nyuuzyou/wikis huggingface dataset:
Continuous iff For Every Element There Exists Ideal Element Precedes Supremum

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be an up-complete meet semilattice.
Then
:$L$ is continuous
:for every element $x$ of $S$ there exists ideal $I$ in $L$:
::$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$
== Proof ==
=== Sufficient Condition ===
Let $L$ be continuous.
By Continuous iff Way Below Closure is Ideal and Element Precedes Supremum:
:$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \map \sup {x^\ll}$ and
::for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$
where $x^\ll$ denotes the way below closure of $x$.
Thus
:for every element $x$ of $S$ there exists ideal $I$ in $L$:
::$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$
=== Necessary Condition ===
Assume that
:for every element $x$ of $S$ there exists ideal $I$ in $L$:
::$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$
Let $x \in S$.
There exists ideal $I$ in $L$:
:$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$
We will prove that
:$I \subseteq x^\ll$
where $x^\ll$ denotes the way below closure of $x$.
Let $y \in I$.
Let $J$ be an ideal in $L$ such that
:$x \preceq \sup J$
Then $I \subseteq J$
Thus by definition of subset:
:$y \in J$
By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
:$y \ll x$
Thus by definition of way below closure:
:$y \in x^\ll$
By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal
:$\forall y \in x^\ll: y \in I$
By definition of subset:
:$x^\ll \subseteq I$
Thus by definition of set equality:
:$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \map \sup {x^\ll}$ and
::for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$
Hence by Continuous iff Way Below Closure is Ideal and Element Precedes Supremum:
:$L$ is continuous
== Sources ==
* 
* 
Category:Continuous Lattices
Category:Up-Complete Semilattices
Continuous iff Mapping at Element is Supremum

== Theorem ==
Let $\left({S, \preceq_1, \tau_1}\right)$ and $\left({T, \preceq_2, \tau_2}\right)$ be complete continuous topological lattices with Scott topologies.
Let $f: S \to T$ be a mapping.
Then $f$ is continuous 
:$\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$
== Proof ==
=== Sufficient Condition ===
Assume that
:$f$ is continuous.
By Continuous iff Directed Suprema Preserving:
:$f$ is preserves directed suprema.
Thus by Directed Suprema Preserving Mapping at Element is Supremum:
:$\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$
=== Necessary Condition ===
Assume that
:$\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$
By Mapping at Element is Supremum implies Way Below iff There Exists Element that Way Below and Way Below:
:$\forall x \in S, y \in T: y \ll f\left({x}\right) \iff \exists w \in S: w \ll x \land y \ll f\left({w}\right)$
Thus by Continuous iff Way Below iff There Exists Element that Way Below and Way Below:
:$f$ is continuous.
== Sources ==
* 
* 
Category:Topological Order Theory
Category:Way Below Relation
Category:Continuous Lattices
Continuous iff Mapping at Element is Supremum of Compact Elements

== Theorem ==
Let $L = \struct {S, \preceq_1, \tau_1}$ and $R = \struct {T, \preceq_2, \tau_2}$ be complete algebraic topological lattices with Scott topologies.
Let $f: S \to T$ be a mapping.
Then $f$ is continuous :
:$\forall x \in S: \map f x = \sup \leftset {\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset {}$
== Proof ==
By Algebraic iff Continuous and For Every Way Below Exists Compact Between:
:$L$ and $R$ are continuous.
=== Sufficient Condition ===
Assume that:
:$f$ is continuous.
By Continuous iff Mapping at Element is Supremum:
:$\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$
Let $x \in S$.
By definitions of image of set and compact closure:
:$\leftset {\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset {} = f \sqbrk {x^{\mathrm{compact} } }$
By Compact Closure is Directed:
:$D := x^{\mathrm{compact} }$ is directed.
By Continuous iff Directed Suprema Preserving:
:$f$ preserves directed suprema.
By definition of mapping preserves directed suprema:
:$f$ preserves the supremum of $D$.
By definition of complete lattice:
:$D$ admits a supremum.
By definition of algebraic:
:$L$ satisfies the axiom of $K$-approximation.
Thus
=== Necessary Condition ===
Assume that
:$\forall x \in S: \map f x = \sup \leftset {\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset{}$
By Mapping at Element is Supremum of Compact Elements implies Mapping at Element is Supremum that Way Below:
:$\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$
Thus by Continuous iff Mapping at Element is Supremum:
:$f$ is continuous.
== Sources ==
* 
* 
Category:Topological Order Theory
Category:Way Below Relation
Category:Continuous Lattices
Continuous iff Meet-Continuous and There Exists Smallest Auxiliary Approximating Relation

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Then:
:$L$ is continuous
:
:$L$ is meet-continuous and there exists the smallest auxiliary approximating relation on $S$
That is:
:$L$ is continuous
:
:$L$ is meet-continuous and there exists an auxiliary approximating relation $\RR$ on $S$
::for every auxiliary approximating relation $\QQ$ on $S$: $\RR \subseteq \QQ$
== Proof ==
=== Sufficient Condition ===
Let $L$ be continuous.
Thus by Continuous Lattice is Meet-Continuous:
:$L$ is meet-continuous.
Thus by Way Below is Approximating Relation and Way Below Relation is Auxiliary Relation:
:$\ll$ is auxiliary approximating relation on $S$.
Thus by Continuous Lattice iff Auxiliary Approximating Relation is Superset of Way Below Relation:
:for every auxiliary approximating relation $\QQ$ on $S$: $\ll \subseteq \QQ$
=== Necessary Condition ===
Let $L$ be meet-continuous.
Assume that:
:there exists an auxiliary approximating relation $\RR$ on $S$
::for every auxiliary approximating relation $\QQ$ on $S$: $\RR \subseteq \QQ$
Let $x \in S$.
By Intersection of Relation Segments of Approximating Relations equals Way Below Closure:
: $\ds \bigcap \set {x^\QQ: \QQ \in \map {\operatorname{App} } L} = x^\ll$
where:
:$x^\QQ$ denotes the $\QQ$-segment of $x$, 
:$\map {\operatorname{App} } L$ denotes the set of all auxiliary approximating relations on $S$.
By Intersection is Subset/General Result:
:$x^\ll \subseteq x^\RR$
By definition of approximating relation:
:$x = \map \sup {x^\RR}$
We will prove that:
:$\forall a \in \set {x^\QQ: \QQ \in \map {\operatorname{App} } L}: x^\RR \subseteq a$
Let $a \in \set {x^\QQ: \QQ \in \map {\operatorname{App} } L}$
Then:
:$\exists \QQ \in \map {\operatorname{App} } L: a = x^\QQ$
By assumption:
:$\RR \subseteq \QQ$
Thus by Relation Segment is Increasing:
:$x^\RR \subseteq a$
By Intersection is Largest Subset/General Result:
:$x^\RR \subseteq x^\ll$
By definition of set equality:
:$x^\RR = x^\ll$
Thus
:$x = \map \sup {x^\ll}$
By definition:
:$L$ satisfies the axiom of approximation.
By Way Below Closure is Directed in Bounded Below Join Semilattice:
:$\forall x \in S: x^\ll$ is directed
By definition of complete lattice:
:$L$ is up-complete.
Hence $L$ is continuous.
== Sources ==
* 
* 
Category:Approximating Relations
Category:Continuous Lattices
Category:Meet-Continuous Lattices
Continuous iff Way Below Closure is Ideal and Element Precedes Supremum

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be an up-complete meet semilattice.
Then
:$L$ is continuous
:$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \map \sup {x^\ll}$ and:
::for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$
where $x^\ll$ denotes the way below closure of $x$.
== Proof ==
=== Sufficient Condition ===
Let $L$ be continuous.
Let $x \in S$.
By definition of continuous:
:$x^\ll$ is directed.
By Way Below Closure is Lower Section:
:$x^\ll$ is a lower section.
Thus by definition:
:$x^\ll$ is an ideal in $L$.
By definition of continuous:
:$L$ satisfies the axiom of approximation.
By the axiom of approximation:
:$x = \map \sup {x^\ll}$
Thus by definition of reflexivity:
:$x \preceq \map \sup {x^\ll}$
Let $I$ be an ideal in $L$ such that
:$x \preceq \sup I$
We will prove that
:$x^\ll \subseteq I$
Let $y \in x^\ll$
By definition of way below closure:
:$y \ll x$
Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
:$y \in I$
=== Necessary Condition ===
Assume that:
:$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \map \sup {x^\ll}$ and:
::for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$
Let $x \in S$.
By assumption:
:$x \preceq \map \sup {x^\ll}$
By Way Below implies Preceding:
:$x$ is upper bound for $x^\ll$
By definition of supremum:
:$\map \sup {x^\ll} \preceq x$
Thus by definition of antisymmetry:
:$x = \map \sup {x^\ll}$
Thus by definition:
:$L$ satisfies the axiom of approximation.
Thus by assumption and definition of ideal:
:$\forall x \in S: x^\ll$ is directed.
Hence $L$ is continuous.
== Sources ==
* 
* 
Category:Continuous Lattices
Category:Way Below Relation
Category:Up-Complete Semilattices
Continuous iff Way Below iff There Exists Element that Way Below and Way Below

== Theorem ==
Let $\struct {S, \preceq_1, \tau_1}$ and $\struct {T, \preceq_2, \tau_2}$ be complete continuous topological lattices with Scott topologies.
Let $f: S \to T$ be a mapping.
Then $f$ is continuous 
:$\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$
== Proof ==
=== Sufficient Condition ===
Assume that
:$f$ is continuous.
By Continuous iff Directed Suprema Preserving:
:$f$ preserves directed suprema.
By Directed Suprema Preserving Mapping at Element is Supremum:
:$\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$
Thus by Mapping at Element is Supremum implies Way Below iff There Exists Element that Way Below and Way Below:
:$\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$
=== Necessary Condition ===
Assume that
:$\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$
Let $U$ be an open subset of $T$.
By definition of Scott topology:
:$U$ is upper.
By definition of interior:
:$\paren {f^{-1} \sqbrk U}^\circ \subseteq f^{-1} \sqbrk U$
We will prove that
:$f^{-1} \sqbrk U \subseteq \paren {f^{-1} \sqbrk U}^\circ$
Let $x \in f^{-1} \sqbrk U$
By definition of preimage of set:
:$p := \map f x \in U$
By Open implies There Exists Way Below Element:
:$\exists u \in T: u \ll p \land u \in U$
By assumption:
:$\exists w \in S: w \ll x \land u \ll \map f w$
We will prove that
:$f \sqbrk {w^\gg} \subseteq U$
Let $h \in f \sqbrk {w^\gg}$
By definition of image of set:
:$\exists z \in w^\gg: h = \map f z$
By definition of way above closure:
:$w \ll z$
By assumption:
:$u \ll h$
By Way Below implies Preceding:
:$u \preceq_2 h$
Thus by definition of upper section:
:$h \in U$
By Preimage of Subset is Subset of Preimage:
:$f^{-1} \sqbrk {f \sqbrk {w^\gg} } \subseteq f^{-1} \sqbrk U$
By Preimage of Image under Left-Total Relation is Superset:
:$w^\gg \subseteq \paren {f^{-1} \circ f} \sqbrk {w^\gg}$
By definition of composition of relations:
:$w^\gg \subseteq f^{-1} \sqbrk {f \sqbrk {w^\gg} }$
By Subset Relation is Transitive:
:$w^\gg \subseteq f^{-1} \sqbrk U$
By Interior is Union of Way Above Closures:
:$\paren {f^{-1} \sqbrk U}^\circ = \bigcup \set {g^\gg: g \in S \land g^\gg \subseteq f^{-1} \sqbrk U}$
By definition of way above closure:
:$x \in w^\gg$
We have:
:$w^\gg \in \set {g^\gg: g \in S \land g^\gg \subseteq f^{-1} \sqbrk U}$
Thus by definition of union:
:$x \in \paren {f^{-1} \sqbrk U}^\circ$
By definition of set equality:
:$\paren {f^{-1} \sqbrk U}^\circ = f^{-1} \sqbrk U$
Thus by definition of interior:
:$f^{-1} \sqbrk U$ is open.
Hence $f$ is continuous.
== Sources ==
* 
* 
Category:Topological Order Theory
Category:Way Below Relation
Category:Continuous Lattices
Continuous implies Increasing in Scott Topological Lattices

== Theorem ==
Let $T_1 = \struct {S_1, \preceq_1, \tau_1}$ and $T_2 = \struct {S_2, \preceq_2, \tau_2}$ be up-complete topological lattices with Scott topologies.
Let $f: S_1 \to S_2$ be a continuous mapping.
Then $f$ is an increasing mapping.
== Proof ==
Let $x, y \in S_1$ such that
:$x \preceq_1 y$
 that
:$\map f x \npreceq_2 \map f y$
By definition of lower closure of element:
:$\map f x \notin \paren {\map f y}^\preceq$
By definition of relative complement:
:$\map f x \in \relcomp {S_2} {\paren {\map f y}^\preceq}$
By definition of reflexivity:
:$\map f y \preceq_2 \map f y$
By definition of lower closure of element:
:$\map f y \in \paren {\map f y}^{\preceq_2}$
By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice:
:$\set {\map f y}^- = \paren {\map f y}^{\preceq_2}$
By definition of closure:
:$\paren {\map f y}^{\preceq_2}$ is a closed set.
By definition of closed set:
:$\relcomp {S_2} {\paren {\map f y}^{\preceq_2} }$ is a open set.
By definition of continuous:
:$f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$ is an open set.
By definition of Definition:Scott Topology:
:$f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$ is an upper section.
By definition of preimage of set:
:$x \in f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$
By definition of upper section:
:$y \in f^{-1} \sqbrk {\relcomp {S_2} {\paren {\map f y}^{\preceq_2} } }$
By definition of preimage of set:
:$\map f y \in \relcomp {S_2} {\paren {\map f y}^{\preceq_2} }$
Thus this contradicts $\map f y \in \paren {\map f y}^{\preceq_2}$
== Sources ==
* 
* 
Category:Topological Order Theory
Continuously Differentiable Curve has Finite Arc Length

== Theorem ==
Let $y = \map f x$ be a real function which is continuous on the closed interval $\closedint a b$ and continuously differentiable on the open interval $\openint a b$.
The definite integral: 
:$s = \ds \int_{x \mathop = a}^{x \mathop = b} \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x$
exists, and is called the arc length of $f$ between $a$ and $b$.
== Proof ==
It intuitively makes sense to define the length of a line segment to be the distance between the two end points, as given by the Distance Formula:
:$\sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$
Similarly, it is reasonable to assume that the actual length of the curve would be approximately equal to the sum of the lengths of each of the line segments, as shown:
To calculate the sum of the length of these line segments, divide $\closedint a b$ into any number of closed subintervals of the form $\closedint {x_{i - 1} } {x_i}$ where:
:$a = x_0 < x_1 < \dotsb < x_{k - 1} < x_k = b$
Define:
:$\Delta x_i = x_i - x_{i - 1}$
:$\Delta y_i = y_i - y_{i - 1}$
As the length of the $i$th line segment is $\sqrt {\paren {\Delta x_i}^2 + \paren {\Delta y_i}^2}$, the sum of all these line segments is given by:
:$\ds \sum_{i \mathop = 1}^k \ \sqrt {\paren {\Delta x_i}^2 + \paren {\Delta y_i}^2}$
Thus the approximate arc length is given by the sum:
:$\ds s \approx \sum_{i \mathop = 1}^k \sqrt {1 + \paren {\frac {\Delta y_i} {\Delta x_i} }^2}\Delta x_i$
Recall that by hypothesis:
:$f$ is continuous on $\closedint a b$
:$f$ is differentiable on $\openint a b$.
Thus the Mean Value Theorem can be applied.
In every open interval $\openint {x_i} {x_{i - 1} }$ there exists some $c_i$ such that:
:$D_x \, \map f {c_i} = \dfrac {\Delta y_i} {\Delta x_i}$
Plugging this into the above sum we have:
:$\ds s \approx \sum_{i \mathop = 1}^k \ \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \Delta x_i$
By hypothesis, $D_x f$ is continuous.
As Square of Real Number is Non-Negative the radicand is always positive.
From Continuity of Root Function and Limit of Composite Function, $\sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$ is continuous as well.
Because Continuous Real Function is Darboux Integrable, there exists a definite integral that confirms the intuitive notion that there is a value that represents the exact length of the curve as the limit of the above Riemann sum.
This definite integral is:
:$\ds s = \int_{x \mathop = a}^{x \mathop = b} \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x$
== Sources ==
* $\S 7.4$
Category:Integral Calculus
Category:Analytic Geometry
Continuously Differentiable Real Function at Removable Discontinuity

== Theorem ==
Let $f : \R \to \R$ be a real function.
Let $a \in \R$ be real number.
Let $f$ be continuous on $\R$ and continuously differentiable in $\R \setminus \set a$.
Suppose that $a$ is a removable discontinuity of $f'$.
That is, suppose the limit $\ds \lim_{x \mathop \to a} \map {f'} x$ exists.
Then $f$ is continuously differentiable at $a$.
=== Corollary ===
== Proof ==
By assumption, $\ds \lim_{x \mathop \to a} \map {f'} x = L$ exists.
By definition:
:$\forall \epsilon \in \R_{>0} : \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map {f'} x - L} < \epsilon$
Let $h := x - a$.
Suppose $0 < h < \delta$.
Consider a closed interval $\closedint 0 h$.
By assumption, $f$ is continuous on $\closedint 0 h$ and continuously differentiable in $\openint 0 h$.
By Mean Value Theorem:
:$\ds \exists \theta \in \openint 0 1 : \frac {\map f h - \map f 0} {h - 0} = \map {f'} {\theta h}$
Thus:
:$0 < \size {\theta h} \le \size h < \delta$ 
and:
Hence:
:$\ds \forall h \in \openint 0 \delta : \size {\frac {\map f h - \map f 0} {h - 0} - L} < \epsilon$
Consider a closed interval $\closedint {-h} 0$.
By assumption, $f$ is continuous on $\closedint {-h} 0$ and continuously differentiable in $\openint {-h} 0$.
By Mean Value Theorem:
:$\ds \exists \theta \in \openint 0 1 : \frac {\map f 0 - \map f {-h} } {0 - \paren {-h} } = \map {f'} {-\theta h}$
Thus:
$0 < \size {- \theta h} \le \size {-h} < \delta$
and:
Hence:
:$\ds \forall y \in \openint {-\delta} 0 : \size {\frac {\map f y - \map f 0} {y - 0} - L} < \epsilon$
Altogether, it holds that:
:$\ds \forall h \in \R : 0 < \size h < \delta : \size {\frac {\map f h - \map f 0} {h - 0} - L} < \epsilon$
Hence, $f$ is differentiable at $a$:
:$\ds \map {f'} a = L = \lim_{x \mathop \to a} \map {f'} x$.
By assumption, $f'$ is continuous on $\R \setminus \set a$.
Hence, $f$ is continuously differentiable on $\R$.
Category:Continuity
Category:Differentiable Real Functions
Continuously Differentiable Real Function at Removable Discontinuity/Corollary

== Corollary to Continuously Differentiable Real Function at Removable Discontinuity ==
Let $f : \R \to \R$ be a real function.
Let $a \in \R$ be real number.
Let $f$ be $n - 1$ times continuously differentiable in $\R$ and $n$ times continuously differentiable in $\R \setminus \set a$.
Suppose that $a$ is a removable discontinuity of $f^{\paren n}$.
That is, suppose the limit $\ds \lim_{x \mathop \to a} \map {f^{\paren n}} x$ exists.
Then $f$ is $n$ times continuously differentiable at $a$.
== Proof ==
Let $y = f^{\paren {n - 1}}$.
Then $y' = f^{\paren n}$.
The result follows from Continuously Differentiable Real Function at Removable Discontinuity.
Category:Continuity
Category:Differentiable Real Functions
Continuum Hypothesis

== Hypothesis ==
There is no set whose cardinality is strictly between that of the integers and the real numbers.
Symbolically, the continuum hypothesis asserts:
:$\aleph_1 = \mathfrak c$
where:
:$\mathfrak c$ denotes the cardinality of the continuum
:$\aleph_1$ denotes Aleph One.
=== Generalized Continuum Hypothesis ===
== Historical Note ==
== Sources ==
* : Appendix $\text{A}.6$: Cardinality
* : $\S 4$: Number systems $\text{I}$: A set-theoretic approach
* 
* : Appendix $\text{A}$: Set Theory: Cardinal Numbers
* 
* : Chapter $1$: General Background: $\S 5$ The continuum problem
Category:Continuum Hypothesis
Category:Cardinality of Continuum
Category:Set Theory
Category:Named Theorems
Category:Open Questions
Continuum Hypothesis/Historical Note

== Historical Note on Continuum Hypothesis ==
The Continuum Hypothesis was originally conjectured by .
In $1940$, showed that it is impossible to disprove the Continuum Hypothesis (CH for short) in Zermelo-Fraenkel set theory (ZF) with or without the Axiom of Choice (ZFC).
In $1963$, showed that it is impossible to prove CH in ZF or ZFC.
These results together show that CH is independent of both ZF and ZFC.
Note, however, that these results do not settle CH one way or the other, nor do they establish that CH is undecidable.
They merely indicate that CH cannot be proved within the scope of ZF or ZFC, and that any further progress will depend on further insights on the nature of sets and their cardinality.
It has been suggested that a key factor contributing towards the difficulty in resolving this question may be the fact that Gödel's Incompleteness Theorems prove that there is no possible formal axiomatization of set theory that can represent the entire spread of possible properties that can uniquely specify any possible set.
== Sources ==
* 
* 
* : Chapter $1$: General Background: $\S 5$ The continuum problem
Category:Historical Notes
Continuum Hypothesis is Independent of ZFC/Historical Note

== Historical Note on Continuum Hypothesis is Independent of ZFC ==
Between $1938$ and $1940$ (accounts differ), showed that the Continuum Hypothesis (CH) is formally consistent with the axioms of Zermelo-Fraenkel set theory (ZF) with or without the Axiom of Choice (ZFC).
Between $1963$ and $1966$, showed that the negation of both the Continuum Hypothesis and the Generalized Continuum Hypothesis are also consistent with the axioms of ZF and ZFC.
Thus, neither ZF and ZFC are strong enough to settle the question of the Continuum Hypothesis either way.
== Sources ==
* : Chapter $1$: General Background: $\S 5$ The continuum problem
Category:Historical Notes
Continuum Property

== Theorem ==
The continuum property (of the set of real numbers $\R$) is a complementary pair of theorems whose subject is the real number line:
=== Least Upper Bound Property ===
=== Greatest Lower Bound Property ===
== Also presented as ==
== Also known as ==
== Also see ==
* Dedekind's Theorem
* Definition:Dedekind Complete
* Continuum Property implies Well-Ordering Principle
Not to be confused with:
* The Continuum Hypothesis
== Sources ==
* : Chapter $\text I$: Functions and Limits: $\S 3$: Bounds of a Function: Theorem $\text{A}$
* : Chapter $2$: Metric Spaces: $\S 5$: Limits
* : $\S 2$: Continuum Property: $\S 2.4$: The Continuum Property
Category:Continuum Property
Category:Real Analysis
Category:Named Theorems
Continuum Property implies Well-Ordering Principle

== Theorem ==
The Continuum Property of the positive real numbers $\R_{\ge 0}$ implies the Well-Ordering Principle of the natural numbers $\N$.
== Proof ==
Suppose that the set of positive real numbers $\R_{\ge 0}$ has the Continuum Property.
 $T \subseteq \N$ is a subset of $\N$ which has no smallest element.
Then:
:$\forall t \in T: \exists u \in T: u < t$
Let $A \in \R_{\ge 0}$.
For every $t \in T$, let $a_t = \dfrac A {2^t}$.
We have that $t \in \N$, so $t \ge 0$.
Therefore $2^t \ge 1$.
This $a_t$ can never be greater than $A$.
Therefore $\set {a_t: t \in T}$ has an upper bound.
As $\R_{\ge 0}$ has the Continuum Property, it follows that $\set {a_t: t \in T}$ has a supremum $B$.
We have that $\dfrac {7 B} 8 < B$.
Thus $\dfrac {7 B} 8$ is not an upper bound of $\set {a_t: t \in T}$.
Therefore there exists $m \in T$ such that:
:$a_m > \dfrac {7 B} 8$
Then there is $n \in T$ such that $n < m$. 
Let $n = m - p$. 
Then:
Therefore $a_n > B$.
This contradicts the supposition that $B$ is the supremum of $\set {a_t: t \in T}$
Thus our initial supposition that $T \subseteq \N$ has no smallest element was false.
As $T$ is arbitrary, it follows that each $T \subseteq \N$ has a smallest element.
Thus the Well-Ordering Principle holds for $\N$, as was to be proved.
== Also see ==
* Equivalence of Well-Ordering Principle and Induction
Category:Number Theory
Category:Natural Numbers
Category:Real Numbers