Topology HW answers (many answers adapted from latexed HW of Colin B. McKinney)

p. 111:

2.) Suppose that f : X ↦ Y is continuous. If x is a limit point of the subset A of X, is it necessarily true that f(x) is a limit point of f(A)?

It is NOT necessarily true that f(x) is a limit point of f(A) if x is a limit point of the subset A of X. The constant function f : R → R, f(x) = 0 is continuous. 5 is a limit point of R, but f(5) = 0 is not a limit point of f(R) = {0} since {0} has no limit points.

9.) Let A_{α} be a finite collection of subsets of X; let X = ∪_{α}A_{α}. Let f : X ↦ Y; suppose that f|A_{α} is continuous for each α.

Since each f|A_{α} is continuous, if we choose any set V closed in Y, (f|A_{α})^{−1}(V) = f^{−1}(V) ∩ A_{α} is closed in A_{α}. Since A_{α} in closed in X, f^{−1}(V) ∩ A_{α} is closed in X. Hence ⋃_{α}(f^{−1}(V) ∩ A_{α}) = f^{−1}(V) ∩ (⋃_{α} A_{α}) = f^{−1}(V) ∩ X = f^{−1}(V) is closed since finite unions of closed sets are closed. Hence f is continuous.

p. 118

3.) Suppose X_{α} is Hausdorff for all α. Show ΠX_{α} is Hausdorff

Take (x_{α})_{α∈J} and (y_{α})_{α∈J} ∈ ΠX_{α} such that (x_{α})_{α∈J} ≠ (y_{α})_{α∈J}. Then there exists α_{0} such that x_{α_{0}} ≠ y_{α_{0}}. Since X_{α_{0}} is Hausdorff there exists U, V open in X_{α_{0}} such that U ∩ V = ∅ and x_{α_{0}} ∈ U, y_{α_{0}} ∈ V. Then (x_{α})_{α∈J} ∈ ΠU_{α} and (y_{α})_{α∈J} ∈ ΠV_{α} are disjoint

open sets in the box and product topology where U_{α} = \{ \begin{matrix}U & α = α_{0} \\ X & α ≠ α_{0} \end{matrix} .

5: Theorem: Let f : A ↦ Π_{a∈J}X_{α} be given by the equation f(a) = (f_{α}(a))_{α∈J}, where f_{α} : A ↦ X_{α} for each α. Let ΠX_{α} have the product topology. Then the function f is continuous if and only if each function f_{α} is continuous. One of the implications stated in the preceeding theorem holds for the box topology. Which one?

The following implication holds: if f is continuous, then each f_{α} is continuous. This is true since each f_{α} is a function that restricts the range of f. The reverse implication, however, is not necessarily true, as shown in example 2, pg. 117.

Prove Theorem 20.4: The uniform topology on R^{J} is finer than the product topology and coarser than the box topology; these three topologies are different if J is infinite.

8.1A Show the uniform topology is finer than the product topology.

Take a basis element in the product topology: ΠU_{α}, where each U_{α} is open in R. Show ΠU_{α} is open in the uniform topology. Take a point x in R^{J}; represent x as x = (x_{α}) for α ∈ J. By definition of the product topology, there only finitely many of the U_{α} ≠ R, so denote let α_{1}, α_{2}, ..., α_{n} be the values of α for which U_{α} ≠ R. Next, for each i, choose ϵ_{i} > 0 so that B\overline{d}(x_{α_{i}}, ϵ_{i}) ⊂ U_{α_{i}}, which is possible since by definition each U_{α_{i}} is open in R.

Claim: if we let ϵ = min\{ϵ_{1}, ..., ϵ_{n}\}, x ∈ B\overline{ρ}(x, ϵ) ⊂ ΠU_{α}.

If a point z ∈ B\overline{ρ}(x, ϵ) then \overline{d}(x_{α}, z_{α}) ≤ \overline{ρ}(x, z) < ϵ ≤ ϵ_{i} for all i and α. Hence z_{α_{i}} ∈ B\overline{ρ}(x_{α_{i}}, ϵ_{i}) ⊂ U_{α_{i}} for all α_{i} and z_{α} ∈ R = U_{α} for all α ≠ α_{1}, ..., α_{n}, so z ∈ ΠU_{α}. Hence the uniform topology is finer than the product topology. 8.1B For J infinite, show that the uniform topology is different than the product topology. B\overline{ρ}(0, 1) is not open in the product topology when J is infinite since B\overline{ρ}(0, 1) ⊂ Π_{α∈J}(−1, 1). Thus, the uniform topology is different than the product topology when J is infinite. Alternatively, you can generalize problem 4 on p. 127. 8.1C Show that the box topology is finer than the uniform topology. Note the book’s proof is incorrect. Begin by letting B be the ϵ_{1}-ball around y in the uniform topology. Show B is open in the box topology. Take x ∈ B. Then there exists a B\overline{ρ}(x, ϵ) ⊂ B\overline{ρ}(y, ϵ_{1}) Then we could define a neighbourhood of x in the box topology as U = ∏(x_{α} − \frac{1}{2}ϵ, x_{α} + \frac{1}{2}ϵ). Claim U is contained in B. For any element z of U, \overline{d}(x_{α}, z_{α}) < \frac{1}{2}ϵ for all α. This implies that \overline{ρ}(x, z) ≤ \frac{1}{2}ϵ < ϵ, implying that the box topology is indeed finer than the uniform topology. 8.1D Show that for J infinite, the uniform topology and the box topology are different. Take I ⊂ J where I is a countable subset of J. Let x_{n_{α}} = \{ \begin{matrix}\frac{1}{n} & α ∈ I \\ 0 & α ∉ I \end{matrix} . If x_{n} = Π_{α∈J}x_{n_{α}}, then \overline{ρ}(x_{n}, 0) = sup\{\frac{1}{n}, 0\} = \frac{1}{n}. Hence x_{n} → 0 in the uniform topology, but x_{n} does not converge to 0 in the box topology. Thus, the uniform topology and the box topology are different when J is infinite. 8.2: Using the metric D defined in Theorem 20.5, describe B_{D}(0, \frac{1}{3}). First off we note that D(0, x) < \frac{1}{3} for x to be an element of B_{D}. But D(0, x) = sup\{\frac{min\{d(0,x_{i}),1\}}{i}\} = sup\{\frac{min\{|x_{i}|,1\}}{i}\} < \frac{1}{3} for each i. For i = 1, the first coordinate, x_{1} must satisfy |x_{1}| < \frac{1}{3}. For i = 2, the second coordinate, x_{2} must satisfy |x_{2}| < \frac{2}{3}. For i = 3, the third coordinate, x_{3} must satisfy |x_{3}| < 1. For 4 ≤ i, the ith coordinate can be any real number. Hence B_{D}(0, \frac{1}{3}) = (−\frac{1}{3}, \frac{1}{3}) × (−\frac{2}{3}, \frac{2}{3}) × (−1, 1) × R × R × ... p. 145: Repeat 4A for the equivalence relation x_{0} × y_{0} x_{1} × y_{1} if x_{0}^{2} + y_{0}^{2} = x_{1}^{2} + y_{1}^{2}. f : X∗ → [0, ∞), f((x, y)) = \sqrt{(x^{2} + y^{2})} is a homeomorphism where [0, ∞) is a subspace of R. Thus X∗ is equivalent to the subspace [0, ∞) of R. pg 152 1.) Let T′ and T be two topologies on X, where T′ is finer than T . What does the connectedness of X in one imply about its connectedness in the other?

R_{l} is finer than R which is finer than R with the indiscrete topology, R and R with the indiscrete topology are connected but R_{l} is not connected by problem 7.

Thus connectedness of X in T does not imply anything about the connectedness of X in T′

Suppose T is not connected, then X = U ∪ V where U, V ∈ T ⊂ T′. Thus, T′ is not connected. Hence connectedness of X in T′ implies connectedness of X in T .

7.) R_{l} is not connected since R_{l} = (∞, 0) ∪ [0, ∞) and (∞, 0) and [0, ∞) are both open in R_{l}.

p. 157

8A: If two spaces X and Y are path connected, is the space X × Y path connected?

Let a × b and c × d be distinct points in X × Y. Since X is path connected, there exists a path γ_{1} : [0, 1] → X from a to c. Similarly, since Y is path connected, there exists a path γ_{2}[0, 1] → Y from b to d, Hence γ = (γ_{1}, γ_{2}) : [0, 1] → X × Y is a path from a × b to c × d, so X × Y is indeed path connected.

8B, If A ⊂ X and A is path connected, is \overline{A} also path connected?

Not necessarily. Take X = R^{2} and A = S, where S is the topologist’s sine curve given in example 7, pg 157. S is path connected, but \overline{S} is not.

8CIf f : X → Y is continuous and X is path connected, is f(X) necessarily path connected?

Fix x, y in f(X). Take c ∈ f^{−1}(x) and d ∈ f^{−1}(y), the inverse images of x and y. Then since X is path connected, there exists a continuous function γ_{X} : [a, b] ↦ X, where [a, b] is a subset of R such that γ_{X}(a) = c and γ_{X}(b) = d. Since both f and γ_{X} are continuous, the composition of f with γ_{X} will also be a continuous function. Let this function be γ. Thus γ = f ∘ γ_{X} : [a, b] ↦ f(X), where γ(a) = f ∘ γ_{X}(a) = f(c) = x and γ(b) = f ∘ γ_{X}(b) = f(d) = y, so f(X) is path connected.

8D If {A_{α}} is a collection of path-connected subspaces of X, and if ∩A_{α} ≠ ∅, is ∪A_{α} necessarily path connected?

Since ∩A_{α} ≠ ∅, by the definition of intersection there exists a point y such that y ∈ A_{α} for all α. Let x, z be two points in ∪A_{α}. x therefore must be an element of A_{α} for some α by the definition of union; similarly for z. Let x ∈ A_{1} and z ∈ A_{2}. Since A_{1} is path connected, and x, y ∈ A_{1}, there exists a path γ_{1} : [0, 1] → A_{1} between x and y. Similarly, there exists a path

γ_{2} : [1, 2] → A_{2} between y and z in A_{2}. Hence γ[0, 2] → ∪A_{α}, γ(x) = \{ \begin{matrix}γ_{1}(x) & x ∈ [0, 1] \\ γ_{2}(x) & x ∈ [1, 2] \end{matrix}

is a path connecting x and z via y, a point in common.