solns2 - Mathematics 185 Intro to Complex Analysis Fall...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

Page1 / 4

solns2 - Mathematics 185 Intro to Complex Analysis Fall...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online