advertisement

Answer to Homework 2 1. 1 A sin n n 1 (1) sin 1, sin 1 has a limit point 0, but 0 A. Moreover, does not interest any other point is A if is small enough. Thus A is neither open nor closed. (2) limit points: 0, 1 ; interior points (0,1). Neither open nor closed. (3) limit points: ; interior points: . Neither open nor closed. (4) Let y A x 2 | d x,0 1. Want to show y is an interior points of A. Let r 1 d y,0 . Consider Br y . For any z Br y , by triangular inequality we know that d z,0 d z, y d y,0 r d y,0 1 d y,0 d y,0 1 . That means z A . That is, any z Br y is in A. We thus find a nbd of y that is contained in A, and y is thus an interior point of A. A is thus an open set because all of its points are interior. (5) A x 2 | d x,0 1 is closed. We show this by proving that Ac is open. Let y Ac , then set r y,0 1 . r 0 by definition of Ac . Consider Br y and let z Br y . By triangular inequality we know d z,0 d y,0 d y, z d y,0 r =1. Thus z Ac . That is, we find a nbd of y which is interior in Ac , and y is thus an interior point. Ac is thus open and A is closed. (6) limit point = interior point = ; N is not open and is closed. (7) interior point = limit point = . is both closed and open.. 2. Let open. A x1 , x2 | p1 x1 p2 x2 m, p1 , p2 0 x1 , x2 0 . Show that Ac is