose (tn)n Z+ to be a sequence in I with y(tn) = ( 1) n and tn 0 f cannot be continuous. Definition 221 ( of a space) Let X be topological space. For each x, y X set x c y if X has a connected subset A with x, y A. A component of X is an equivalence class of c. Why is c transitive? Definition 222 (Path components of a space) Let X be topological space. For each x, y X set x p y if X if there is a path in X from x to y. A path component of X is an equivalence class of p. Why is p transitive?

4 Definition 223 (Concatenation of paths) Let f, g : I X be paths in X such that f (1) = g(0). The concatenation of f and g is the path f g : I X where { f (2t), 0 t 1 f g(t) = 2 1 g(2t 1), 2 t 1 Proposition 224 (Path components of a space) If f, g : I X are paths in X such that f (1) = g(0) then f g is a path in X from f (0) to g(1). Proof. f cont. and t 2t cont. so t f (2t) cont. g cont. and t 2t 1 cont. so t g(2t 1) cont. Gluing lemma gives f g cont. Examples 225 ( and path components) 1. of Q are single points. Suppose a, b A Q and a b. Let m = 2 a + (1 2 )b 2 2 Let U = {x A x < m} and V = {x A x > m}. Then U and V give a separation of A. 2. Path components of Q are single points. Each path component of a space X must be contained in a component of X Single points are path connected. 3. of topologists s sine curve X from Example 220 are the space X since X is connected. 4. Path components of topologists s sine curve X are the space are the sets U and V from Example 220.

5 Definition 226 (Locally connected) A topological space X is locally connected if for all x X and every open neighborhood Ux of x there is a connected neighborhood Vx of x with Vx Ux. Definition 227 (Locally path connected) A topological space X is locally path connected if for all x X and every open neighborhood Ux of x there is a path connected neighborhood Vx of x with Vx Ux. Examples 228 (Local connectivity) 1. Q is not locally connected or locally path connected. 2. [ ] 1 n Z+ 2n+1, 1 2n R is locally path connected. 3. {0} [ ] 1 n Z+ 2n+1, 1 2n R is not locally path connected. 4. Topologist s sine curve is not locally connected (hence not locally path connected). 5. R Q is not locally connected or locally path connected. 6. R 2 Q 2 is locally path connected.

6 Proposition 229 A space X is locally connected if and only if for every open subset U X each component of U is open in X. Proof. Suppose X is locally connected. Let U X be open. Let C U be a component of U. Let x C. x has a connected open nbdh Vx U. By definition of component Vx C. Thus C = x C Vx is open. Suppose for every open subset U X each component of U is open in X. Let U X be open. Let C U be the component of x U. C is a connected open neighborhood of x contained in U. Proposition 230 A space X is locally path connected if and only if for every open subset U X each path component of U is open in X. Proposition 231 If X is locally path connected then its components are also path components.

7 Intermediate Value Theorem for R lead us to notion of connectedess. Another classic theorem for continuous functions on R is the Maximum Theorem: If f : [a, b] R is continuous then there is y [a, b] s.t. for all x [a, b] we have f (y) f (x). What topological property of [a, b] ensures that continuous images achieve a maximum value? We want a property P such that if X has property P then for any continuous f : X R there is y X such that x X, f (y) f (x). Note that this property is not shared by R or (a, b). On the other hand it should (probably) be shared by any finite set with the discrete topology. Definition 232 (Cover) A cover of a topological space X is a collection A of subsets of X such that A = X. Definition 233 (Subcover) A subcover of a cover A of a topological space X is a collection S A such that S = X. Definition 234 (Open cover) An open cover of a topological space X is a collection U of open sets such that U = X. Definition 235 (Compact) A topological space X is compact if every open cover of X has a finite subcover.

8 Examples 236 (Noncompact Spaces) Showing that X is not compact only requires an infinite open cover of X which has no finite subcover: 1. R is not compact. For each n Z let Un = (n, n + 4 ). 3 Let U = {Un n Z}. U is an open cover of R since U = R. Let S U be a subcover. For all n Z n + 1 Un and n + 1 / Um for m n Hence for each n Z we have Un S. Thus the only subcover of U is U which is infinite. 2. (0, 1) is not compact. For each n Z+ let Un = ( 1, 1 ). n Let U = {Un n Z+}. 3. Q and Z are not compact. 4. R n is not compact. n+ 4 3 Definition 237 (Cover of a subspace) If A is a subspace of X a collection of set S covers A if A S. Proposition 238 A subspace A X is compact if and only if any covering of A by open sets of X has a finite subcovering.

9 Proposition 239 (Continuous images of compact spaces are compact) If X and Y are spaces and X is compact and f : X Y is continuous then f (X ) is a compact subspace of Y. Proof. Suppose X is compact and f : X Y is continuous Let V be a cover of f (X ) by open sets in Y. Let U = {f 1 (V ) V V}. U is an open cover of X so it has a finite subcover {f 1 (V1),, f 1 (Vn)} So {V1,, Vn} is a finite subcover of f (X ).