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Unformatted text preview: Solutions to some problems from HW1 Math H264 1.3.3. Let f ( x, y ) = x- 1 sin( xy ) , x negationslash = 0 . How would youde fine f (0 , y ) , y ∈ R , so as to make f a continuous function on R 2 ? Solution. For each y , if we fix it and find the limit of f at (0 , y ) with respect to x , we get lim x → f ( x, y ) = lim x → x- 1 sin( xy ) = y , so we have to define f (0 , y ) = y . Let’s check that so defined f is, indeed, continuous on R 2 . For any x negationslash = 0 and any y we have f ( x, y ) = x- 1 sin( xy ) = ( xy )- 1 sin( xy ) y = ϕ ( xy ) y, where ϕ is the continuous function ϕ ( t ) = braceleftbigg t- 1 sin t, t negationslash = 0 1 , t = 0 . For x = 0 and any y we also have f (0 , y ) = y = ϕ ( xy ) y . Since the function μ ( x, y ) = xy is continuous, ϕ ( xy ) is continuous, and so is f . By d ( x, y ) I will denote the distance between points x and y of a metric space X ; in the case X = R n , d ( x, y ) = | x − y | . 1.3.9. Let U and V be open sets in R n and let f : U −→ V be a homeomorphism, that is, a one-to-one mapping from U onto V such that both f and f- 1 are continuous. Let S ⊂ U be such that S ⊂ U and f ( S ) ⊂ V . Prove that f ( ∂S ) = ∂f ( S ) ....
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This note was uploaded on 04/30/2011 for the course MECHANICAL MSC taught by Professor Dr.rahula during the Spring '10 term at University of Moratuwa.
- Spring '10