solns2

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

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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 γ i : [0 , 1] → S such that γ i (0) = z and γ i (1) = z i . Out of these we can manufacture a path from z 1 to z 2 . Define γ : [- 1 , 1] → S by γ ( t ) = γ 1 (- t ) for t ∈ [- 1 , 0], and γ ( t ) = γ 2 ( t ) for all t ∈ [0 , 1]. It’s easy to show that γ is continuous 1 , since γ 1 (0) = γ 2 (0), and the range of γ is contained the union of the ranges of γ 1 , γ 2 , hence in S . It’s tempting to simply declare victory now and assert that we’ve proved S to be path connected, but we haven’t 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 we’ve said is that our path γ lies within S . To see that its range is indeed in S , note that for any t > 0, the restriction of γ to the interval [0 , t ] defines a path in S from z to γ ( t ). Thus γ ( t ) ∈ S for all t ∈ [0 , 1], by definition of S . The same applies for t ∈ [- 1 , 0], of course. So γ 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 γ : [0 , 1] → S with γ (0) = z and γ (1) = z . Since S is open, there exists δ > 0 such that N δ ( z ) ⊂ S . Any point w ∈ N δ ( z ) can be joined to z via a path in S , as follows. Define r = 1 + | w- z | . Define Γ : [0 , r ] → S by Γ( t ) = γ ( t ) for all t ∈ [0 , 1], and Γ( t ) = z + ( t- r ) w- z | w- z | for t ∈ [1 , r ]. That is, Γ follows γ 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 δ ( z ) ⊂ S , hence in S , so Γ is a path in S joining z to w . Thus N δ ( z ) ⊂ S , so S is open. This last paragraph is correct, but there’s a more elegant solution. We’ve already shown that ∼ is an equivalence relation. If N δ ( z ) ⊂ S then w ∼ z for any w ∈ N δ ( z ), for the line segment from z to w lies entirely in N δ ( 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; that’s why I write “corresponding”.) We’ve already proved that if z ∼ z and z ∼ w , then z ∼ w . Thus N δ ( z ) ⊂ S ....
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This note was uploaded on 07/31/2009 for the course MATH 185 taught by Professor Lim during the Spring '07 term at Berkeley.

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

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