Week 8 answers to exercises

Week 8 Exercises — solutions

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Exercise 8.0. This is an unseen exercise on closure, boundary and dense sets. Consider the sets $A = {0, 1} \subset R$ and $B = R ∖ A = (- \infty, 0) \cup (0, 1) \cup (1, + \infty)$ as a subsets of four different topological spaces, given in the table below. Complete the table.

.
	The space $X$
	$(R,$ antidiscrete $)$	$(R,$ cofinite $)$	$(R,$ Euclidean $)$	$(R,$ discrete $)$

$\overset{―}{A}$ (closure in $X$ )	$R$	$A$	$A$	$A$
Is $A$ dense in $X$ ? (yes/no)	yes	no	no	no
$\overset{―}{B}$	$R$	$R$	$R$	$B$
Is $B$ dense in $X$ ? (yes/no)	yes	yes	yes	no
$\partial A$	$R$	$A$	$A$	$\emptyset$

Hint. You may wish to recall that $\overset{―}{A} =$ the smallest closed set in $X$ which contains $A$ = ${z \in X :$ all open neighbourhoods of $z$ meet $A}$ and that $\partial A = \overset{―}{A} \cap \overset{―}{(X ∖ A)} .$

Answer to E8.0. [These exercises without answers]

Explanation of the above entries in the table: let $X$ be $(R,$ antidiscrete topology $) .$ The only closed sets in $X$ are $\emptyset$ and $R .$ Of these, only $R$ contains $A .$ Hence $\overset{―}{A},$ which is the smalest closed set containing $A,$ is $R .$ In the same way $\overset{―}{B} = R .$ Since $R ∖ A = B,$ we have $\partial A = \overset{―}{A} \cap \overset{―}{B} = R \cap R = R .$

Now let $X$ be $(R,$ cofinite topology $) .$ Note that the closed sets in the cofinite topology are finite sets and $R .$ Since $A$ is finite, $A$ is closed and $\overset{―}{A} = A .$ Since $B$ is infinite, the smallest closed set which contains $B$ is $R,$ hence $\overset{―}{B} = R .$ We have $\partial A = A \cap R = A .$

Next, let $X$ be $(R,$ Euclidean topology $);$ this is a Hausdorff space, so singletons ${0}$ and ${1}$ are closed, and $A = {0} \cup {1}$ is closed as a finite union of closed sets. Hence $\overset{―}{A} = A .$ Every open neighbourhood of $0$ contains points from $B,$ because an open set cannot consists just of points of the finite set $A;$ hence $0 \in \overset{―}{B} .$ Similarly, $1 \in \overset{―}{B} .$ Hence $R = {0} \cup {1} \cup B \subseteq \overset{―}{B} .$ We thus have $\overset{―}{B} = R,$ and $\partial A = A \cap R = A .$

Finally, if $X$ is $(R,$ discrete topology $),$ then every subset of $X$ is closed and is equal to its own closure. So $\overset{―}{A} = A,$ $\overset{―}{B} = B$ and $\partial A = A \cap B = \emptyset .$

In each case, a set is dense in $R$ if its closure is $R .$

Exercise 8.1. (a) Use the following two results,
- • a connected component of a topological space is a connected set (Lemma 7.4),
- • if the space $X$ has a connected dense subset then $X$ is connected (Lemma 7.11),
to show that each connected component of a topological space is a closed set.

(b) Deduce from (a) that if a topological space $X$ has finitely many connected components, then each connected component is both closed and open in $X .$

(c) Give an example of a topological space where connected components are closed but not open.

Answer to E8.1. [These exercises without answers]

(a) Let $x \in X,$ and denote the connected component of $x$ by $C .$ By Lemma 7.4, $C$ is connected. By definition of “dense”, $C$ is dense in the subspace $\overset{―}{C}$ of $X .$ Therefore, by Lemma 7.11, $\overset{―}{C}$ is connected. Since $\overset{―}{C}$ contains $x,$ this means that $\overset{―}{C}$ must lies in the connected component of $x,$ that is, $\overset{―}{C} \subseteq C .$ On the other hand, $C \subseteq \overset{―}{C}$ for all sets $C,$ see Claim 7.7. So $C = \overset{―}{C},$ which is equivalent to $C$ being closed.

(b) Assume that $X$ is the union of finitely many connected components $C_{1}, \dots, C_{n} .$ In part (a), we proved that $C_{1}, \dots, C_{n}$ are closed, and now we will show that $C_{1}$ is open. Recall that $C_{1}, \dots, C_{n}$ are equivalence classes, hence they are disjoint, and

$C_{1} = X ∖ (C_{2} \cup \dots \cup C_{n}) .$

Finite unions of closed sets are closed, so $C_{2} \cup \dots \cup C_{n}$ is closed, and its complement $C_{1}$ is open, as claimed.

(c) The set $Q$ of rational numbers, viewed as the subspace of the Euclidean line $R,$ is totally disconnected, as shown in E7.1(3): every non-empty connected subset of $Q$ is a singleton. Since connected components of $Q$ are connected sets, it follows that connected components of $Q$ are singletons. But a singleton (a one-point set) is not open in $Q :$ we showed exactly that in E7.1(2).

Exercise 8.2. (a) Suppose that $X$ is a topological space, points $x, y \in X$ are joined by a path in $X,$ and points $y, z \in X$ are also joined by a path in $X .$ Show that $x, z$ are joined by a path in $X .$

(b) Furthermore, show that “ $x \sim y$ $⟺$ $x, y$ are joined by a path in $X$ ” is an equivalence relation on $X .$

Equivalence classes defined by the relation $\sim$ from (b) are called path-connected components of $X .$ In general, a path-connected component does not need to be open or closed in $X .$ Nevertheless:

(c) Show that if $X$ is an open subset of a Euclidean space $R^{n},$ then each path-connected component of $X$ is open. Deduce that an open connected subset of $R^{n}$ is path-connected.

Answer to E8.2. [These exercises without answers]

(a) Let $𝜙$ be a path joining $x$ and $y$ in $X .$ By definition, this means that $𝜙 : [0, 1] \to X$ is a continuous function, $𝜙 (0) = x$ and $𝜙 (1) = y .$ Similarly, a path joining $y$ and $z$ is a continuous function $𝜓 : [0, 1] \to X$ with $𝜓 (0) = y$ and $𝜓 (1) = z .$

We need to construct a continuous function $𝜒 : [0, 1] \to X$ such that $𝜒 (0) = x$ and $𝜒 (1) = z .$ Intuitively, we will construct a path which will first trace the path from $x$ to $y,$ then trace the path from $y$ to $z;$ this is called the concatenation of paths in the Figure, this will look like the continuous curve which is the union of the blue curve from $x$ to $y$ and the red curve from $y$ to $z .$

The functions $\hat{𝜙} : [0, \frac{1}{2}] \to X,$ $\hat{𝜙} (t) = 𝜙 (2 t),$ and $\hat{𝜓} : [\frac{1}{2}, 1] \to X,$ $\hat{𝜓} (t) = 𝜓 (2 t - 1),$ are continuous as compositions of continuous functions. Define $𝜒 : [0, 1] \to X$ by

$𝜒 (t) = {\begin{cases} \hat{𝜙} (t), & t \in [0, \frac{1}{2}], \\ \hat{𝜓} (t), & t \in [\frac{1}{2}, 1] . \end{cases}$

Note that $𝜒 (\frac{1}{2})$ is well-defined as $\hat{𝜙} (\frac{1}{2}) = \hat{𝜓} (\frac{1}{2}) = y .$ Moreover, $𝜒 (0) = 𝜙 (2 \times 0) = x$ and $𝜒 (1) = 𝜓 (2 \times 1 - 1) = z .$

We prove that $𝜒$ is continuous using the closed set criterion of continuity, Proposition 2.5. If $F \subseteq X$ is closed in $X,$ $𝜒^{- 1} (F) = {\hat{𝜙}}^{- 1} (F) \cup {\hat{𝜓}}^{- 1} (F) .$ Since $\hat{𝜙}$ is continuous, ${\hat{𝜙}}^{- 1} (F)$ is closed in $[0, \frac{1}{2}],$ hence compact. Similarly, ${\hat{𝜓}}^{- 1} (F)$ is compact. A union of two compact sets is compact by E5.2, so $𝜒^{- 1} (X)$ is compact, hence closed in the Hausdorff space $[0, 1] .$ We have verified the closed set criterion of continuity for $𝜒 .$ Thus, $x$ and $z$ are joined by the path $𝜒 .$

(b) Part (a) shows that the relation $\sim$ is transitive. We note that $\sim$ is reflexive: $x \sim x$ because the constant path, $𝜙 (t) = x$ for all $t \in [0, 1],$ joins $x$ and $x .$ Moreover, $\sim$ is symmetric: if $𝜙$ is a path joining $x$ and $y,$ then the path $\bar{𝜙} : [0, 1] \to X,$ $\bar{𝜙} (t) = 𝜙 (t - 1),$ joins $y$ and $x .$ We conclude that $\sim$ is an equivalence relation.

(c) Suppose that $X \subseteq R^{n}$ is an open set. If $x \in X,$ take $r > 0$ such that $B_{r} (x) \subseteq X .$ Note that in an open ball $B_{r} (x)$ in a Euclidean space, every point is joined to $x$ by a path — in fact, by a straight line segment. Hence $B_{r} (x)$ lies inside the path-connected component of $x .$

Thus, the path-connected component of each point of $X$ contains an open ball centred at that point. This is the definition of an open set in a metric space.

Now, if $X$ is a connected open set in $R^{n},$ then $X$ cannot have more than one path-connected component: otherwise $X$ would be a union of disjoint open path-connected components, and a union of disjoint non-empty open sets is disconnected. Thus, $X$ consists of one path-connected component, and is path-connected.

References for the exercise sheet

E8.1(a), a connected component is closed, is [Willard, Theorem 26.12]. The example in E8.1(c) is both [Armstrong, Example 3 in Section 3.5] and [Willard, Example 26.13a].

E8.2(a) (concatenation of paths) is based on [Sutherland, Lemma 12.2]. E8.2(c), a connected open set in Euclidean space is path-connected, is [Sutherland, Proposition 12.25] and [Willard, Corollary 27.6].

Version 2024/11/20. These exercises in PDF To other course material