Exercise 6.2.4

Question

% Custom definitions (IntroToSetTheory)
% Source section: sec_6_2.tex

% Common sets
\def
ats{{\boldsymbol{N}}}
\def\ints{{\boldsymbol{Z}}}
\defats{{\boldsymbol{Q}}}
\def\prats{ats^+}
\defeals{{\boldsymbol{R}}}
\def\es{\varnothing}

% Common variables/symbols
\def\vphi{\varphi}
\def\a{\alpha}
\def\b{\beta}
\def\g{\gamma}
\def\d{\delta}
\def\e{\varepsilon}
\def\z{\zeta}
\def\k{\kappa}
\def\l{\lambda}
\def
{
u}
\def{ho}
\def\s{\sigma}
\def	{	au}
\def\x{\xi}
\def\w{\omega}
\def\W{\Omega}
\def\al{\aleph}

% Logic
\def\bic{\leftrightarrow}

ewcommand{\propP}[1]{\prop{P}(#1)}
\def\lxor{\veebar}

% For set-buider notation
\def\where{\mid}

% Other stuff
\def\dom{\mathrm{dom}\,}
\defan{\mathrm{ran}\,}

% Cardinality shortcuts
\def\cnats{{\al_0}}
\def\ccont{{2^\cnats}}

% Equivalence classes

ewcommand\eclass[2]{\squares{#1}_{#2}}

% Shortcuts that make writing easier

ewcommand\parens[1]{\left( #1 ight)}

ewcommand\squares[1]{\left[ #1 ight]}

ewcommand\braces[1]{\left\{ #1 ight\}}

ewcommand\angles[1]{\left\langle #1 ightangle}

ewcommand\ceil[1]{\left\lceil #1 ightceil}

ewcommand\floor[1]{\left\lfloor #1 ightfloor}

ewcommand\abs[1]{\left| #1 ight|}

ewcommand\dabs[1]{\left\| #1 ight\|}

ewcommand\vect[1]{\mathrm{\mathbf{#1}}}

ewcommand\conj[1]{\overline{#1}}

ewcommand\pset[1]{\mathcal{P}\left(#1ight)}

ewcommand\inv[1]{#1^{-1}}

ewcommand\prop[1]{\mathbf{#1}}
\defest{estriction}

ewcommand	et[2]{{^{#1}#2}}

% Families of set
\def\famF{\mathcal{F}}

% These are needed so that half-open intervals do not cause auto-indentation issues due to unmatched brackets

ewcommand\clop[1]{[#1)}

ewcommand\ilab[1]{#1)}

% Other miscellaneous stuff
\def\ss{\subseteq}
\def\pss{\subset}
\def\Seq{\mathrm{Seq}}
\def\prece{\preccurlyeq}
\def\sd{\bigtriangleup}

% Environment shortcuts

ewcommand\gath[1]{\begin{gather*} #1 \end{gather*}}

ewcommand\ali[1]{\begin{align*} #1 \end{align*}}

ewcommand\qproof[1]{\begin{proof} #1 \end{proof}}

ewcommand\bicproof[2]{
  $(	o)$ #1

$(\leftarrow)$ #2
}

ewcommand\seteqproof[2]{
  $(\ss)$ #1

$(\supseteq)$ #2
}

% Environment for indenting nested paragraphs (useful in case trees)

ewenvironment{indpar}
{
  \begin{adjustwidth}{1cm}{}
    }{
  \end{adjustwidth}
}

% Exercise, Theorem, and solution shortcuts

ewcommand\exercise[2]{{
      enewcommand\label[1]{} % Needed to suppress multiply-defined label warning since the same question numbers are used in different sections
      \setcounter{subsubsection}{#1-1} % Subsubsections are used to that lemmas have the full nested numbering
      \stepcounter{subsubsection} % Need to increment this so that the lemma counter gets reset
      \setcounter{question}{#1-1} % Manually set the question number since we always know the exercise number
      \question{#2} % The actual exam class question
    }}

ewcommand\exerciseapp[3]{
  \setqf{#2}
  \exercise{#1}{#3}
  \setqf{}
}

ewcommand	heorem[2]{{
      enewcommand\label[1]{} % Needed to suppress multiply-defined label warning since the same question numbers are used in different sections
      \setcounter{subsubsection}{#1-1} % Subsubsections are used to that lemmas have the full nested numbering
      \stepcounter{subsubsection} % Need to increment this so that the lemma counter gets reset
      \setcounter{question}{#1-1} % Manually set the question number since we always know the theorem number
      \question{#2} % The actual exam class question
    }}

ewcommand	heoremapp[3]{
  \setqf{#2}
  	heorem{#1}{#3}
  \setqf{}
}

ewcommand\sol[1]{
  \begin{solution}

#1
  \end{solution}
}

% Main problem and theorem labels
\def\mainprob{	extbf{Main Problem.}}
\def\mainthrm{	extbf{Main Theorem.}}

% Saves some typing
\def\cbthrm{Cantor-Bernstein Theorem}

% Book title
\def\booktitle{
  Introduction to Set Theory \
  Third Edition, Revised and Expanded \
  by Karel Hrbacek and Thomas Jech
}

% Environment aliases used in the source sections

ewenvironment{lem}{\begin{lemma}}{\end{lemma}}

ewenvironment{defin}{\begin{definition}}{\end{definition}}

ewenvironment{cor}{\begin{corollary}}{\end{corollary}}

ewenvironment{thrm}{\begin{theorem}}{\end{theorem}}

% Override indpar to avoid changepage/adjustwidth dependency
enewenvironment{indpar}{\begin{quote}}{\end{quote}}
Which of the following statements are true?

(a) If $X$ and $Y$ are transitive, then $X\cup Y$ is transitive.

(b) If $X$ and $Y$ are transitive, then $X \cap Y$ is transitive.

(c) If $X \in Y$ and $Y$ is transitive, then $X$ is transitive.

(d) If $X \ss Y$ and $Y$ is transitive, then $X$ is transitive.

(e) If $Y$ is transitive and $S \ss \pset{Y}$, then $Y \cup S$ is transitive.

Dan Whitman · Accepted Answer

% Custom definitions (IntroToSetTheory)
% Source section: sec_6_2.tex

% Common sets
\def
ats{{\boldsymbol{N}}}
\def\ints{{\boldsymbol{Z}}}
\defats{{\boldsymbol{Q}}}
\def\prats{ats^+}
\defeals{{\boldsymbol{R}}}
\def\es{\varnothing}

% Common variables/symbols
\def\vphi{\varphi}
\def\a{\alpha}
\def\b{\beta}
\def\g{\gamma}
\def\d{\delta}
\def\e{\varepsilon}
\def\z{\zeta}
\def\k{\kappa}
\def\l{\lambda}
\def
{
u}
\def{ho}
\def\s{\sigma}
\def	{	au}
\def\x{\xi}
\def\w{\omega}
\def\W{\Omega}
\def\al{\aleph}

% Logic
\def\bic{\leftrightarrow}

ewcommand{\propP}[1]{\prop{P}(#1)}
\def\lxor{\veebar}

% For set-buider notation
\def\where{\mid}

% Other stuff
\def\dom{\mathrm{dom}\,}
\defan{\mathrm{ran}\,}

% Cardinality shortcuts
\def\cnats{{\al_0}}
\def\ccont{{2^\cnats}}

% Equivalence classes

ewcommand\eclass[2]{\squares{#1}_{#2}}

% Shortcuts that make writing easier

ewcommand\parens[1]{\left( #1 ight)}

ewcommand\squares[1]{\left[ #1 ight]}

ewcommand\braces[1]{\left\{ #1 ight\}}

ewcommand\angles[1]{\left\langle #1 ightangle}

ewcommand\ceil[1]{\left\lceil #1 ightceil}

ewcommand\floor[1]{\left\lfloor #1 ightfloor}

ewcommand\abs[1]{\left| #1 ight|}

ewcommand\dabs[1]{\left\| #1 ight\|}

ewcommand\vect[1]{\mathrm{\mathbf{#1}}}

ewcommand\conj[1]{\overline{#1}}

ewcommand\pset[1]{\mathcal{P}\left(#1ight)}

ewcommand\inv[1]{#1^{-1}}

ewcommand\prop[1]{\mathbf{#1}}
\defest{estriction}

ewcommand	et[2]{{^{#1}#2}}

% Families of set
\def\famF{\mathcal{F}}

% These are needed so that half-open intervals do not cause auto-indentation issues due to unmatched brackets

ewcommand\clop[1]{[#1)}

ewcommand\ilab[1]{#1)}

% Other miscellaneous stuff
\def\ss{\subseteq}
\def\pss{\subset}
\def\Seq{\mathrm{Seq}}
\def\prece{\preccurlyeq}
\def\sd{\bigtriangleup}

% Environment shortcuts

ewcommand\gath[1]{\begin{gather*} #1 \end{gather*}}

ewcommand\ali[1]{\begin{align*} #1 \end{align*}}

ewcommand\qproof[1]{\begin{proof} #1 \end{proof}}

ewcommand\bicproof[2]{
  $(	o)$ #1

$(\leftarrow)$ #2
}

ewcommand\seteqproof[2]{
  $(\ss)$ #1

$(\supseteq)$ #2
}

% Environment for indenting nested paragraphs (useful in case trees)

ewenvironment{indpar}
{
  \begin{adjustwidth}{1cm}{}
    }{
  \end{adjustwidth}
}

% Exercise, Theorem, and solution shortcuts

ewcommand\exercise[2]{{
      enewcommand\label[1]{} % Needed to suppress multiply-defined label warning since the same question numbers are used in different sections
      \setcounter{subsubsection}{#1-1} % Subsubsections are used to that lemmas have the full nested numbering
      \stepcounter{subsubsection} % Need to increment this so that the lemma counter gets reset
      \setcounter{question}{#1-1} % Manually set the question number since we always know the exercise number
      \question{#2} % The actual exam class question
    }}

ewcommand\exerciseapp[3]{
  \setqf{#2}
  \exercise{#1}{#3}
  \setqf{}
}

ewcommand	heorem[2]{{
      enewcommand\label[1]{} % Needed to suppress multiply-defined label warning since the same question numbers are used in different sections
      \setcounter{subsubsection}{#1-1} % Subsubsections are used to that lemmas have the full nested numbering
      \stepcounter{subsubsection} % Need to increment this so that the lemma counter gets reset
      \setcounter{question}{#1-1} % Manually set the question number since we always know the theorem number
      \question{#2} % The actual exam class question
    }}

ewcommand	heoremapp[3]{
  \setqf{#2}
  	heorem{#1}{#3}
  \setqf{}
}

ewcommand\sol[1]{
  \begin{solution}

#1
  \end{solution}
}

% Main problem and theorem labels
\def\mainprob{	extbf{Main Problem.}}
\def\mainthrm{	extbf{Main Theorem.}}

% Saves some typing
\def\cbthrm{Cantor-Bernstein Theorem}

% Book title
\def\booktitle{
  Introduction to Set Theory \
  Third Edition, Revised and Expanded \
  by Karel Hrbacek and Thomas Jech
}

% Environment aliases used in the source sections

ewenvironment{lem}{\begin{lemma}}{\end{lemma}}

ewenvironment{defin}{\begin{definition}}{\end{definition}}

ewenvironment{cor}{\begin{corollary}}{\end{corollary}}

ewenvironment{thrm}{\begin{theorem}}{\end{theorem}}

% Override indpar to avoid changepage/adjustwidth dependency
enewenvironment{indpar}{\begin{quote}}{\end{quote}}
(a) We claim that this is true.
  \qproof{
    Consider any $x \in X \cup Y$.
    If $x \in X$ then $x \ss X$ since $X$ is transitive.
    Since also $X \ss X \cup Y$ we clearly have that $x \ss X \ss X \cup Y$.
    We can make the same argument if it is the case that $x \in Y$.
    Hence since $x$ was arbitrary this shows that $X \cup Y$ is indeed transitive.
  }

(b) We claim that this is true.
  \qproof{
    Consider any $x \in X \cap Y$.
    Then $x \in X$ and $x \in Y$.
    Since $X$ and $Y$ are transitive this means that $x \ss X$ and $x \ss Y$.
    So consider any $y \in x$ then $y \in X$ and $y \in Y$ so that $y \in X \cap Y$.
    Hence since $y$ was arbitrary it follows that $x \ss X \cap Y$ so that $X \cap Y$ is transitive since $x$ was arbitrary.
  }

(c) We claim that this is \emph{not} true.
  \qproof{
    It was shown in Exercise~6.2.3 part (a) that $Y = \braces{\es, \braces{\es}, \braces{\braces{\es}}}$ is transitive.
    So let $X = \braces{\braces{\es}}$ so that clearly $X \in Y$.
    If then $x = \braces{\es}$ then $x \in X$ but $x$ is not a subset of $X$ since $\es 
otin X$.
    Hence the original hypothesis is not true.
  }

(d)  We claim that this is \emph{not} true.
  \qproof{
    Again $Y = \braces{\es, \braces{\es}, \braces{\braces{\es}}}$ is transitive so let $X = \braces{\braces{\es}, \braces{\braces{\es}}}$ so that clearly $X \ss Y$.
    Then if $x = \braces{\es}$ then $x \in X$ but $x$ is not a subset of $X$ because $\es 
otin X$.
    Thus the original hypothesis is false.
  }

(e) We claim that this is true.
  \qproof{
    Consider any $x \in Y \cup S$.
    If $x \in Y$ then since $Y$ is transitive $x \ss Y$.
    Hence $x \ss Y \ss Y \cup S$.
    On the other hand if $x \in S$ then $x \in \pset{Y}$ since $S \ss \pset{Y}$
    Hence $x \ss Y \ss Y \cup S$.
    Since in all cases $x \ss Y \cup S$ and $x$ was arbitrary this shows that $Y \cup S$ is transitive by definition.
  }

Exercise 6.2.4

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Exercise 6.2.4

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