Math 500: Topology

Homework 6
Lawrence Tyler Rush
<me@tylerlogic.com>

January 12, 2013
http://coursework.tylerlogic.com/courses/math500/homework06

Problems

1


The quotient space, X*, induced by the equivalence relation on X = 2
               2   2    2   2
(u,v) ~ (x,y) if u + v = x + y

is homeomorphic to +. At a bird’s-eye view, the equivalence classes will be the radii of concentric circles centered at the origin of 2. To convince ourselves of this homeomorphism, we will make use of on Munkres’ Corollary 22.3 by defing a satisfying g and proving that it is a quotient map.

Defining a g Let g : 2 + be defined by (x,y)↦→x2 + y2, that is, g maps a point to the square of its distance from the origin.

g is a quotient map The map g is surjective as any x + is mapped to by (0,√ --
  x). It is also continuous since it’s the composition of two continuous functions: the addition function and the component-wise squaring function, both on 2. In other words g is (+ ) (()2,()2) with the adddition function being continuous by Munkres’ Lemma 21.4 and the component-wise squaring function being continuous by Munkres’ Theorem 18.4. We these two properties, we are left only to prove U + is open whenever g-1(U) is open in order to know that g is a quotient map.

So assume that g-1(U) is an open set of 2. Let r2 be an element of U for some r 0. Then as g is surjective, there is a (x,y) g-1(U) which maps to r2, i.e. x2 + y2 = r2. Since g-1(U) is open then there is a ball B, say with radius d, centered at (x,y) which is contained in g-1(U). Therefore B= (r2 -d,r2 + d) is an open interval containing r2. Thus any element z2 of Bwith z 0 has

  2   2
|r - z | < d
(1.1)

Letting θ be the angle at which (x,y) is from the x-axis, we see that the point (z cosθ,z sinθ) is contained within B by the following

∘ --------------------------------     ∘ --------------------
  (rcosθ - z cos θ)2 +(rsinθ- z sinθ)2 =    (r- z)2(cos2θ+ sin2θ)
                                       ∘ ------2
                                    =    (r- z)
                                    =  |r- z|
                                    ≤  |r2 - z2|
                                    <  d
making use of Equation 1.1 for the last deduction. Thus since g(z cosθ,z sinθ) = (z cosθ)2 + (z sinθ)2 = z2(cos2θ + sin2θ) = z2, then we conclude that for every point in Bthere is a point of B which maps to it, i.e. Bis completely contained in U and thus the openess of U is implied.

With this and the facts that g is surjective and continuous, then g is a quotient map. Finally since g-1(r) is the set of all points of distance √r- from the orgin then X* = {g-1(r)|r +}. Combining these results, we appeal to Corollary 22.3 to give us that g induces a homeomorphism between X* and +.

2


Assume that Y is a closed subspace of a normal space X. Thus closed subsets A and B of Y are also closed in X. By the normalcy of X, we can therefore separate A and B by disjoint open sets of X, say U and V , respectively. So because A and B are contained in Y , then the open sets U Y and V Y of Y separate A and B and are disjoint since (U Y ) (V Y ) = Y (U V ) = Y ∩∅ = . Therefore Y is normal.

3 Prove Urysohn’s Lemma for metric space (X,d)


Define the function dA : X +, where A is some subset of X, by
x ↦-→  inf{d(x,a) | a ∈ A }.

This definition yields the following lemma.

Lemma 3.1 For a subset A of (X,d), the function dA is continuous.

Proof. We must consider both forms of basis elements of +, (u,v) and [0,v). If x dA-1(u,v), then for y B(x,δ) where δ = min(dA(x) - u,v - dA(x)) we have

dA (y) ≥ dA(x)- d(x,y) > dA(x)- δ ≥ u

and

dA (y) ≤ dA(x)+ d(x,y) < dA(x)+ δ ≤ v

which implies that dA(y) is also in (u,v). Hence the entirety of B(x,δ) is contained within dA-1(u,v). On the other if x dA-1[0,v), then the previous argument will hold for dA(x)0. So when dA(x) = 0 then any element of B(x,v) is within a distance of v from an element of A, namely x, so B(x,v) dA-1([0,v)).

Thus the preimage of dA for any basis element of + will be open since in any case we can find a ball contained within the preimage. Therefore dA is continuous. __

Now we can define fAB : X [0,1] by

         dA(x)
x ↦-→ d-(x)+-d-(x)-
      A      B

for two disjoint closed subsets of X, A and B. The denominator will not be zero since if that were true, an element of X would have to be contained in both A and B or at least be a limit point of them both; but this cannot be the case as the sets are both closed and disjoint. Thus this fact, combined with both Lemma 3.1 and Munkres’ Theorem 21.5 informs us that fAB is continuous. Therefore fAB : X [a,b] defined by (b - a)fAB + a is a continuous function which satisfies the requirements specified in Urysohn’s Lemma.