Week 8 Exercises — solutions

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

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

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

Next, let \(X\) be \((\RR ,\) 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 \(\overline A=A.\) Every open neighbourhood of \(0\) contains points from \(B,\) because an open set cannot consists just of points of the finite set \(A;\) hence \(0\in \overline B.\) Similarly, \(1\in \overline B.\) Hence \(\RR =\{0\}\cup \{1\}\cup B \subseteq \overline B.\) We thus have \(\overline B=\RR ,\) and \(\partial A = A \cap \RR = A.\)

Finally, if \(X\) is \((\RR ,\) discrete topology\(),\) then every subset of \(X\) is closed and is equal to its own closure. So \(\overline A=A,\) \(\overline B=B\) and \(\partial A = A \cap B = \emptyset .\)

In each case, a set is dense in \(\RR \) if its closure is \(\RR .\)

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 \(\overline C\) of \(X.\) Therefore, by Lemma 7.11, \(\overline C\) is connected. Since \(\overline C\) contains \(x,\) this means that \(\overline C\) must lies in the connected component of \(x,\) that is, \(\overline C\subseteq C.\) On the other hand, \(C\subseteq \overline C\) for all sets \(C,\) see Claim 7.7. So \(C=\overline C,\) 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

\[ C_1 = X \setminus (C_2 \cup \dots \cup C_n). \]

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

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

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

We need to construct a continuous function \(\chi \colon [0,1]\to X\) such that \(\chi (0)=x\) and \(\chi (1)=z.\) 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 \(\widehat \phi \colon [0,\frac 12]\to X,\) \(\widehat \phi (t) = \phi (2t),\) and \(\widehat \psi \colon [\frac 12,1]\to X,\) \(\widehat \psi (t)= \psi (2t-1),\) are continuous as compositions of continuous functions. Define \(\chi \colon [0,1]\to X\) by

\[ \chi (t) = \begin {cases} \widehat \phi (t), & t\in [0,\frac 12],\\ \widehat \psi (t), & t\in [\frac 12, 1]. \end {cases} \]

Note that \(\chi (\frac 12)\) is well-defined as \(\widehat \phi (\frac 12) = \widehat \psi (\frac 12) = y.\) Moreover, \(\chi (0) = \phi (2\times 0)=x\) and \(\chi (1)= \psi (2\times 1-1) = z.\)

We prove that \(\chi \) is continuous using the closed set criterion of continuity, Proposition 2.5. If \(F\subseteq X\) is closed in \(X,\) \(\chi ^{-1}(F) = {\widehat \phi }^{-1}(F)\cup {\widehat \psi }^{-1}(F).\) Since \(\widehat \phi \) is continuous, \({\widehat \phi }^{-1}(F)\) is closed in \([0,\frac 12],\) hence compact. Similarly, \({\widehat \psi }^{-1}(F)\) is compact. A union of two compact sets is compact by E5.2, so \(\chi ^{-1}(X)\) is compact, hence closed in the Hausdorff space \([0,1].\) We have verified the closed set criterion of continuity for \(\chi .\) Thus, \(x\) and \(z\) are joined by the path \(\chi .\)

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

(c) Suppose that \(X\subseteq \RR ^n\) 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 \(\RR ^n,\) 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