Topology Preliminary Exam August 2014

Here is the exam.

1] Let $X$ be a topological space and let $f,g : X \to \Bbb{R}$ be two continuous functions. Define $h : X \to \Bbb{R}$ by $h(x) = \min \{f(x),g(x)\}$ . Is $h$ continuous? Prove it or construct a counterexample.

Proof: Perhaps the simplest way to prove this is to observe that $\min \{f,g\} = \frac{f+g - |f-g|}{2}$ . Such an expression reveals that $\min \{f,g\}$ is a continuous function, because $f+g$ , $f-g$ , and $| \cdot |$ are continuous and the expression for $\min \{f,g\}$ given above involves sums, differences, scalar multiples, and compositions of continuous functions, all of which result in continuous functions.

To see that $f + g : X \to \Bbb{R}$ is continuous, there are at least two ways. First, recall that addition $+ : \Bbb{R} \times \Bbb{R}$ , and so $f + g$ is merely addition composed with $x \mapsto (f(x),g(x))$ Alternatively, one can do it “by hand.” Let $x \in X$ be arbitrary, and let $\epsilon > 0$ . Since $f,g$ are continuous functions, there are open sets $U_f$ and $U_g$ such that $y \in U_f$ ( $y \in U_g)$ implies $|f(x)-f(y)| < \frac{\epsilon}{2}$ ( $|g(x)-g(y)| < \frac{\epsilon}{2}$ ). Hence, if $y$ is in the open set $U_f \cap U_g$ , then

$|(f(x)+g(x)) - (f(y)+g(y))| \le |f(x)-f(y)| + |g(x)-g(y)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

(Note: for the second proof, I used the “continuity at a point” formulation of continuity.)

That $a \mapsto af(x)$ is a continuous map for each $a \in \Bbb{R}$ from $X$ to $\Bbb{R}$ similarly can be shown in two different ways. From this we get that $-g$ is continuous and therefore by the above $f + (-g) = f-g$ is continuous. $\blacksquare$

Remark: This easily generalizes to any finite collection of continuous functions However, it does not generalize to infinite collections of continuous functions, where the minimum is replaced by the infinimum and appropriate boundedness assumptions are made. This I erroneously believed for quite some time, but the following example was just the remedy I needed. Let $f_n : [0,1] \to \Bbb{R}$ be given by $f_n(x) = x^n$, which is certainly continuous for each $n \in \Bbb{N}$ , and define $h(x) := \inf_{n \in \Bbb{N} } f_n(x) = \inf_{n \in \Bbb{N}} x^n$ . If $x \in [0,1)$ , then $0 \le x^n < 1$ for every $n \in \Bbb{N}$ , and since the sequence $\{x_n\}$ converges to $0$ , it follows that $\inf_{n \in \Bbb{N} } x^n = 0$ or h(x)=0$. And because $h(1) = 1$ , it is clear that

$h(x) = \begin{cases} 0, & x \in [0,1) \\ 1, & x = 1 \\ \end{cases}$

is without a modicum of doubt a discontinuous function.My reason for thinking that this was true was influenced by knowing that $x \mapsto d(x,A) := \inf \{d(x,a) \mid a \in A\}$ is continuous, where $A$ is a subset in some metric space $(X,d)$ , which is an infimum of continuous functions. Interestingly, when the indexing set is a compact Topological space, the infimum will be a continuous function (see this blog post by Willie Wong).

2] Consider the space $\{0,1\}$ with the discrete topology, and let $X$ be the product of countably many copies of $\{0,1\}$ . Let $A$ be the subset of $X$ consisting of points with an infinite number of $0$ ‘ and an infinite number of $1$ ‘s. What is the closure of $A$ in $X$ . Justify your answer,

Proof: We will argue that $A$ is dense in $X$ (i.e., $\overline{A} = X$ ). Let $\prod_{n=1}^\infty U_n$ be some non-empty basis element, and let $F \subseteq \Bbb{N}$ be the finite set (possibly empty) for which $U_n = \{0,1\}$ if $n \notin F$ . Now consider the following point $(x_n)_{n=1}^\infty$ in $X$ . If $n \in F$ , choose $x_n$ to be any point in $U_n$ . As for the indices in $\Bbb{N} \setminus F$ , since $\Bbb{N} \setminus F$ is infinite, then it must contain an infinite number of even and odd integers, for if either was finite, then, by the pigeonhole principle, the rest of them would have to be in $F$ (because $\Bbb{N} = F \cup (\Bbb{N} \setminus F)$ ). But this cannot be, as $F$ is finite, so $\Bbb{N} \setminus F$ must contain an infinite number of even and odd integers. Thus, if $n \in \Bbb{N} \setminus F$ is even, $x_n=0$ , and if odd, $x_n = 1$ . This will insure that $(x_n)_{n=1}^\infty$ is a point in $A \cap \prod_{n=1}^\infty U_n$ , thereby proving $A$ dense in $X$ . $\blacksquare$

3] Let $X$ and $Y$ be topological spaces, and let $p_X : X \times Y \to X$ be the projection onto the first factor

Prove that if $Y$ is compact, then $p_X$ is a closed map.
Give an example where $p_X$ is not closed.

Proof: Let $C$ be some closed set in $X \times Y$ , and let $x \in X p_X(C)$ . Then $(x,y) \notin C$ for every $y \in Y$ , and since $C$ is closed, there is for each $y \in Y$ some basis element $U_y \times V_y$ about the point $(x,y)$ such that $U_y \times V_u \cap C = \emptyset$ . Since $\{V_y\}_{y \in Y}$ is a cover of the compact space $Y$ , there a finite number of points $y_1,...,y_n \in Y$ such that $Y = V_{y_1} \cup ... \cup V_{y_n}$ . Let $U := U_{y_1} \cap ... \cap U_{y_n}$ . We claim that $x \in U \subseteq X \setminus p_X(C)$ . Certainly $x \in U$ , since $x \in U_y$ for every $y \in Y$ . By way of contradiction, suppose $x_0 \in U$ but $x_0 \in p_X(C)$ . Then there is some $y_0 \in Y$ such that $(x_0,y_0) \in C$ . But $Y = V_{y_1} \cup ... \cup V_{y_n}$ , so $y_0 \in V_{y_i}$ for some $i \in \{1,...,n\}$ . But this means that $(x_0,y_0) \in (U_{y_i} \times V_{y_i}) \cap C$ , which is a contradiction. Hence $x_0 \in X \setminus p_X(C)$ and therefore $x \in U \subseteq X \setminus p_X(C)$ . This shows that $X \setminus p_X(C)$ is open which entails $p_X(C)$ is closed.

For the second part of the problem, consider $C = \{(x,\frac{1}{x}) \mid x \in \Bbb{R} \setminus \{0\} \}$ . This set is closed in $\Bbb{R} \times \Bbb{R}$ , because it is by definition the graph of the continuous function $x \mapsto \frac{1}{x}$ and $\Bbb{R} \setminus \{0\}$ is Hausdorff. But $\pi_1(C) = (-\infty, 0) \cup (0,\infty)$ , which is not closed. $\blacksquare$

Remark: For part two of the above problem, we used the following theorem (See problem one on the June 2013 prelim for a proof):

Let $X$ be some topological space, $Y$ some Hausdorff space, and $f : X \to Y$ some continuous function. Then the graph of $f$ defined as $G_f := \{(x,f(x)) \mid x \in X \}$ is closed in $X \times Y$ .

For fun, find a counterexample to the proposition obtained by dropping the Hausdorff condition on $Y$ ; also, find an example of a continuous function whose graph is closed, even though $Y$ is not Hausdorff.

5] Let $S^1$ be the unit circle in $\Bbb{R}^2$ , and let $X$ be the space obtained from $S^1$ by identifying antipodal points. I.e., $X = S^1 / \sim$ where $(x,y) \sim (-x,-y)$ for all $(x,y) \in S^1$ . Show that $S^1$ is homeomorphic to $S^1$ .

Proof: In this problem, we will view $S^1$ as the subset $\{z \in \Bbb{C} \mid |z| = 1 \}$ of the complex plane. Let $p : S^1 \to S^1 / \sim$ denote the canonical projection map generating the quotient topology on $S^1 / \sim$ , and consider the map $g : S^1 \to S^1$ defined by $g(z) =z^2$ . For each $[z] \in S^1 / \sim$ , we have $p^{-1}(\{[z]\}) = \{-z,z\}$ and $g(z) = z^2 = (-z)^2 = g(-z)$ , proving that $g$ is constant on the fibers of $p$ . By the universal property of quotient maps, there is a unique map $f : S^1 / \sim \to S^1$ such that $f \circ p = g$ , and this map must be continuous because $g$ is continuous. Moreover, it is injective since $f([z]) = f([w])$ implies $z^2 = w^2$ and therefore either $z = -w$ or $z =w$ , which means $[z] = [w]$ ; and it is surjective since $f(e^{i \frac{\theta}{2}}) = (e^{i \frac{\theta}{2}})^2 = e^{i \theta}$ for every $e^{i \theta} \in S^1$ . This means that $f$ is a continuous bijection from a compact space to a Hausdorff which entails that it is a homeomorphism. $\blacksquare$ .

6] Let $q : X \to Y$ be a quotient map. Assume that $Y$ is connected and that, for all $y \in Y$ , $q^{-1}{\{y\})$ is connected. Prove that $X$ is connected.

Proof: See problem 3 on the June 2016 prelim.

7] Let $p : E \to B$ be a covering map. Assume that $B$ is Hausdorff. Prove that $E$ is Hausdorff.

Proof: Let $x,y \in E$ be distinct points. If $p(x) \neq p(y)$ , then, since $B$ is Hausdorff, there are disjoint, open sets $V,W$ about $p(x)$ and $p(y)$ , respectively; and because

$\emptyset = p^{-1}(\emptyset) = p^{-1}(V \cap W) = p^{-1}(V) \cap p^{-1}(W),$

and since $p^{-1}(V)$ and $p^{-1}(W)$ open sets containing $x$ and $y$ , respectively, the conclusion that $E$ is Hausdorff follows. Now suppose that $p(x) = p(y)$ . Because $p$ is a covering map, there is an open neighborhood $U$ of $p(x)$ evenly covered by $p$ . Since $x,y \in p^{-1}(U)$ , there are open sets $V_x$ and $V_y$ containing $x$ and $y$ , respectively. If $V_x = V_y$ , then this fact will contradict with the fact that $p$ carries $V_x$ homeomorphically onto $U$ , since a homeomorphism is, in particular, an injection and we assumed that $\neq y$ with $p(x) = p(y)$ . Since $V_x$ and $V_y$ are not equal, they must be disjoint, which proves that $E$ is Hausdorff. $\blacksquare$

Remark: Recall that $U$ being evenly covered by $p$ means there is a pairwise disjoint collection of open sets $\{V_i\}_{i \in I}$ in $E$ such that $p^{-1}(U) = \bigcup_{i \in I} V_i$ and $p \big|_{V_i} : V_i \to U$ is a homeomorphism (i.e., $p$ carries $V_i$ homeomorphically onto $U$ ) for each $i \in I$ .

Compendium of Solutions

Topology Preliminary Exam August 2014

2 thoughts on “Topology Preliminary Exam August 2014”

Leave a comment Cancel reply

Share this:

Related

2 thoughts on “Topology Preliminary Exam August 2014”

Leave a comment Cancel reply