8/19/2019 190w16HW8soln

1/3

Math 190: Winter 2016Homework 8 Solutions

Due 5:00pm on Wednesday 3/9/2016

Problem 1:  Prove that the one-point compactification of  Rn is homeomorphic to S n.Solution:  We know that  Rn is a locally compact Hausdorff space (which is not com-pact) and that  S n is a compact Hausdorff space. By the uniqueness of the one-pointcompactification, it is enough to show that there is a subspace   X   of   S n such thatS n − X   is a single point and X   is homeomorphic to  Rn. But we know that

X  = S n − {N },

where   N   = (1, 0, . . . , 0) is such a subspace (where the homeomorphism with   Rn isafforded by stereographic projection).

Problem 2:  Prove that the one-point compactification of  S Ω  is homeomorphic to  S Ω.

Solution:   Both of the spaces  S Ω   and  S Ω  are order topologies, and hence Hausdorff.The sets S Ω and  S Ω are both well orders. In particular, they have the least upper boundproperty. If  a0  ∈ S Ω  is the smallest element, we have that  S Ω  = [a0, Ω] is compact. If x ∈  S Ω  and  x < y <  Ω, we have that  x  ∈  [a0, y) ⊂  [a0, y] with [a0, y] compact, so thatS Ω  is locally compact. Finally, we have that S Ω − S Ω  = {Ω}   is a single point. By theuniqueness of the one-point compactification of a locally compact Hausdorff space, itfollows that the one-point compactification of  S Ω  is homeomorphic to  S Ω.

Problem 3:   Let (X, d) be a metric space and assume that  X  has a countable densesubset (i.e., X  is separable). Prove that  X   is second-countable.

Solution:   Let D =  {x1, x2, . . . } be a countable dense subset of  X  and letB  :=  {Bd(xn, 1/m) :   n, m ∈  Z>0}.

Then B  is a countable collection of open sets in  X .We claim that B  is a basis for the topology of  X . To see this, let U  ⊂ X  be open and

let y  ∈  U . It is enough to show that there exists  B  ∈ B  such that  y  ∈  B  ⊂  U . Indeed,let 0  < 0   suchthat  xn  ∈  Bd(y,/4). Choose  m  ∈  Z>0  so that  /4  <  1/m < /2. Since   /4  <  1/m,we have that  y  ∈  Bd(xn, 1/m). On the other hand, if  z  ∈  Bd(xn, 1/m), we have thatd(y, z ) ≤  d(y, xn) + d(xn, z ) < /4 + 1/m <

  34

  < . It follows that

y ∈  Bd(xn, 1/m) ⊂  Bd(y, ) ⊂  U,

so that B  is a countable basis for the topology on  X  and  X   is second-countable.

Problem 4:   Suppose  X   is a second-countable space and  A  ⊂  X   is an uncountablesubset. Prove that uncountably many points in  A  are limit points of  A.

Solution:   Let   B   =   {B1, B2, . . . }   be a countable basis for the topology on   X . Wedefine a function

ϕ : (A − A) −→  Z>01

8/19/2019 190w16HW8soln

2/3

2

as follows. For any a  ∈  (A − A), there exists a basic open set  Bna such that A ∩ Bna ={a}. Choose one such na  and set  ϕ(a) := na. We claim that  ϕ  is an injection. Indeed,for   a1   =   a2   in   A −  A, we have that   A ∩  Bϕ(a)   =   {a} =   {a

}   =   A ∩  Bϕ(a). Theinjectivity of  ϕ  implies that A − A is countable. Since A  is uncountable, this forces  A

to be uncountable.Problem 5:  Which of the four countability axioms (first-countable, second-countable,Lindelöf, separable) does  S Ω  satisfy? What about  S Ω?

Solution:   We claim that  S Ω   is not separable. Indeed, if  A  ⊂  S Ω   is countable, thenthere is an upper bound b  for  A. Choosing b > b, we get that (b, ∞) is a neighborhoodof   b which does not meet   A. Thus,   b /∈   A. It follows that   S Ω   is also not second-countable.

S Ω   is not Lindelöf because  {[a0, x) :   x  ∈  S Ω}  (where  a0   is the smallest element of S Ω) is an uncountable open cover without a proper subcover.

S Ω   is  first-countable. Indeed, for any  x  ∈  S Ω, let  x be the immediate successor of 

x. Provided x  = a0, we have that {(z, x

) :   z < x} is a countable basis at x. If  x  =  a0,then {[x, x)}  is a singleton basis at  x.

We claim that  S Ω   is not first-countable. Indeed, there does not exist a countablebasis at Ω. If {B1, B2, . . . } were such a basis, for all n ≥  0 there would be xn ∈ S Ω suchthat (xn, Ω] ⊂  Bn. Let b be an upper bound in  S Ω of the countable set {x1, x2, . . . } andlet  b be the immediate successor of  b. Then (b, Ω] is a neighborhood of Ω containingnone of the Bn. This completes the proof that S Ω  is not first-countable. It follows thatS Ω  also fails to be second-countable.

S Ω is not separable. If  A ⊂  S Ω is countable, let b  ∈  S Ω be an upper bound for  A∩S Ω.If  b is the immediate successor of  b, we get that (b, Ω) is a neighborhood of  b whichdoes not meet  A. Therefore, b /∈ A.

S Ω   is  Lindelöf. To see this, let U  be an open cover of  S Ω. There exists  U  ∈ U   suchthat Ω ∈ U . Choose  x  ∈  S Ω   such that (x, Ω] ⊂ U . We have that [a0, x] is countable.Write [a0, x] = {y1, y2, . . . }. For every  n, there exists  U n  ∈ U   such that  yn ∈ U n. Now{U, U 1, U 2, . . . }  is a countable subcover of  U .

Problem 6:   Let X  be a regular space and let  x, y ∈  X  be distinct points. Prove thatthere exist neighborhoods U, V   of  x, y  such that  U  ∩ V   = ∅.

Solution:  Since X   is regular, X  is Hausdorff. Choose neighborhoods  U  of  x  and  W   of y such that U ∩W   = ∅. Now X −W  is a closed set with y /∈ X −W . By regularity, thereis a neighborhood V   of  y  and an open set V  containing X  − W   such that  V  ∩ V  = ∅.Now X  − V  is a closed set with V   ⊂ X  − V . In particular, we get that  V   ⊂ X  − V .

On the other hand, we have   U   ⊂   X  − W . But (X  − V ) ∩  (X  − W ) =   ∅, forcingU  ∩ V   = ∅.

Problem 7:  Prove that every order topology is regular.

Solution:   Let X  be an order topology, let  x  ∈  X , and let A  ⊂  X  be a closed set withx /∈ A.

8/19/2019 190w16HW8soln

3/3

3

Suppose x  is the smallest element of  X . If  x is the immediate successor of  x, then[x, x) and (x, ∞) are disjoint neighborhoods of   x   and   A, respectively. If   x   has noimmediate successor, choose  y > x   such that [x, y) ∩ A  =  ∅. Now choose y ∈  (x, y).We have that [x, y) and (y∞) are disjoint neighborhoods of  x and A, respectively.

If  x   is the largest element of  X , we reason as in the previous paragraph, with theorder <   reversed.If  x  is neither smallest nor largest, consider the closed subspaces  X 1 = (−∞, x] and

X 2   = [x, ∞) of   X   and let   Ai   =   X  ∩ X i   for   i   = 1, 2. Then  Ai   is closed in   X i   and,by the previous arguments, there exist neighborhoods  U i  of  Ai  and  V i  of  x  in  X i  suchthat  U i ∩ V i  = ∅. We get that  U  ∩ V   = ∅, where  U   =  U 1 ∪ U 2   and  V   =  V 1 ∪ V 2. Weclaim that  U   and  V   are open in  X . Indeed, we have that  V 1   is open in the open ray(−∞, x) and V 2   is open in the open ray (x, ∞). Similarly, U  contains an open intervalcontaining x and  U i − {x}  is open in  X   for  i  = 1, 2.