Math 5345          Fall semester 2000

Selected solutions to exercises due Wednesday, October 11

Posted October 24, 2000

Page 42, #1    One way to approach this problem systematically is to divide it into cases, depending on how many subsets of the form  {point}  there are in the topology. Let's call the basic set  {a,b,c}.

• If there are  3  such open subsets, then we have the discrete topology.     [1 possibility]
• If there are  0  such open sets, then we have the indiscrete topology, and then {Ø,{a,b},{a,b,c}} and 2 similar possibilities by permutation.     [4 possibilities]
• If there is  1  such open set, let's see what happens when it is  {a}. We'll divide this into two subcases:
• {b,c} is also open. Then the topology is {Ø,{a},{b,c}}.    [By permutation, 3 possibilities]
• {b,c} is not open. We get the following possibilities:    [By permutation, a total of 12 possibilities]
{Ø,{a},{a,b,c}},   {Ø,{a},{a,b},{a,b,c}},   {Ø,{a},{a,c},{a,b,c}}, {Ø,{a},{a,b},{a,c},{a,b,c}}
• There are  2  such open sets. If they are  {a}  and  {b},  then the topology must include  Ø, {a}, {b}, {a,b}  and  {a,b,c}.  Then we have to decide which sets to include that contain  c.  Hewever,  {c}  itself must not be one of them. After this is done and we have counted the number of possibilities, we multiply by 3 to take permutations into account.

Page 42, #2    We're given that  A = (0,1){2}[3,4].
Ai = (0,1)(3,4)     (union of open intervals)
Ab = {0, 1, 2, 3, 4}     (endpoints of the intervals, and also the isolated point  2
Ae = (-,0)(1,2)(2,3)(4,)
A' = [0,1][3,4]     (Note that  2  is not an accumulation point.)

Page 43, #6
(i) By definition, the closure is the union of  Ai  and  Ab.  Also, it was proved that the closure is the smallest closed set containing  A.  So, if  A  is closed, then this smallest closed subset is  A  itself. It follows that is  A  is closed, then  A  must contain  Ab.

(ii) Recall that  Ai  is always a subset of  A.  So, if  Ab  is also a subset of  A,  then  A  contains its closure. Since the closure always contains  A,  we conclude that  A  is equal to its closure under the given hypothesis. Since the closure is always a closed set, we conclude that  A  is closed.

Page 43, #8    We're given that  A = {rational numbers}
Proof that Ae is empty.   In an earlier exercise, we proved that there is a rational number between any two real numbers. So, if  a is any irrational number, and  e  is a [small] positive real number, then there is a rational number between  ae  and  ae. It follows that  a cannot be a point of  Ae.

Proof that  Ai  is empty.   This time, let  a be a rational number. If  e  is a [small] positive number, we know that there is an integer  n such that  0 < 1/n < e. So,  a - 1/n  and a + 1/n  are rational numbers. One thing to say is that we can mimic what was done in that previous exercise to show that there is an irrational number between  a - 1/n  and a + 1/n,  so that  a cannot be a point of  Ai. Somewhat more explicitly, we can observe that  a + 1/(n)  is such an irrational number. So, it follows in the same way that  a cannot be a point of  Ai.

Page 47 #2    The key observation is that if  a1, …, an  are points of  X,  then:
(X - {a1})(X - {an}) = X - {a1, …, an}.
So, every nontrivial open set is the intersection of finitely many sets from the sub-basis.

Page 47 #3   .The hypothesis is that every set of the form  {a}  is closed in the given topology. So, it follows that  X - {a}  is open in the topology, and therefore (by the equality in problem #2 above) that every complement of a finite set is open in the given topology. We conclude that the topology of finite complements is coarser than [or equivalent to] any topology that satisfies this hypothesis.

Page 47, #4    The first step is to show that the set of open rectangles is a basis, i.e. that the intersection of finitely many open rectangles is either empty or an open rectangle. . Thus, suppose that the kth rectangle  Rk  is described by the inequalities  ak < x < bk  and  ck < y < dk  for  k = 1,…,n. Then the intersection of the rectangles  R1, …, Rn  is described by the inequalities  inequalities  a < x < b  and  c < y < d,  where,  a = max(a1,…,an);  b = min(b1,…,bn);  c = max(c1,…,cn);  and  d = min(d1,…,dn).

Next, we show that every open rectangle is open in the metric topology:   given a point  p  in an open rectangle  R, an open disk centered at  p with radius  r is contained in  R, if we take  r to be the minimum of the perpendicular distances from  p to the sides of  R.

Finally, we show that every open disk is open in the "rectangle topology". Namely, if  p is contained in the open disk  D, we can find a rectangle containing  p and contained in  D as follows:   (1) draw a radius of the disk, passing through  p,  (2) draw a rectangle with a corner at the point where this radius meets the boundary of  D, and the opposite corner located on the same radius, just large enough so that  p is in the interior of the rectangle.

Comments and questions to:  roberts@math.umn.edu

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