Skip to main content
Contents Dark Mode Prev Up Next \(\DeclareMathOperator{\sgn}{sgn}
\DeclareMathOperator{\Inn}{Inn}
\DeclareMathOperator{\lcm}{lcm}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\Perm}{Perm}
\DeclareMathOperator{\Stab}{Stab}
\DeclareMathOperator{\Orb}{Orb}
\DeclareMathOperator{\Rot}{Rot}
\DeclareMathOperator{\re}{Re}
\DeclareMathOperator{\im}{Im}
\DeclareMathOperator{\img}{image}
\DeclareMathOperator{\conj}{conj}
\DeclareMathOperator{\Id}{Id}
\def\expi{E}
\def\wrap{W}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Quat}{\mathbb{H}}
\newcommand{\extC}{\hat{\C}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\extR}{\hat{\R}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\extF}{\hat{\F}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Proj}{\mathbb{P}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\B}{\mathbb{B}}
\newcommand{\M}{{\rm MOB}}
\newcommand{\E}{{\rm EUC}}
\renewcommand{\H}{{\rm HYP}}
\newcommand{\HU}{\rm HYP_{\U}}
\renewcommand{\S}{{\rm ELL}}
\newcommand{\D}{\mathbb{D}}
\newcommand{\closedD}{\hat{\D}}
\newcommand{\U}{\mathbb{U}}
\newcommand{\spacer}{\rule[0cm]{0cm}{0cm}}
\newcommand{\MOD}{\mathbin{\text{MOD}}}
\newcommand{\twotwo}[4]{\left[ \begin{array}{cc} #1 \amp #2 \\
#3 \amp #4 \end{array} \right]}
\let\oldsection\section
\renewcommand{\section}{\clearpage\oldsection}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Section 7.1 Shroeder-Bernstein Theorem
Theorem 7.1.1 . (Schroeder-Bernstein).
If \(f\colon A\to B\) and \(g\colon B\to A\) are one-to-one maps, then \(A\sim B\text{.}\)
Proof outline.
For each
\(a\in A\text{,}\) form a sequence to the right and to the left of
\(a\text{.}\)
\begin{equation*}
\cdots \to g^{-1}(f^{-1}(g^{-1}(a))) \to f^{-1}(g^{-1}(a)) \to g^{-1}(a) \to a \to f(a) \to g(f(a)) \to f(g(f(a))) \to \cdots
\end{equation*}
Observe that theses sequences may or may not terminate on the left.
Observe that each of these sequences is one of two types: (i) terminates on the left with an element \(y\in B\) (that is, \(y\) is not in the image of \(f\) ); or (ii) does not have the property that defines type (i).
Define \(h\colon A\to B\) by \(h(a)=g^{-1}(a)\) if \(a\) lies in a sequence of type (i), and \(h(a) =
f(a)\) if \(a\) lies in a sequence of type (ii). Now show \(h\) is well-defined and is a bijection.
Checkpoint 7.1.2 .
Complete the proof of the Schroeder-Berstein Theorem following the outline given above.
