※ 引述《jing7 (jing)》之銘言：
: 證明
: A＝{(x,y):x^2+y^2>1}
: is open in R2
: 完全不會
: 實在麻煩各位幫我解答 謝謝
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Given p=(x,y) in A. We can choose ε=√(x + y ) - 1 to form an open ball
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B(p,ε) = {(x',y') | (x'-x) + (y'-y) < ε }.
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Then we must show that x' + y' > 1 (ie. (x',y') in A).
Note that we can reparameterize (x',y') in B(p,ε):
x' = x + ε'cosθ, y' = y + ε'sinθ, 0≦θ≦2π, 0 ≦ε'<ε
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Then, x' + y' = (x + y ) + ε' + ε'(xsinθ+ycosθ)
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> (x + y ) + 0 - (√(x + y ) - 1)√(x + y )
= 1.
So, (x', y') in A and B(p,ε) is contained in A.
Then, p is an interior point. Futhermore, p is arbitrary.
Therefore, A is open.
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※ 編輯: yueayase (36.236.229.213), 03/20/2015 21:22:27