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Unformatted text preview: Mathematics 185 Intro to Complex Analysis Fall 2005 M. Christ Solutions to Selecta from Problem Set 2 2.7 Let S C be a nonempty open set. Define z w to mean that there exists a path from z to w within S . Show that each component of S (that is, each equivalence class) is a path connected open set. Solution. Let z S be any point, and let S = { w S : w z } = [ z ]. We must show that S is open and path connected. If z 1 , z 2 S , then there exist paths i : [0 , 1] S such that i (0) = z and i (1) = z i . Out of these we can manufacture a path from z 1 to z 2 . Define : [ 1 , 1] S by ( t ) = 1 ( t ) for t [ 1 , 0], and ( t ) = 2 ( t ) for all t [0 , 1]. Its easy to show that is continuous 1 , since 1 (0) = 2 (0), and the range of is contained the union of the ranges of 1 , 2 , hence in S . Its tempting to simply declare victory now and assert that weve proved S to be path connected, but we havent quite said it properly. To show that S is path connected, we need to construct a path in S from z 1 to z 2 . All weve said is that our path lies within S . To see that its range is indeed in S , note that for any t > 0, the restriction of to the interval [0 , t ] defines a path in S from z to ( t ). Thus ( t ) S for all t [0 , 1], by definition of S . The same applies for t [ 1 , 0], of course. So is indeed a path in S , not merely in S . To show that S is open, consider any point z S . There exists some path : [0 , 1] S with (0) = z and (1) = z . Since S is open, there exists > 0 such that N ( z ) S . Any point w N ( z ) can be joined to z via a path in S , as follows. Define r = 1 +  w z  . Define : [0 , r ] S by ( t ) = ( t ) for all t [0 , 1], and ( t ) = z + ( t r ) w z  w z  for t [1 , r ]. That is, follows from z to z , then proceeds in a straight line from z to w . The line segment from z to w is contained in N ( z ) S , hence in S , so is a path in S joining z to w . Thus N ( z ) S , so S is open. This last paragraph is correct, but theres a more elegant solution. Weve already shown that is an equivalence relation. If N ( z ) S then w z for any w N ( z ), for the line segment from z to w lies entirely in N ( z ), hence in S . Clearly there is a corresponding path from z to w within S . (Remember, a path is a function, not a set such as a line segment; thats why I write corresponding.) Weve already proved that if z z and z w , then z w . Thus N ( z ) S ....
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 Spring '07
 Lim
 Math

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