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