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<h1 id="a-solutions-manual-for-topology-by-james-munkres"><a href="README.html">A solutions manual for Topology by James Munkres</a></h1>
<h2 id="chapter-1.-set-theory-and-logic">Chapter 1. Set Theory and Logic</h2>
<h3 id="fundamental-concepts">1. Fundamental Concepts</h3>
<p><span id="1"></span><strong>1.</strong> Check the distributive laws for <span class="math inline">\(\cup\)</span> and <span class="math inline">\(\cap\)</span> and DeMorgan’s laws.</p>
<p><strong><em>Proof.</em></strong> <span class="math inline">\(\quad\)</span><em>Distributive laws:</em> <span class="math inline">\(x\in A\cap(B\cup C)\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> and (<span class="math inline">\(x\in B\)</span> or <span class="math inline">\(x\in C\)</span>) <span class="math inline">\(\Leftrightarrow\)</span> (<span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\in B\)</span>) or (<span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\in C\)</span>) <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in (A\cap B)\cup (A\cap C)\)</span>. Similarly, <span class="math inline">\(x\in A\cup(B\cap C)\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> or (<span class="math inline">\(x\in B\)</span> and <span class="math inline">\(x\in C\)</span>) <span class="math inline">\(\Leftrightarrow\)</span> (<span class="math inline">\(x\in A\)</span> or <span class="math inline">\(x\in B\)</span>) and (<span class="math inline">\(x\in A\)</span> or <span class="math inline">\(x\in C\)</span>) <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in(A\cup B)\cap(A\cup C)\)</span>.<br />
<span class="math inline">\(\quad\)</span><em>DeMorgan’s laws:</em> <span class="math inline">\(x\in A-(B\cup C)\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin B\cup C\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin B\)</span> and <span class="math inline">\(x\notin C\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin B\)</span> and <span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin C\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in (A-B)\cap(A-C)\)</span>. Similarly, <span class="math inline">\(x\in A-(B\cap C)\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin B\cap C\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> and (<span class="math inline">\(x\notin B\)</span> or <span class="math inline">\(x\notin C\)</span>) <span class="math inline">\(\Leftrightarrow\)</span> (<span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin B\)</span>) or (<span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin C\)</span>) <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in (A-B)\cup(A-C)\)</span>. <span class="math inline">\(\quad\square\)</span></p>
<p><span id="2"></span><strong>2.</strong> Determine which of the following statements are true for all sets <span class="math inline">\(A, B, C\)</span>, and <span class="math inline">\(D\)</span>. If a double implication fails, determine whether one or the other of the possible implications holds. If an equality fails, determine whether the statement becomes true if the “equals” symbol is replaced by one or the other of the inclusion symbols <span class="math inline">\(\subset\)</span> or <span class="math inline">\(\supset\)</span>.<br />
<span class="math inline">\(\quad\)</span>(a) <span class="math inline">\(A\subset B\)</span> and <span class="math inline">\(A\subset C\Leftrightarrow A\subset (B\cup C)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) <span class="math inline">\(A\subset B\)</span> or <span class="math inline">\(A\subset C\Leftrightarrow A\subset (B\cup C)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(c) <span class="math inline">\(A\subset B\)</span> and <span class="math inline">\(A\subset C\Leftrightarrow A\subset (B\cap C)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(d) <span class="math inline">\(A\subset B\)</span> or <span class="math inline">\(A\subset C\Leftrightarrow A\subset (B\cap C)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(e) <span class="math inline">\(A-(A-B)=B\)</span>.<br />
<span class="math inline">\(\quad\)</span>(f) <span class="math inline">\(A-(B-A)=A-B\)</span>.<br />
<span class="math inline">\(\quad\)</span>(g) <span class="math inline">\(A\cap (B-C)=(A\cap B)-(A\cap C)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(h) <span class="math inline">\(A\cup (B-C)=(A\cup B)-(A\cup C)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(i) <span class="math inline">\((A\cap B)\cup (A-B)=A\)</span>.<br />
<span class="math inline">\(\quad\)</span>(j) <span class="math inline">\(A\subset C\)</span> and <span class="math inline">\(B\subset D\Rightarrow (A\times B)\subset (C\times D)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(k) The converse of (j).<br />
<span class="math inline">\(\quad\)</span>(l) The converse of (j), assuming that <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> are nonempty.<br />
<span class="math inline">\(\quad\)</span>(m) <span class="math inline">\((A\times B)\cup (C\times D)=(A\cup C)\times (B\cup D)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(n) <span class="math inline">\((A\times B)\cap (C\times D)=(A\cap C)\times (B\cap D)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(o) <span class="math inline">\(A\times (B-C)=(A\times B)-(A\times C)\)</span>.<br />
<span class="math inline">\(\quad\)</span>(p) <span class="math inline">\((A-B)\times (C-D)=(A\times C-B\times C)-A\times D\)</span>.<br />
<span class="math inline">\(\quad\)</span>(q) <span class="math inline">\((A\times B)-(C\times D)=(A-C)\times (B-D)\)</span>.</p>
<p><strong><em>Solution.</em></strong> <span class="math inline">\(\quad\)</span>(a) False. ‘<span class="math inline">\(\Rightarrow\)</span>’ holds. (b) False. ‘<span class="math inline">\(\Rightarrow\)</span>’ holds. (c) True. (d) False. ‘<span class="math inline">\(\Leftarrow\)</span>’ holds. (e) False. ‘<span class="math inline">\(\subset\)</span>’ holds. (f) False. ‘<span class="math inline">\(\supset\)</span>’ holds. (g) True. (h) False. ‘<span class="math inline">\(\supset\)</span>’ holds. (i) True. (j) True. (k) False. (l) True. ‘<span class="math inline">\(\Leftrightarrow\)</span>’ holds. (m) False. ‘<span class="math inline">\(\subset\)</span>’ holds. (n) True. (o) True. (p) True. (q) False. ‘<span class="math inline">\(\supset\)</span>’ holds.</p>
<p><span id="3"></span><strong>3.</strong> <span class="math inline">\(\quad\)</span>(a) Write the contrapositive and converse of the following statement: “If <span class="math inline">\(x < 0\)</span>, then <span class="math inline">\(x^2 - x > 0\)</span>,” and determine which (if any) of the three statements are true.<br />
<span class="math inline">\(\quad\)</span>(b) Do the same for the statement “If <span class="math inline">\(x >0\)</span>, then <span class="math inline">\(x^2 -x >0\)</span>.”</p>
<p><strong><em>Solution.</em></strong> <span class="math inline">\(\quad\)</span>(a) <em>Contrapositive.</em> “If <span class="math inline">\(x^2 - x\le 0\)</span>, then <span class="math inline">\(x\le 0\)</span>”. <em>Converse.</em> “If <span class="math inline">\(x^2 -x >0\)</span>, then <span class="math inline">\(x >0\)</span>”. The contrapositive is true.<br />
<span class="math inline">\(\quad\)</span>(b) <em>Contrapositive.</em> “If <span class="math inline">\(x^2 -x\le 0\)</span>, then <span class="math inline">\(x\le 0\)</span>”. <em>Converse.</em> “If <span class="math inline">\(x^2 -x >0\)</span>, then <span class="math inline">\(x >0\)</span>”. None is true.</p>
<p><span id="4"></span><strong>4.</strong> Let <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> be sets of real numbers. Write the negation of each of the following statements:<br />
<span class="math inline">\(\quad\)</span>(a) For every <span class="math inline">\(a \in A\)</span>, it is true that <span class="math inline">\(a^2 \in B\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) For at least one <span class="math inline">\(a \in A\)</span>, it is true that <span class="math inline">\(a^2 \in B\)</span>.<br />
<span class="math inline">\(\quad\)</span>(c) For every <span class="math inline">\(a \in A\)</span>, it is true that <span class="math inline">\(a^2 \notin B\)</span>.<br />
<span class="math inline">\(\quad\)</span>(d) For at least one <span class="math inline">\(a \notin A\)</span>, it is true that <span class="math inline">\(a^2 \in B\)</span>.</p>
<p><strong><em>Solution.</em></strong> <span class="math inline">\(\quad\)</span>(a) For at least one <span class="math inline">\(a \in A\)</span>, it is true that <span class="math inline">\(a^2 \notin B\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) For every <span class="math inline">\(a \in A\)</span>, it is true that <span class="math inline">\(a^2 \notin B\)</span>.<br />
<span class="math inline">\(\quad\)</span>(c) For at least one <span class="math inline">\(a \in A\)</span>, it is true that <span class="math inline">\(a^2 \in B\)</span>.<br />
<span class="math inline">\(\quad\)</span>(d) For every <span class="math inline">\(a \notin A\)</span>, it is true that <span class="math inline">\(a^2 \notin B\)</span>.</p>
<p><span id="5"></span><strong>5.</strong> Let <span class="math inline">\(\mathcal{A}\)</span> be a nonempty collection of sets. Determine the truth of each of the following statements and of their converses:<br />
<span class="math inline">\(\quad\)</span>(a) <span class="math inline">\(x\in \bigcup_{A\in \mathcal{A}}A\Rightarrow x\in A\)</span> for at least one <span class="math inline">\(A\in \mathcal{A}\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) <span class="math inline">\(x\in \bigcup_{A\in \mathcal{A}}A\Rightarrow x\in A\)</span> for every <span class="math inline">\(A\in \mathcal{A}\)</span>.<br />
<span class="math inline">\(\quad\)</span>(c) <span class="math inline">\(x\in \bigcap_{A\in \mathcal{A}}A\Rightarrow x\in A\)</span> for at least one <span class="math inline">\(A\in \mathcal{A}\)</span>.<br />
<span class="math inline">\(\quad\)</span>(d) <span class="math inline">\(x\in \bigcap_{A\in \mathcal{A}}A\Rightarrow x\in A\)</span> for every <span class="math inline">\(A\in \mathcal{A}\)</span>.</p>
<p><strong><em>Solution.</em></strong> <span class="math inline">\(\quad\)</span>(a) True. True. (b) False. True. (c) True. False. (d) True. True.</p>
<p><span id="6"></span><strong>6.</strong> Write the contrapositive of each of the statements of Exercise 5.</p>
<p><strong><em>Solution.</em></strong> <span class="math inline">\(\quad\)</span>(a) <span class="math inline">\(x\notin A\)</span> for every <span class="math inline">\(A\in \mathcal{A} \Rightarrow x\notin \bigcup_{A\in \mathcal{A}}A\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) <span class="math inline">\(x\notin A\)</span> for at least one <span class="math inline">\(A\in \mathcal{A} \Rightarrow x\notin \bigcup_{A\in \mathcal{A}}A\)</span>.<br />
<span class="math inline">\(\quad\)</span>(c) <span class="math inline">\(x\notin A\)</span> for every <span class="math inline">\(A\in \mathcal{A} \Rightarrow x\notin \bigcap_{A\in \mathcal{A}}A\)</span>.<br />
<span class="math inline">\(\quad\)</span>(d) <span class="math inline">\(x\notin A\)</span> for at least one <span class="math inline">\(A\in \mathcal{A} \Rightarrow x\notin \bigcap_{A\in \mathcal{A}}A\)</span>.</p>
<p><span id="7"></span><strong>7.</strong> Given sets <span class="math inline">\(A, B\)</span>, and <span class="math inline">\(C\)</span>, express each of the following sets in terms of <span class="math inline">\(A, B\)</span>, and <span class="math inline">\(C\)</span>, using the symbols <span class="math inline">\(\cup, \cap\)</span>, and <span class="math inline">\(-\)</span>. <span class="math display">\[
\begin{gathered}
D = \{x \mid x \in A\text{ and }(x \in B\text{ or }x \in C)\}, \\
E = \{x \mid (x \in A\text{ and }x \in B)\text{ or }x \in C\}, \\
F = \{x \mid x \in A\text{ and }(x \in B \Rightarrow x \in C)\}.
\end{gathered}
\]</span></p>
<p><strong><em>Solution.</em></strong> <span class="math inline">\(\quad\)</span><span class="math inline">\(D=A\cap(B\cup C)\)</span>. <span class="math inline">\(E=(A\cap B)\cup C\)</span>. <span class="math inline">\(F=A\cap((A-B)\cup C)=(A\cap(A-B))\cup(A\cap C)\)</span> <span class="math inline">\(=\)</span> <span class="math inline">\((A-B)\cup(A\cap C)\)</span> <span class="math inline">\(=A-(B-C)\)</span>.</p>
<p><span id="8"></span><strong>8.</strong> If a set <span class="math inline">\(A\)</span> has two elements, show that <span class="math inline">\(\mathcal{P}(A)\)</span> has four elements. How many elements does <span class="math inline">\(\mathcal{P}(A)\)</span> have if <span class="math inline">\(A\)</span> has one element? Three elements? No elements? Why is <span class="math inline">\(\mathcal{P}(A)\)</span> called the powerset of <span class="math inline">\(A\)</span>?</p>
<p><strong><em>Proof.</em></strong> <span class="math inline">\(\quad\)</span><span class="math inline">\(|A|=1\Leftrightarrow |\mathcal{P}(A)|=2\)</span>. <span class="math inline">\(|A|=0\Leftrightarrow |\mathcal{P}(A)|=1\)</span>. <span class="math inline">\(|A|=3\Leftrightarrow |\mathcal{P}(A)|=8\)</span>.<br />
<span class="math inline">\(\quad\)</span>For every <span class="math inline">\(X\subset A\)</span>, let <span class="math inline">\(\chi_X\)</span> be the function given by <span class="math inline">\(x\mapsto 1\)</span> if <span class="math inline">\(x\in X\)</span>; otherwise <span class="math inline">\(x\mapsto 0\)</span>. The mapping of <span class="math inline">\(\mathcal{P}(A)\)</span> onto <span class="math inline">\(\{0,1\}^A\)</span> given by <span class="math inline">\(S\mapsto \chi_S\)</span> is bijective. Thus the cardinality of <span class="math inline">\(\mathcal{P}(A)\)</span> is identical to that of <span class="math inline">\(\{0,1\}^A\)</span>.<span class="math inline">\(\quad\square\)</span></p>
<p><span id="9"></span><strong>9.</strong> Formulate and prove DeMorgan’s laws for arbitrary unions and intersections.</p>
<p><strong><em>Proof.</em></strong> <span class="math inline">\(\quad\)</span>Let <span class="math inline">\(X=\{X_i\mid i\in I\}\)</span> be the family of sets indexed by a nonempty set <span class="math inline">\(I\)</span>. <span class="math inline">\(x\in A-\bigcup_{i\in I} X\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin \bigcup_{i\in I} X\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin X_i\)</span> for every <span class="math inline">\(i\in I\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in \bigcap_{i\in I}(A-X)\)</span>. Similarly, <span class="math inline">\(x\in A-\bigcap_{i\in I} X\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin \bigcap_{i\in I} X\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in A\)</span> and <span class="math inline">\(x\notin X_i\)</span> for at least one <span class="math inline">\(i\in I\)</span> <span class="math inline">\(\Leftrightarrow\)</span> <span class="math inline">\(x\in \bigcup_{i\in I}(A-X)\)</span>. <span class="math inline">\(\quad\square\)</span></p>
<p><span id="10"></span><strong>10.</strong> Let <span class="math inline">\(\mathbb{R}\)</span> denote the set of real numbers. For each of the following subsets of <span class="math inline">\(\mathbb{R}\times \mathbb{R}\)</span>, determine whether it is equal to the cartesian product of two subsets of <span class="math inline">\(\mathbb{R}\)</span>.<br />
<span class="math inline">\(\quad\)</span>(a) <span class="math inline">\(\{(x, y) \mid x\text{ is an integer}\}\)</span>.<br />
<span class="math inline">\(\quad\)</span>(b) <span class="math inline">\(\{(x,y)\mid 0<y\le 1\}\)</span>.<br />
<span class="math inline">\(\quad\)</span>(c) <span class="math inline">\(\{(x,y)\mid y>x\}\)</span>.<br />
<span class="math inline">\(\quad\)</span>(d) <span class="math inline">\(\{(x, y)\mid x \text{ is not an integer and } y \text{ is
an integer}\}\)</span>.<br />
<span class="math inline">\(\quad\)</span>(e) <span class="math inline">\(\{(x,y)\mid x^2+y^2<1\}\)</span>.</p>
<p><strong><em>Solution.</em></strong> <span class="math inline">\(\quad\)</span>(a) <span class="math inline">\(\mathbb{Z}\times\mathbb{R}\)</span>. (b) <span class="math inline">\(\mathbb{R}\times\{y\in\mathbb{R}\mid 0<y\le 1\}\)</span>. (c) No. (d) <span class="math inline">\((\mathbb{R}-\mathbb{Z})\times\mathbb{Z}\)</span>. (e) No.</p>
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