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

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

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern