Math 500: Topology

Homework 3
Lawrence Tyler Rush
<me@tylerlogic.com>
January 12, 2013
http://coursework.tylerlogic.com/courses/math500/homework03

Problems

P-1

In each of the following subproblems, let X and Y be the orginal topological spaces on which f is defined and X or Y be the respective spaces which are alterted as per the subproblem description.

(a) Make X finer.

Making the domain finer won’t affect the continuity. Let U be an open set of Y . Then f-1(U) is open in X. But since X is finer than X, then TX ⊂ TX and therefore f-1(U) is also open in X.

(b) Make X courser.

Making the domain courser can, but will not necessarily, result in f not being continuous. As an example of a function “losing its continuity”: if X and Y are both the discrete topologies on ℝ, and f is the identity map, then making X courser by changing it to the indiscrete topology will make f-1(U) non-open if U is any proper, nontrivial subset of Y . On the other hand, f can retain its continuity, as exampled by the following scenario. Let X be the discrete topology on ℝ and Y be the indiscrete, with f again being the identity map. In this case, no matter how course X is made, f will always be continuous.

(c) Make Y finer.

Again making the topology of Y finer can, but will not necessarily, cause f to lose its continuity. An example when it does is if f is the identity map an X and Y are the same topological spaces, then adding any set to the topology on Y (and any other sets necessary to maintain it topological status) will cause f to no longer be continuous. An example of where f does not lose continuity is if X and Y are the same sets, X has the discrete topology, Y has any other except for the discrete, and f is the idenity map. Then in this case X has the “finest” topology for the set X = Y , so no matter what sets are added to the topology on Y to make it finer, no set added will be added that isn’t already in the discrete topology.

(d) Make X courser.

Making Y courser will not affect the continuity of f. This is so since TY ⊂ TY and every UTY has that f-1(U) is open in X, so any set of TY will have the same.

P-2

PIC (a) cup [1] PIC (b) saucer [?] PIC (c) glass [3] PIC (d) spoon [4]
PIC (e) fork [5] PIC (f) plate [6] PIC (g) coin [7] PIC (h) nail [8]
PIC (i) bolt [9] PIC (j) nut [10] PIC (k) wedding ring [11] PIC (l) flower pot [12]
PIC (m) key [13]

Figure 1: Images of items to partition into homeomorphic equivalency classes.

Using the objects in the images of Figure 1 we have the following homeomorphism classes.
 
saucer ≡ glass ≡ spoon ≡ fork ≡ plate ≡ coin ≡ nail ≡ bolt
 
cup ≡ nut ≡ wedding ring ≡ flower pot ≡ key

P-3

Here we can use polar coordinates to convert between the disk and the square. Basically, a point (r,θ) in the square will be the point in the disk of radius r away from the origin, and at an angle θ from the positive x-axis. Given this we define our map f : D2 → I2 as follows1
∘ ------------- f(x,y) = ( x2 + y2,arctan(y,x))

begetting an inverse function of

-1 f (r,θ) = (rcosθ,rsin θ)

Since each of the composite functions which make up f are individually continuous for x + y ≤ 1 then the indivial components of f are each continuous by Munkres Theorem 18.2 (c) which in turn gives us, by Munkres Theorem 18.4, that f itself is continuous. An identical argument holds for f-1. Because f-1 is continuous, then f is open. So because f is an invertible, open, continuous map, than it is a homeomorphism, and thus D2 and I2 are homeomorphic.

P-4 Munkres §18 exercise 1

Let f : ℝ → ℝ be continuous according to the open set definition. Let x ∈ ℝ and ϵ > 0, then f-1((f(x) -ϵ,f(x) + ϵ)) is open. This means that there exists some interval contained inside it which contains x, i.e. there exists a δ > 0 such that (x-δ,x + δ) ⊂ f-1((f(x) -ϵ,f(x) + ϵ)). Thus we have that any y in (x-δ,x + δ) will also be in f-1((f(x) - ϵ,f(x) + ϵ)), and hence f(y) ∈ (f(x) - ϵ,f(x) + ϵ). Thus f is continuous according to the ϵ - δ definition.

Conversely assume that the ϵ-δ definition of continuity holds for f. Let V be open in ℝ, then for each x ∈ f-1(V ) there is an ϵ such that (f(x) -ϵ,f(x) + ϵ) ⊂ V . From the ϵ-δ property of f we get that f((x-δ,x + δ)) ⊂ (f(x) -ϵ,f(x) + ϵ) ⊂ V , which implies that (x - δ,x + δ) ⊂ f-1(V ), i.e. f-1(V ) is open. Thus f is continuous according to the set definition.

P-5 Prove Munkres’ §18 Theorem 1

To prove the equivalency of this theorem, we will proceed by proving
(a)
(1)=⇒(3)
(b)
(3)=⇒(2)
(c)
(2)=⇒(1)
(d)
(1)=⇒(4)
(e)
(4)=⇒(1)

and in each case f : X → Y will be a function with X and Y topological spaces.

(a) (1)=⇒(3)

Assume that f is a continuous function. Let B ∈ Y be closed. Then Y \ B is open. Therefore f-1(Y \ B) is as well, but this is equal to X \ f-1(B), and so f-1(B) must be closed.

(b) (3)=⇒(2)

Let f : X → Y be a function such that for all closed sets B in Y , f-1(B) is closed. For A ⊂ X, A ⊂ f-1(f(A)) is always true and since sets are subsets of their own closure, then A ⊂ f-1(f(A)). Since f(A) is a closed set, then by assumption f-1(f(A)) is closed, but because it contains A, then it contains A since the closure of A is the union of closed supersets of A. So we have A ⊂ f-1(f(A)), implying f(A) ⊂f(A).

(c) (2)=⇒(1)

Assume that for all A ⊂ X, f(A) = f(A). Let U be an open set of Y . Let x ∈X \ f-1(U), which implies that f(x) ∈ f(----------) X \ f-1(U). Since Y \ U is closed we have
( ---------) ------------- ----------------- ------------- ----- f X \f-1(U) ⊂ f(X \f- 1(U )) = f(f-1(Y)\ f-1(U)) = f(f-1(Y \U )) ⊂ Y \U = Y \U

and from it we get x ∈ f-1(Y \U), but f-1(Y \U) = X \f-1(U) and so x ∈ X \f-1(U). Thus X \ f-1(U) ⊂ X \f-1(U), and therefore, since a set is a subset of its own closure, X \ f-1(U) = X \f-1(U), so X \f-1(U) is closed. By this f-1(U) is open, which yields that f is continuous.

(d) (1)=⇒(4)

Assume that f : X → Y is a continuous function. Let V ⊂ Y be a neighborhood of f(x) for some x ∈ X. Then x ∈ f-1(V ) and f-1(V ) is open. Since f(f-1(V )) ⊂ V is always true, then f-1(V ) is a neighborhood U of x with f(U) ⊂ V .

(e) (4)=⇒(1)

Assume that for all neighborhoods V of f(x), there exists a neighborhood U of x such that f(U) ⊂ V . Let V be an open set of Y . For each x ∈ f-1(V ) let Ux denote a neighborhood of x such that f(Ux) ⊂ V , i.e. Ux ⊂ f-1(V ). Therefore ∪x∈f-1(V )Ux ⊂ f-1(V ), but since each Ux contains x then f-1(V ) ⊂∪x∈f-1(V )Ux ⊂ f-1(V ). Therefore f-1(V ) equals ∪x∈f-1(V )Ux which, as a union of open sets, is open. Hence f is continuous.

P-6 Munkres §19 exercise 7

By ℝn denote the set
ℝ× ℝ × ⋅⋅⋅×(ℝ \{0})× {0}× {0}⋅⋅⋅ ◟-------◝◜--------◞ n terms

Note that with this notation ℝ0 is the product containing only singletons of zero. So then, we can represent ℝ∞ by

⋃ ℝ ∞ = ℝn n∈ℕ0

So in light of Munkres Theorem 19.5,

--- -- -- (------) --- --- ℝn = ℝ × ℝ × ⋅⋅⋅× ℝ \{0} × {0}× {0}⋅⋅⋅ = ℝ◟-×-ℝ×◝⋅◜⋅⋅×-ℝ◞ ×{0}× {0}⋅⋅⋅ n times

for both the box and product topologies. This gives us that

--- ⋃ --- ℝ ∞ = ℝn n∈ℕ0

which simply implies that the closure of ℝ∞ is ℝω for both the box and product topologies.

References

[1]   Cup image Figure 1a http://img1.123freevectors.com/wp-content/uploads/objects_big/067_objects_coffee-cup-free-vector.jpg

[2]   saucer image figure 1b: http://www.bryanchina.com/Mugs/BWE-066%20Cappuccino.Espresso%20Cappuccino%20Saucer%20White.JPG

[3]   glass image figure 1c: http://party.rainbow-rental.com/dinnerware/dinnerware_images/highball.jpg

[4]   spoon image figure 1d: http://iblogwhatihear.com/wp-content/uploads/2010/01/spoon.jpg

[5]   fork image figure 1e: http://www.ccesonline.com/images/fork260.jpg

[6]   plate image figure 1f: http://9pin.in/images/designer-photo-plate-room-tea.jpg

[7]   coin image figure 1g: http://www.marshu.com/articles/images-website/articles/presidents-on-coins/quarter-coin-head.jpg

[8]   nail image figure 1h: http://image.tradevv.com/2010/03/26/zjlongtong1_1080859_600/stainless-steel-nail.jpg

[9]   bolt image figure 1i: http://us.123rf.com/400wm/400/400/moori/moori0803/moori080300178/2744907-used-metal-bolt-on-a-white-background.jpg

[10]   nut image figure 1j: http://www.portlandbolt.com/image/products/full/heavy_hex_nut1.jpg

[11]   wedding ring image figure 1k: http://images.pictureshunt.com/pics/w/wedding_ring-2165.jpg

[12]   flower pot image figure 1l: http://cchs.usd224.com/Classes09/Flowersforless/FlowerPot.jpg

[13]   key image figure 1m: http://www.feelnumb.com/wp-content/uploads/2009/03/keyHorizontal.jpg