\documentclass{article}
\usepackage{amssymb,amstext,amsfonts,amsmath,xspace}
\usepackage{enumerate}
\input{fitch.sty}
\title{The Independence of the General Continuum Hypothesis}
\author{Ryan Flannery}
\date{}
\newcommand{\sN}{\ensuremath{\mathbb{N}}\xspace}
\newcommand{\sZ}{\ensuremath{\mathbb{Z}}\xspace}
\newcommand{\sQ}{\ensuremath{\mathbb{Q}}\xspace}
\newcommand{\sR}{\ensuremath{\mathbb{R}}\xspace}
\newcommand{\sC}{\ensuremath{\mathbb{C}}\xspace}
\newcommand{\Godel}{G\"{o}del\xspace}
\newcommand{\ZF}{ZF\xspace}
\newcommand{\CH}{CH\xspace}
\newcommand{\GCH}{GCH\xspace}
\newcommand{\AC}{AC\xspace}
\newcommand{\Const}{(\ensuremath{V=L})\xspace}
\newcommand{\ZFConst}{\ZF+\Const\xspace}
\newcommand{\liff}{\ensuremath{\leftrightarrow}}
\newcommand{\lthen}{\ensuremath{\rightarrow}}
\newcommand{\limplies}{\ensuremath{\models}}
\newcommand{\lproves}{\ensuremath{\vdash}}
\newcommand{\intersect}{\ensuremath{\cap}}
\newcommand{\union}{\ensuremath{\cup}}
\newcommand{\modA}{\ensuremath{\mathfrak{A}}\xspace}
% Vn zermelo hierarchy of sets, V0 - V3
\newcommand{\VO}{\emptyset}
\newcommand{\VI}{\{\VO\}}
\newcommand{\VII}{\{\VO, \VI\}}
\newcommand{\VIII}{\{\VO, \VI, \{\VI\}, \VII\}}
\begin{document}
\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
In this paper, the independence of the generalized continuum hypothesis of
the Zermelo-Fraenkel set-theory axioms will be discussed in a manner that
individuals with only a mild background in set theory can follow. This is
not, however, an exhaustive trace through the details of the proof. Rather,
most of details will be omitted.
\begin{flushleft}
{\bf NOTE:} Currently, this paper only covers \Godel's half of the proof,
stating that \ZF set theory cannot disprove \GCH. The other half of the
proof by Cohen, that \ZF cannot prove \GCH, will (hopefully) be filled in
later (I make no promises as to when).
\end{flushleft}
This paper is divided into four sections. The first section (this section)
briefly reviews topics that will be required for the rest of the paper. The
second section covers the proof of \Godel's theorem that the generalized
continuum hypothesis cannot be {\em disproved} in Zermelo-Fraenkel set theory.
The third section simply states the proof of Cohen's theorem that the
generalized continuum hypothesis cannot be {\em proved} in Zermelo-Fraenkel
set theory. The final section uses these two theorems to conclude that the
generalized continuum hypothesis is {\em undecidable} in Zermelo-Fraenkel set
theory.
%-----------------------------------------------------------------------------
\subsection{The Zermelo-Fraenkel Axioms}\label{zfaxioms}
The following set of axioms were constructed by by Ernst Zermelo, Adolf
Fraenkel, and Thoralf Skolem for the purpose of axiomatizing set theory, and
are considered the standard axioms for set theory. The axiom of choice is
omitted for the purpose of this paper.
The language of Zermelo-Fraenkel set theory is very close to that of
first-order logic. In fact, it contains all of first-order logic with the
addition of a new relation symbol '$\in$', which is interpreted in the
following manner:
\begin{center}
$x\in y$ means that $x$ is a member of the set $y$
\end{center}
\begin{description}
\item[Axiom of Extensionality.] Two sets are equal if and only if they
have identical elements. In the language of set theory, two sets $x$
and $y$ are equal if and only if:
\begin{equation*} \forall z (z\in x \liff z\in y) \end{equation*}
\item[Null-Set Axiom.] There exists a set with no members (a.k.a. the
``empty set'', and denoted by $\emptyset$). In the language of set
theory, the $\emptyset$ can be expressed as the set $y$ in the sentence,
\begin{equation*} \forall x (\lnot(x\in y)) \end{equation*}
\item[Axiom of Infinity.] There exists a set $x$ such that $\emptyset\in x$
and such whenever $a\in x$, $\left\{a\right\}\in x$ also.\footnote{Note
that this axiom guarantees the existence of an infinite set. Since
$\emptyset\in x$, so too does $\left\{\emptyset\right\}\in x$, and
$\left\{\left\{\emptyset\right\}\right\}\in x$, and so on...} In the
language of set theory,
\begin{equation*} \exists x (\emptyset\in x \land
\forall a(a\in x \liff \{a\}\in x)) \end{equation*}
\item[Power Set Axiom.] If $x$ is a set, then there is a set consisting of
all and only the subsets of $x$. In the language of set theory, the
power set of $x$ can be expressed as the set $y$ in the sentence
\begin{equation*}
\forall x\exists y\forall z(z\in y\liff \forall a(a\in z\lthen a\in x))
\end{equation*}
\item[Axiom of Union.] If $x$ is a set then there is a set whose members are
precisely the members of the members of $x$ (often denoted by
$\bigcup x$). In the language of set theory,
\begin{equation*}
\forall x\exists y\forall z(z\in x\liff \forall a(a\in z\liff a\in y))
\end{equation*}
\item[Axiom of Replacement.] Let $\phi(x,y)$ be any formula in the language
of set theory such that for every set $\alpha$ there is a unique set
$\beta$ satisfying $\phi(\alpha,\beta)$. Let $x$ be any set. Then there
exists a set $y$ consisting of just those $b$ such that $\phi(a,b)$ for
some $a\in x$. Notice that since there is no way of working with whole
formulas in the language of set theory, this is really an axiom schema
given a set of formula's for $\phi$.
\item[Axiom of Subset Selection.] (a.k.a. the Axiom of Separation) Given
any set $x$ and a proposition $\phi(y)$, there exists a subset of $x$
containing only the elements where $\phi(y)$ holds. As in the Axiom of
Replacement, this requires an axiom schema to express in the language of
set theory.
\item[Axiom of Foundation.] (Sometimes called the Axiom of Regularity.) For
every nonempty set $x$, there is a set $a\in x$ such that
$a\intersect x=\emptyset$. Notice, this is equivalent to stating that
$\in$ is a well-founded relation. In the language of set theory,
\begin{equation*}
\forall x(x\not= \emptyset \lthen \exists a(a\in x\land
a\intersect x=\emptyset))
\end{equation*}
\end{description}
Collectively, these set of axioms of the collections of theorems they prove
are called \ZF set theory, and this abbreviation will be used for the
remainder of this paper.
Within these axioms it is possible to define the universe of sets, usually
denoted $V$. The null-set axiom asserts that the empty set ($\emptyset$)
exists, and the power set axiom can be used to build successive levels in this
hierarchy. For example,
\begin{center}\begin{tabular}{lcll}
{\bf Level} & & {\bf Set} & \\
$V_0$ & = & $\VO$ & By the null-set axiom. \\
$V_1$ & = $P(V_0)$ = & $\VI$ & By the power set axiom. \\
$V_2$ & = $P(V_1)$ = & $\VII$ & By the power set axiom. \\
$V_3$ & = $P(V_2)$ = & $\VIII$ & By the power set axiom. \\
\vdots & & \vdots &
\end{tabular}\end{center}
where $P(x)$ denotes the power set of a set $x$.
This isn't quite the entire universe of sets, however. The axiom of infinity
asserts there exists at least one infinite set, and although the $V$ hierarchy
defined above grows exponentially, the axiom of infinity is needed to assert
the existence of infinite sets.
For an infinite set of cardinality $\alpha$ ($\alpha$ is a limit ordinal),
$V_\alpha$ is defined as the union of all lesser levels $V_i$ of the hierarchy.
This union exists by the axiom of union, and
\begin{equation*} V_\alpha = \bigcup_{i<\alpha} V_i \end{equation*}
The universe $V$ is then defined as the union of all levels $V_i$ of this
hierarchy, and this universe constitutes the domain which $\forall$ and
$\exists$ quantify over in the language of \ZF set theory.
\begin{flushleft}
{\bf NOTE:} The above definition of $V_\alpha$, if taken literally, is
an infinitely long formula. {\em Such a formula is not allowed in set theory!}
The set $V_\alpha$ is actually defined using the axiom of replacement. The
``$\bigcup$'' notation is used because it conveys the meaning more clearly than
the actual formula.
\end{flushleft}
%-----------------------------------------------------------------------------
\subsection{The Axiom of Constructibility}\label{aconst}
The following axiom that is discussed is not a standard part of \ZF set
theory. However, it plays a critical role in \Godel's proof of the
consistency of \GCH with \ZF.
In the \ZF axioms listed in Section \ref{zfaxioms}, it may be noticed that
nowhere was a set formally defined. As such, the notion of subset and
power set are quite vague. Indeed, the version of power set adopted in Section
\ref{zfaxioms} is often referred to as the {\em unrestricted power set} (the
meaning of this name will become clear in a moment). The idea behind the
axiom of constructibility is to remove this vagueness, and formalize the
notion of a set, a subset, and the power set of a set.
First, let a set be any {\em describable} collection, where ``describable'' is
meant as ``definable in the language of set theory'' (which is first-order
logic along with the `$\in$' relation). Next, replace the
previous definition of the power set operation with a ``describable power set''
operation (or a {\em restricted power set} operation). With this new power set
operation, the power set of a set $x$ is the set of all describable subsets of
$x$. Finally, we build up the universe of sets just as was before only using
this new notion of power set. The resulting universe is called the
{\em constructible hierarchy}, and usually denoted by $L$.
\begin{center}\begin{tabular}{lcll}
{\bf Level} & & {\bf Set} & \\
$L_0$ & = & $\VO$ & By the null-set axiom. \\
$L_1$ & = $P(L_0)$ = & $\VI$ & By the new power set axiom. \\
$L_2$ & = $P(L_1)$ = & $\VII$ & By the new power set axiom. \\
$L_3$ & = $P(L_2)$ = & $\VIII$ & By the new power set axiom. \\
\vdots & & \vdots &
\end{tabular}\end{center}
And for a limit ordinal $\alpha$,
\begin{equation*} L_\alpha = \bigcup_{i<\alpha} L_i \end{equation*}
This may appear to be identical to the $V$ hierarchy, but the describable
power set operation does in fact produce a different hierarchy. Notice, for
a given level $L_\alpha$, $a\in L_{\alpha+1}$ if and only if there is some
formula $\phi(x)$ in the language of set theory with one free variable, $x$,
such that the elements of $a$ satisfy $\phi(x)$\footnote{Here, the formula
$\phi$ may contain quantifiers. If it does, keep in mind that they do not
quantify over all of $L$ since that has not been constructed yet. Rather,
they quantify over the previous levels of the hierarchy.}. For infinite
sets, this is where the $L$ hierarchy separates and the $V$ hierarchy
separate. Namely, $L$ becomes a sub-hierarchy of $V$, since for each level
of the hierarchy $L_\alpha$ is contained in $V_\alpha$.
When adopting the axiom of constructibility, the universe of sets is
considered to be precisely $L$, and replaces $V$. Thus, the usual notation
for \ZF set theory with the axiom of constructibility is $\ZF+(V=L)$, and this
notation is adopted for the remainder of this paper.
One important theorem in \ZFConst is that $L$ can be well-ordered, even if
the axiom of choice is not adopted. This fact will also play a role in
\Godel's proof of the relative consistency of \GCH in \ZF set theory.
%-----------------------------------------------------------------------------
\subsection{The Continuum Hypothesis}\label{ch}
The continuum hypothesis is a statement first formulated by George Cantor
regarding the sizes of infinite sets. Cantor had shown that the set of
natural numbers \sN was smaller than the set of real number \sR. Using the
notion of cardinality (degrees of infinity) introduced by Cantor, the
cardinality of \sN is less than the cardinality of \sR. Cantor's original
statement of the continuum hypothesis is the following:
\begin{center}
{\em There is no set whose size is strictly between that of \sN and that of
\sR.}\footnote{The set of real numbers \sR is also called the {\em continuum}.
Hence the name of this hypothesis.}
\end{center}
In modern notation, the cardinality of \sN is $\aleph_0$ and the cardinality
of \sR is $\aleph_1$. Using this notation, the continuum hypothesis is:
\begin{equation*} \lnot\exists a(\aleph_0 < |a| < \aleph_1) \end{equation*}
where $|a|$ denotes the cardinality of the set $a$. An equivalent statement
is that
\begin{equation*} 2^{\aleph_0} = \aleph_1. \end{equation*}
The generalized continuum hypothesis (abbreviated from here on as \GCH) is, as
it sounds, a generalization of the continuum hypothesis. It states that for
any infinite set of cardinality $\alpha$, the power set of this set has
cardinality $\alpha+1$, and there are no sets with cardinality greater than
$\alpha$ but less than $\alpha+1$. Formally,
\begin{equation*}
\lnot\exists a(\aleph_\kappa < |a| < \aleph_{\kappa+1})
\end{equation*}
or,
\begin{equation*} 2^{\aleph_\kappa} = \aleph_{\kappa+1} \end{equation*}
for $\kappa$ a cardinal.
There is, however, other notation used for stating \CH and \GCH. For an
infinite set $s$ with cardinality $\alpha$, let $s^*$ denote a set with
cardinality $\alpha+1$ (the next cardinality). Using this notation, the
generalized continuum hypothesis simply states that $P(s)=s*$, where $s$ is
an infinite set. The continuum hypothesis is simply a special case where
$s$ has cardinality $\aleph_0$, and $s^*$ has cardinality $\aleph_1$.
Cantor was able to prove that the for any infinite set $s$, $P(s)$ has
cardinality of {\em at least} $s^*$. This follows since Cantor showed that
$P(s)$ has greater cardinality than $s$, and $s^*$ is the least cardinality
greater than $s$. What \Godel shows, is that for a set $s$, $P(s)$ has
cardinality of {\em at most} $s^*$.
%-----------------------------------------------------------------------------
\subsection{Undecidability}
As a result of \Godel's first incompleteness theorem, any formal system capable
of representing the natural numbers is {\em incomplete}, meaning there are
statements in that system that cannot be proved true or false. Such
statements are considered {\em undecidable} since their truth or falsity
cannot be ascertained. Formally, the notation $\Sigma\lproves\phi$ is used to
express that a sentence $\phi$ is provable from a set of sentences $\Sigma$
(usually a set of axioms). In this notation, a statement $\phi$ is
undecidable from a set of sentences $\Sigma$ if and only if
\begin{equation*} \Sigma\not\lproves\phi \end{equation*}
and
\begin{equation*} \Sigma\not\lproves\lnot\phi. \end{equation*}
Note that $\Sigma\not\lproves\lnot\phi$ is also required for undecidability.
If this were not the case, and $\Sigma\lproves\lnot\phi$, then it is simply
the case that $\phi$ is false, not undecidable.
For the purpose of this paper, the set of sentences $\Sigma$ will always be
either the \ZF axioms or \ZFConst.
%-----------------------------------------------------------------------------
\subsection{Consistency of Sentences}\label{constsent}
This section covers a theorem in mathematical logic that allows \Godel to
prove the relative consistency of \GCH. Formally stated, the theorem
states the following:
\begin{quote}
Let $\Sigma$ be a consistent set of sentences. If $\Sigma$ plus some
additional sentence $\sigma$ is also consistent, and if
$\Sigma+\sigma\lproves\phi$, then it is impossible for
$\Sigma\lproves\lnot\phi$. More succinctly,
{\em if $\Sigma$ and $\Sigma+\sigma$ are consistent, then}
\begin{equation*}
\Sigma+\sigma\lproves\phi\lthen\Sigma\not\lproves\lnot\phi.
\end{equation*}
\end{quote}
Given a $\Sigma$, $\sigma$, and $\phi$ as described above, knowing that
$\Sigma+\sigma\lproves\phi$ only allows $\Sigma\not\lproves\lnot\phi$ to
be inferred. It does {\em not} allow $\Sigma\lproves\phi$ to be inferred.
By showing that $\Sigma\not\lproves\lnot\phi$, one can ``safely adopt''
$\phi$ without introducing any contradictions since it is impossible for
$\Sigma$ to prove $\lnot\phi$. In such a case, it said that $\phi$ is
consistent with $\Sigma$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\Godel's Proof}
In 1940, Kurt \Godel proved that \GCH cannot be disproved from the \ZF axioms,
even if the axiom of choice is included. Specifically, \Godel's theorem
states the following:
\begin{center}
{\em If \ZF is consistent then it remains consistent if \GCH is added.}
\end{center}
Which is equivalent to both of the following statements:
\begin{itemize}
\item If \ZF is consistent then \GCH is not disprovable.
\item If \ZF is consistent then $\lnot\GCH$ is not provable.
\end{itemize}
This is proved in the following manner. We first consider the non-standard
axiom of constructibility (\Const) and add it to our list of axioms for
set theory (thus, we are working with \ZFConst). First, \Godel shows that
if \ZF is consistent then \ZFConst is also consistent. Next, \Godel proves
\GCH from \ZFConst. He then uses the result from mathematical logic
discussed in Section \ref{constsent} to conclude that
$\ZF\not\lproves\lnot\GCH$.
The detailed proof below is broken into two parts. First, the proof that if
\ZF is consistent then \ZFConst is consistent is discussed. Then, the
proof that \ZFConst proves \GCH is discussed.
%-----------------------------------------------------------------------------
\subsection{The Relative Consistency of \ZFConst}\label{relconst}
As was repeatedly stated above, a proof of \GCH from \ZFConst is useful
only if \ZFConst is consistent. If it were not the case that \ZFConst
were consistent, then anything would be provable (via a contradiction), and
the system wouldn't be sound. So, it needs to be shown that \ZFConst
is consistent.
However, as a result of \Godel's second incompleteness theorem and since \ZF
is capable of defining the concept of natural numbers, if \ZF is consistent
then it is {\em impossible} to prove so within \ZF itself. As such, proving
the consistency of \ZFConst (within \ZFConst) is also impossible
(indeed, proving the consistency of $\ZF+Anything$ is impossible, unless it
is inconsistent).
Instead, the {\em relative consistency} of \ZFConst is proved. That is,
the consistency of \ZF is simply accepted, and it is then shown that
\ZFConst {\em must} also be consistent.
\begin{equation*}
\ZF\text{ is Consistent} \lthen \ZFConst\text{ is Consistent}
\end{equation*}
Initially, one might think that adding an additional axiom to \ZF may make the
system more susceptible to being inconsistent, since \Const may contradict the
other axioms. \Godel shows, however, that if \ZF is consistent it remains so
when \Const is accepted. He does so in the following manner.
Consider any arbitrary model \modA of \ZF. Since we accept that \ZF is
consistent, such a model must exist. In Section \ref{zfaxioms} it was shown
that by using the \ZF axioms, one could construct the $V$ hierarchy. Now, in
Section \ref{aconst} a more strict version of the power set operation was
defined, and the $L$ (constructible) hierarchy was constructed. It was
briefly mentioned that the for corresponding levels of the $L$ and $V$
hierarchy's, $L$ is always a sub-hierarchy of $V$ since it is {\em restricted}
in the way it grows using the power set operation. As such, {\em \modA must
contain a submodel that is a model of \ZFConst!} Thus, any model of \ZF
also constitutes a model of \ZFConst. That is, for all models \modA,
\begin{equation*}
\models_\modA \ZF \lthen \models_\modA \ZFConst
\end{equation*}
where $\models_\modA \Sigma$ is read as ``\modA is a model of $\Sigma$''.
As such, in any interpretation where \ZF is consistent, \ZFConst is
automatically consistent, and it is said that {\em \ZFConst is consistent
relative to \ZF.}
%-----------------------------------------------------------------------------
\subsection{$\ZFConst\lproves\GCH$}\label{gchproof}
Perhaps the reason by $\ZF\not\lproves\GCH$ is because in \ZF set theory, the
notion of set was never formally defined. Also, the power set operation was
{\em unrestricted}. That is, $V_{\alpha+1}$ is defined as $P(V_\alpha)$,
which is to say that $V_{\alpha+1}$ is the set of all subsets of $V_\alpha$.
Without a precise definition of a set, there is no precise definition of a
subset, and therefore no precise definition for the power set of a set. How
can one prove any statement regarding the relative sizes of a set and its
power set (which is exactly what \GCH states) without such precision?
As discussed in Section \ref{aconst}, by adopting the axiom of
constructibility, an entirely different hierarchy is constructed, called
the {\em constructible hierarchy} (or $L$) in which sets have a precise
definition. This definition also gives a precise definition of a {\em
subset} and the {\em power set} of a set.
Let $S$ be any set. In this case, let $S$ be infinite. Then there exists
a functions $\phi$ with one free variable in normal form\footnote{By
{\em normal form}, it is meant that $\phi$ contains variables
($x_1,x_2,\ldots$), constants ($c_1,c_2,\ldots$), the `$\in$' relation
symbol, $\lnot$ and $\lor$, and quantifiers ($\forall$ and $\exists$)
which quantify all variables over $S$.}, such that $\phi(x)$ defines a set
$S'$ containing all subsets of $S$. Roughly, $\phi(x)$ then looks
something like the following:
\[\begin{array}{c}
\forall x_1,\ldots,x_a \exists y_1,\ldots,y_b \forall z_1,\ldots,z_c
\exists u_1,\ldots,u_d \ldots \\
L(\ldots x\ldots x_1\ldots x_a\ldots y_1\ldots y_b\ldots z_1\ldots z_c\ldots
u_1\ldots u_d)\footnote{The ordering of variables in $L$ is arbitrary. They
may be arranged in whatever order is needed.}
\end{array}\]
Let $f$ be a function in $S$. That is, $f$ has $n$ variables
($x_1,\ldots,x_n$) and for any elements $s_1,\ldots,s_n$ of $S$,
$f(s_1,\ldots,s_n)$ defines an element of $S$.
Next, {\em Skolemize} $\phi$, by replacing all universally quantified
variables with Skolem constants (for this example, the name of the constants
are left as the name of the variables) and all existentially quantified
variables with Skolem functions $f_1,\ldots,f_m$, where each $f_i$ is a
function in $S$.
\[\begin{array}{c}
L(\ldots x\ldots x_1\ldots x_a\ldots f_1(x_1,\ldots,x_a)\ldots
f_b(x_1,\ldots,x_a)\ldots z_1\ldots z_c\ldots \\
f_k(x_1,\ldots,x_a,z_1,\ldots z_c)\ldots
f_{k+d}(x_1,\ldots,x_a,z_1,\ldots z_c)\ldots
\end{array}\]
Using this approach, it is possible to construct functions defining each
level of the constructible hierarchy. By definition of the $L$ hierarchy,
if two ordinals $\alpha$ and $\beta$ are such that $\alpha < \beta$, then
it's easily seen that $L_\alpha\in L_\beta$ and $L_\alpha\subset L_\beta$.
Further, it can be proven that for any sets $x$ and $y$, if $x\in y$ then the
size (or cardinality in the infinite case) of $x$ is less than the size
(cardinality) of $y$.
This, in the constructible hierarchy, proves for any constructible subset
$s$ of a set $L_\alpha$ ($\alpha$ an infinite ordinal), $s$ {\em must} have a
cardinality less than $\alpha$. An equivalent (and more relevant) statement
of this is that for a set $S$ all of whose subsets have cardinality less than
$\alpha$, the cardinality of $S$ must be less than $\alpha+1$. Now, consider
any infinite cardinal $\alpha$, and its associated level of the $L$ hierarchy
$L_\alpha$. This means that every (constructible) subset of $L_\alpha$ is
an element of $L_{\alpha^*}$, and that the cardinality of $L_{\alpha^*}$ must
be less than the cardinality of $L_{\alpha^{**}}$. That is, the cardinality
of $P(S)$ is at most the cardinality of $S^*$.
As discussed in Section \ref{ch}, Cantor proved
\begin{equation*} |P(x)| \ge |x^*| \end{equation*}
and \Godel just showed that
\begin{equation*} |P(x)| \le |x^*|. \end{equation*}
Thus, it must be the case that
\begin{equation*} |P(x)| = |x^*| \end{equation*}
which is the statement of the generalized continuum hypothesis.
%-----------------------------------------------------------------------------
\subsection{\Godel's Conclusion}
All of the pieces are now present to prove that \GCH is consistent with \ZF.
By accepting that \ZF is consistent, it was shown that \ZFConst is also
consistent. Then, it was shown that $\ZFConst\lproves\GCH$.\\
\begin{fitch}
\fh \ZF\text{ is Consistent.} & Premise \\
\fa \ZFConst\text{ is Consistent.} & by Section \ref{relconst} \\
\fa \ZFConst\lproves\GCH & by Section \ref{gchproof} \\
\fa \ZF\not\lproves\lnot\GCH & by Section \ref{constsent} \\
\end{fitch} \\
Thus, as long as \ZF is consistent, $\ZF\not\lproves\lnot\GCH$, or
\begin{center}
{\em \GCH is relatively consistent with \ZF set theory.}
\end{center}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Cohen's Proof}
In 1963, Paul Cohen proved that \GCH cannot be proven from the \ZF axioms.
Although Cohen's proof is not discussed in this paper, it is the second piece
of the proof required for proving the independence of \GCH from the \ZF axioms.
Formally, Cohen proves the following:
\begin{equation*} \ZF\not\lproves\GCH \end{equation*}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Conclusion \& Remarks}
By \Godel's proof,
\begin{equation*} \ZF\not\lproves\lnot\GCH, \end{equation*}
and by Cohen's proof,
\begin{equation*} \ZF\not\lproves\GCH. \end{equation*}
In other words, within \ZF set theory it is impossible to prove either \GCH
or $\lnot\GCH$. Thus, \GCH is considered {\em undecidable}, since it's truth
or falsity cannot be ascertained.
With this result, one may wonder why \GCH isn't simply adopted as an
additional axiom for set theory. There are two problems with this.
\begin{enumerate}
\item {\em \GCH isn't intuitive.} Although for finite sets it is
easy to see that for a set $x$, the $P(x)$ contains precisely $2^{|x|}$
elements, it isn't clear for case that $x$ is infinite. As such, \GCH
has no place as an axiom.
\item {\em Adopting \GCH still leaves some problems undecidable.} For
example, Whitehead's problem with abelian groups and Souslin's problem of
Dedekind complete sets with no end points still remain undecidable
(neither of these problems are discussed in detail here). By adopting
\GCH as an axiom, very little is gained.
\end{enumerate}
Ideally, one would want to find some other axiom to adopt which is intuitive
and, like the axiom of constructibility, proves \GCH. Although the axiom of
constructibility proves \GCH, it (like \GCH) isn't intuitive.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newpage
\addcontentsline{toc}{section}{References}
\begin{thebibliography}{99}
\bibitem{Cohen} Cohen, Paul J. \underline{Set Theory and the Continuum
Hypothesis}. New York: W. A. Benjamin, Inc., 1966.
\bibitem{Devlin} Devlin, Keith. \underline{The Joy of Sets}. New York:
Springer, 1979.
\bibitem{Enderton} Enderton, Herbert B. \underline{A Mathematical
Introduction to Logic}. San Diego: Academic Press, 2002.
\bibitem{Godel1} \Godel, Kurt. ``What is Cantor's Continuum Problem?''
\underline{The American Mathematical Monthly} Vol. 54, No. 9
(November, 1947).
\bibitem{Godel2} \Godel, Kurt. ``Consistency-Proof for the Generalized
Continuum-Hypothesis.'' \underline{Proceedings of the National Academy of
Sciences of the United States of America} Vol. 25, No. 4
(April, 1939).
\bibitem{Godel3} \Godel, Kurt. ``The Consistency of the Axiom of Choice and
the Generalized Continuum Hypothesis.'' \underline{Proceedings of the
National Academy of Sciences of the United States of America} Vol. 24, No. 12
(December, 1938).
\bibitem{Kunen} Kunen, Kenneth. \underline{Set Theory: An Introduction to
Independence Proofs}. New York: North-Holland Publishing Company, 1980.
\bibitem{Smullyan} Smullyan, Raymond M. and Melvin Fitting. \underline{Set
Theory and the Continuum Problem}. Oxford: Clarendon Press, 1996.
\end{thebibliography}
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