Math 500: Topology

Homework 6

Lawrence Tyler Rush

<me@tylerlogic.com>

January 12, 2013

http://coursework.tylerlogic.com/courses/math500/homework06

Problems

The quotient space, X^*, induced by the equivalence relation on X = ℝ²

2 2 2 2 (u,v) ~ (x,y) if u + v = x + y

is homeomorphic to ℝ₊. At a bird’s-eye view, the equivalence classes will be the radii of concentric circles centered at the origin of ℝ². 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 : ℝ² → ℝ₊ 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 ∈ ℝ₊ 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 ℝ². 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 ∈ ℝ₊ 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 ℝ². 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 B′ with 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 B′ there is a point of B which maps to it, i.e. B′ is 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 ∈ ℝ₊}. Combining these results, we appeal to Corollary 22.3 to give us that g induces a homeomorphism between X^* and ℝ₊.

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 d_A : X → ℝ₊, 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 d_A is continuous.

Proof. We must consider both forms of basis elements of ℝ₊, (u,v) and [0,v). If x ∈ d_A^-1(u,v), then for y ∈ B(x,δ) where δ = min(d_A(x) - u,v - d_A(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 d_A(y) is also in (u,v). Hence the entirety of B(x,δ) is contained within d_A^-1(u,v). On the other if x ∈ d_A^-1[0,v), then the previous argument will hold for d_A(x)≠0. So when d_A(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) ∈ d_A^-1([0,v)).

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

Now we can define f_AB : 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 f_AB is continuous. Therefore f_AB : X → [a,b] defined by (b - a)f_AB + a is a continuous function which satisfies the requirements specified in Urysohn’s Lemma.