Math 5345          Fall semester 2000

Selected solutions to exercises due Monday, October 2

Posted: Oct 19, 2000

Page 20, #3    The Archimedean property of the ordering says that there is a positive integer  n with  ne > 1.   So,  0 < 1/n< e.

Page 20, #4    Step 1:   By the previous problem, let  n be an integer such that  0 < 1/nb - a.
Step 2:   Apply the result of the previous problem again, taking  m to be the smallest integer such that  m·(1/n) > a.  Doing a few calculations with inequalities, one checks that that  a < m/n < b.

Page 21, #9    The idea is to take two (2-dimensional linear) planes in  R4, taking some care to choose planes which actually intersect. The most immediate possibility is for one of them to have the equations  x1 = x2 = 0,  and for the other to have the equations  x3 = x4 = 0.  Thus, the intersection is exactly the origin. More generally, if you have a system of 4 linear equations in 4 unknowns, which has a unique solution, then you can take the first plane to be defined by two of the equations and the second plane to be defined by the other 2 equations.

Page 23, #4    The "cheapest possible" solution involves exhibiting 2 points  A and  B on  SRn-1(C)  such that the line segment joining them does not lie completely inside  SRn-1(C). We can do this by taking  A and  B to be at opposite ends of a diameter. (Formally, this means that  A = C + X and  B = C - X for some  X in  Rn  with  ||X|| = 1.) The midpoint of the line segment is  C, which is obviously not on the  (n- 1)-sphere.

Pp. 33-34 #2B)    We'll discuss the equivalence of  d  and  d2. For  e < 1,  an open ball of radius e in one metric is obviously the same thing as an open ball of radius e in the other metric. For  e > 1,  an open ball of radius e in the metric  d  obviously an open ball of radius 1  [and the same center … ]  in the metric  d  and hence also in the metric  d2. Finally, for  e > 1,  an open ball of radius e in the metric  d2  is the entire space, so that it contains an open ball  [with whatever center is under consideration]  in the metric  d.

Page 34, #3    Let  P = (a,b)  be a point of the region. We have to show that there is a ball of some positive radius  d  centered at  P and lying  (i) to the right of the y-axis,  (ii) above the x-axis,  (iii) below the hyperbola  xy= 1.  So, we would like to take:

d = min(d1, d2, d3),
where  d1 = a,     d2 = b,  and   d3  is the minimum distance from  P to the hyperbola.

Unfortunately, we don't yet have the means to prove that this distance has a nonzero minimum (or positive greatest lower bound) without using some stuff that we haven't proved yet.   This situation will change when we study compactness in the next chapter, but for now we have to do it the hard way. It's easier to find an open rectangle containing  P that misses the hyperbola than it is to directly find an open disk centered at  P with this property. Once we have found the rectangle, then we can easily find an open disk centered at  P with the desired property.

To find our rectangle explicitly, we'll set  r = ab. Then  r< 1, since  P is below the hyperbola. So, we look at points  (a+dx,b+dy).  We can tolerate positive  dx  and  dy,  provided that  a+dx  and  b+dy  aren't too large. For instance, we can notice [perhaps after some trial and error] that:
a = r/b  and  b = r/a,
so that the following bounds should work:
a+
dx <  and  b+dy < .
Indeed, since  r< 1, we have   and  ,  so that our rectangle actually contains  P.  Moreover:
(a+
dx)(b+dy) < · = ab/r= 1.

There are a few more details, but these are the main points.

Page 34, #8B)    According to the instructions, we need to show that the inverse image of an open interval is open. The most interesting and important case is an open interval  (a,b),  where  0 < a < b.  In the case, the inverse image is the set of all points  (x,y)  where  a < x2 + y2 < b.  This is the (ring-shaped) region consisting of all points  P that are inside a circle of radius  b  [centered at the origin]  but outside a circle of radius  a. So, if  P is in this region, at a distance = r from the origin, then a circle of radius = min(r-a,b-r)  will lie entirely inside the region. This claim should follow easily from the triangle inequality ...