sol1 - Solutions to some problems from HW1 Math H264 1.3.3...

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

View Full Document Right Arrow Icon
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 0 f ( x,y ) = lim x 0 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 ) . Solution. Let y f ( ∂S ), that is, y = f ( x ) where x ∂S . Let ε > 0. Find δ > 0 such that d ( x,z ) < δ implies d ( y,f ( z )) < ε . Since x ∂S , there exist z 1 S
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern