avewright/proofwiki-math huggingface dataset:
Theorem: Prime Gap Size is Unbounded

Let $n \in \N$ be a natural number, arbitrarily large.
Then there exist consecutive prime numbers $p_1$ and $p_2$ such that:
:$p_2 - p_1 > n$
That is, the size of prime gaps is unbounded.
That is, there are blocks of consecutive composite numbers whose length exceeds any given $n \in \N$.

Proof:
Let $p$ be a prime number greater than $n + 1$.
Consider the primorial:
:$q = 2 \times 3 \times 5 \times \cdots \times p$
All of the $p - 1$ numbers:
:$q + 2, q + 3, q + 4, \ldots, q + p$
are composite.
Hence the result.
{{qed}}
Theorem: Linear Bound between Complex Function and Derivative

Let $f: D \to \C$ be a complex-differentiable function, where $D \subseteq \C$ is an open set.
Let $z_0 \in D$.
Let $\epsilon \in \R_{>0}$.
Then there exists $r \in \R_{>0}$ such that for all $z \in \map {B_r} {z_0}$:
:$\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } < \epsilon \size {z - z_0}$
where $\map {B_r} {z_0}$ denotes the open ball with center $z_0$ and radius $r$.

Proof:
{{begin-eqn}}
{{eqn | l = \map {f'} {z_0} 
      | r = \ds \lim_{z \mathop \to z_0} \dfrac{ \map f z - \map f {z_0} } {z - z_0}
      | c = {{Defof|Complex-Differentiable at Point}}
}}
{{eqn | ll = \leadsto
      | l = 0
      | r = \lim_{z_0 \mathop \to z} \dfrac {\map f z - \map f {z_0} - \map {f'} {z_0} \paren{ z - z_0 } } {z - z_0}
      | c = Combination Theorem for Limits of Complex Functions
}}
{{eqn | r = \lim_{z_0 \mathop \to z} \size {\dfrac {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } {z - z_0} }
      | c = Modulus of Limit
}}
{{eqn | r = \lim_{z_0 \mathop \to z} \dfrac {\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } } {\size {z - z_0} }
      | c = Complex Modulus of Quotient of Complex Numbers
}}
{{end-eqn}}
Given $\epsilon > 0$, we can find $r > 0$ by definition of limit such that for all $z \in \map {B_r} {z_0}$:
:$\dfrac {\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } } {\size {z - z_0} } < \epsilon$
We rearrange to:
:$\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } < \epsilon \size {z - z_0}$
{{qed}}
Theorem: Supremum Operator Norm of Linear Transformation is Bounded Above by Hilbert-Schmidt Norm

Let $T_A : \R^n \to \R^m$ be the linear transformation such that:
:$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $\norm {\, \cdot \,}_{HS}$ be the Hilbert Schmidt norm.
Then:
:$\norm {T_A} \le \norm A_{HS}$

Proof:
We have that Norms on Finite-Dimensional Real Vector Space are Equivalent.
Choose the Euclidean norm.
Let $X = \struct {\R^n, \norm {\, \cdot \,}_2}$ and $Y = \struct {\R^m, \norm {\, \cdot \,}_2}$ be normed vector spaces.
Let the matrix $A \in \R^{m \times n}$ be given by:
:$A = \begin {bmatrix}
a_{1 1} & \cdots & a_{1 n} \\
 \vdots & \ddots & \vdots \\
a_{m 1} & \cdots & a_{m n} \\
\end{bmatrix}$
We have that Set of Linear Transformations is Isomorphic to Matrix Space.
Let $T_A : \R^n \to \R^m$ be the linear transformation such that:
:$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
From Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous:
:$\ds \norm{T_A \mathbf x}_2 \le \norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 }$
Take the supremum of both sides with the condition $\forall \mathbf x \in \R^n : \norm {\mathbf x}_2 \le 1$.
On the left we are left with $\norm {T_A}$.
On the right we have:
{{begin-eqn}}
{{eqn | l = \norm {\norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 } }
      | r = \sup \set {\norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 } : \mathbf x \in \R^n : \norm {\mathbf x}_2 \le 1 }
}}
{{eqn | r = \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 } \sup \set {\norm{\mathbf x}_2 : \mathbf x \in \R^n : \norm {\mathbf x}_2 \le 1 }
}}
{{eqn | r = \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 } \cdot 1
}}
{{eqn | r = \norm {A}_{HS}
      | c = {{Defof|Hilbert-Schmidt Norm}}
}}
{{end-eqn}}
Hence:
:$\norm {T_A} \le \norm {A}_{HS}$
{{qed}}
Theorem: Continuous Image of Path-Connected Set is Path-Connected/Metric Space

Let $M_1, M_2$ be metric spaces whose metrics are $d_1, d_2$ respectively.
Let $f: M_1 \to M_2$ be a continuous mapping.
Let $S \subseteq M_1$ be a path-connected subspace of $M_1$.
Then $f \sqbrk S$ is a path-connected subspace of $M_2$.

Proof:
Let $\map f s, \map f {s'} \in f \sqbrk S$, for some $s, s' \in S$.
Let $\mathbb I$ denote the closed unit interval:
:$\mathbb I = \closedint 0 1$
Let $p: \mathbb I \to S$ be a continuous mapping such that:
:$\map p 0 = s, \map p 1 = s'$
Such a $p$ exists since $S$ is a path-connected subspace of $M_1$.
Now define $q: \mathbb I \to f \sqbrk S$ by:
:$\map q t := f \circ \map p t$
Then $q$ is a continuous mapping, being the composition of such, and:
:$\map q 0 = \map f {\map p 0} = \map f s$
:$\map q 1 = \map f {\map p 1} = \map f {s'}$
Thus $\map f s$ and $\map f {s'}$ are connected by a path in $f \sqbrk S$.
Since these points were arbitrary in $f \sqbrk S$, the result follows.
{{qed}}
Theorem: Path-Connectedness is Preserved under Homeomorphism

Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be topological spaces.
Let $\phi: T_1 \to T_2$ be a homeomorphism.
Let $S \subseteq T_1$ be a subset of $T_1$.
Let $S$ be path-connected in $\struct {T_1, \tau_1}$.
Then $\phi \sqbrk S$ is path-connected in $\struct {T_2, \tau_2}$.
That is, path-connectedness is a topological property.

Proof:
By definition of homeomorphism, $\phi$ is a continuous mapping.
The result now follows from Continuous Image of Path-Connected Set is Path-Connected.
{{qed}}
Category:Topological Properties
Category:Path-Connected Sets
7susxuru3awz4i4gh6u86ag6wha5h0r
Theorem: Holomorphic Function is Continuously Differentiable

Let $U \subseteq \C$ be an open set.
Let $f : U \to \C$ be a holomorphic function.
Then $f$ is continuously differentiable in $U$.

Proof:
Let $z \in U$.
The derivatives $\map {f'} z$ and $\map {f' '} z$ exist by Cauchy's Integral Formula for Derivatives.
Therefore $f'$ is continuous at $z$ by Complex-Differentiable Function is Continuous.
{{qed}}
Theorem: Simple Connectedness is Preserved under Homeomorphism

Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be topological spaces.
Let $\phi: T_1 \to T_2$ be a homeomorphism.
Let $S \subseteq T_1$ be a subset of $T_1$.
Let $S$ be simply connected in $\struct {T_1, \tau_1}$.
Then $\phi \sqbrk S$ is simply connected in $\struct {T_2, \tau_2}$.
That is, simple connectedness is a topological property.

Proof:
By {{Defof|Simply Connected}}, $S$ is path-connected.
Path-Connectedness is Preserved under Homeomorphism shows that $\phi \sqbrk S$ is path-connected.
Let $x \in S$.
Let:
:$\gamma_1 : \closedint 0 1 \to \phi \sqbrk S$
:$\gamma_2 : \closedint 0 1 \to \phi \sqbrk S$
be two loops in $\phi \sqbrk S$.
Let $\gamma_1, \gamma_2$ both have base point $x$.
Composite of Continuous Mappings is Continuous shows that $\phi^{-1} \circ \gamma_1$ and $\phi^{-1} \circ \gamma_2$ are continuous mappings.
So $\phi^{-1} \circ \gamma_1 , \phi^{-1} \circ \gamma_2$ are loops in $S$ with base point $\map {\phi^{-1} } x$.
By {{Defof|Simply Connected}}, there exists a path homotopy $H: \closedint 0 1 \times \closedint 0 1 \to S$ such that:
{{begin-eqn}}
{{eqn | q = \forall s \in \closedint 0 1
      | l = \map H {s, 0}
      | r = \map {\phi^{-1} \circ \gamma_1} s
      | c = 
}}
{{eqn | q = \forall s \in \closedint 0 1
      | l = \map H {s, 1}
      | r = \map {\phi^{-1} \circ \gamma_2} s
      | c = 
}}
{{eqn | q = \forall t \in \closedint 0 1
      | l = \map H {0, t}
      | r = \map {\phi^{-1} \circ \gamma_1} 0
      | rr= = \map {\phi^{-1} } x
      | c = 
}}
{{eqn | q = \forall t \in \closedint 0 1
      | l = \map H {1, t}
      | r = \map {\phi^{-1} \circ \gamma_1} 1
      | rr= = \map {\phi^{-1} } x
      | c = 
}}
{{end-eqn}}
Composite of Continuous Mappings is Continuous shows that $ \phi \circ H : \closedint 0 1 \times \closedint 0 1 \to \phi \sqbrk S$ is a continuous mapping.
Then:
{{begin-eqn}}
{{eqn | q = \forall s \in \closedint 0 1
      | l = \map {\phi \circ H} {s, 0}
      | r = \map {\phi \circ \phi^{-1} \circ \gamma_1} s
      | rr= = \map {\gamma_1} s
      | c = 
}}
{{eqn | q = \forall s \in \closedint 0 1
      | l = \map {\phi \circ H} {s, 1}
      | r = \map {\phi \circ \phi^{-1} \circ \gamma_2} s
      | rr= = \map {\gamma_2} s
      | c = 
}}
{{eqn | q = \forall t \in \closedint 0 1
      | l = \map {\phi \circ H} {0, t}
      | r = \map {\phi \circ \phi^{-1} \circ \gamma_1} 0
      | rr= = x
      | c = 
}}
{{eqn | q = \forall t \in \closedint 0 1
      | l = \map {\phi \circ H} {1, t}
      | r = \map {\phi \circ \phi^{-1} \circ \gamma_1} 1
      | rr= = x
      | c = 
}}
{{end-eqn}}
It follows that $\phi \circ H$ is a path homotopy between $\gamma_1$ and $\gamma_2$.
By {{Defof|Simply Connected}}, $\phi \sqbrk S$ is simply connected.
{{qed}}
Theorem: Equivalence of Definitions of Continuous Complex Function

{{TFAE|def = Continuous Complex Function}}
Let $A_1, A_2 \subseteq \C$ be subsets of the complex plane.
Let $f: A_1 \to A_2$ be a complex function from $A_1$ to $A_2$.
Let $a \in A_1$.

Proof:
=== Definition using Limit implies Epsilon-Delta Definition ===
Let $\epsilon > 0$.
By the Epsilon-Delta condition of limit, there exists $\delta > 0$, so for all $z \in A_1$:
:$0 < \cmod {z - a} < \delta \implies \cmod {\map f z - \map f a} < \epsilon$
In the case of $0 = \cmod {z - a}$, we must have $z = a$, so
:$\cmod {\map f z - \map f a} = 0 < \epsilon$
{{qed|lemma}}
Theorem: Function f is Big-O of g iff g is Big-Omega of f

Let $f: \N \to \R, g: \N \to \R$ be two real sequences, expressed here as real-valued functions on the set of natural numbers $\N$.
Then:
:$\map f n \in \map \OO {\map g n}$
{{iff}}:
:$\map g n \in \map \Omega {\map f n}$ 
where:
:$\OO$ denotes big-$\OO$ notation
:$\Omega$ denotes big-$\Omega$ notation.

Proof:
Recall the definitions:
:$\map \Omega f = \set {g: \N \to \R: \exists c_1 \in \R_{>0}: \exists n_1 \in \N: \forall n > n_1: 0 \le c_1 \cdot \size {\map f n} \le \size {\map g n} }$
:$\map \OO g = \set {f: \N \to \R: \exists c_2 \in \R_{>0}: \exists n_2 \in \N: \forall n > n_2: 0 \le \size {\map f n} \le c_2 \cdot \size {\map g n} }$
for some $n_1, n_2 \in \N$.
{{finish|Note that the above are the canonical definitions which are to be used during the course of the below. Failing to take into account the sign of the function may render this proof incomplete.}}
Hence:
:If $g \in \map \Omega f$, then $\forall n > n_1: \exists c_1 > 0 : \map g n \ge c_1 \cdot \map f n$
:If $f \in \map \OO g$, then $\forall n > n_2: \exists c_2 > 0 : \map f n \le c_2 \cdot \map g n$
First let us assume that $n \ge \max \set {n_1, n_2}$ in all the below work.
Theorem: Noetherian Hausdorff Space is Finite Discrete Space

Let $\struct {S, \tau}$ be a Noetherian Hausdorff space.
Let $S \ne \O$.
Then $\struct {S, \tau}$ is a finite discrete space.

Proof:
{{ProofWanted}}
Category:Hausdorff Spaces
Category:Noetherian Topological Spaces
Category:Discrete Topologies
Category:Finite Topological Spaces
3hu53kqjr3kn79vdbik404icicva5uq
Theorem: Sum of Additive Function Values is Well-Defined

Let $\SS$ be an algebra of sets.
Let $f: \SS \to \overline \R$ be an additive function on $\SS$.
Let $A, B \in \SS$.
Then the sum
:$\map f A + \map f B$
is well-defined in the extended real numbers $\overline \R$.

Proof:
Suppose the sum $\map f A + \map f B$ is void.
By {{Defof|Extended Real Addition}}, this happens when the sum is $\paren{ +\infty } + \paren{ -\infty }$, or $\paren{ -\infty } + \paren{ +\infty }$.
{{WLOG}}, assume that $\map f A = +\infty$, and $\map f B = -\infty$.
Set Difference and Intersection are Disjoint shows that $A \setminus B$ and $A \cap B$ are disjoint.
Set Difference and Intersection are Disjoint shows that $B \setminus A$ and $A \cap B$ are disjoint.
By definition of additive function and Set Difference Union Intersection, it follows that:
:$ \map f { A \setminus B } + \map f { A \cap B } = \map f A = +\infty$
:$ \map f { B \setminus A } + \map f { A \cap B } = \map f B = -\infty$
Suppose $\map f { A \cap B } = +\infty$.
It follows that:
:$\map f { B \setminus A } + \paren{ +\infty } = -\infty$
which is a contradiction.
Suppose $\map f { A \cap B } = -\infty$.
It follows that:
:$\map f { A \setminus B } + \paren{ -\infty } = +\infty$
which is a contradiction.
Finally, suppose $\map f { A \cap B } \in \R$.
As $\map f { A \setminus B } + \map f { A \cap B } = +\infty$, it follows that $\map f { A \setminus B } = +\infty$.
As $\map f { B \setminus A } + \map f { A \cap B } = -\infty$, it follows that $\map f { B \setminus A } = -\infty$.
By definition of additive function and Set Difference is Disjoint with Reverse, it follows that the sum:
:$\map f { A \setminus B } + \map f { B \setminus A }$
is well-defined.
As this sum is equal to $\paren{ +\infty } + \paren{ -\infty}$, this is a contradiction.
{{qed}}
Category:Set Systems
Category:Additive Functions
r1wwpy68jjr7v6njr8wq68aoeh469tj
Theorem: Path-Homotopic Paths have Path Homotopy

Let $X$ be a topological space.
Let $f, g: \closedint 0 1 \to X$ be paths.
Then $f$ and $g$ are path-homotopic, {{iff}} there exists a path homotopy between $f$ and $g$.

Proof:
=== Necessary Condition ===
By {{Defof|Path-Homotopic}}, there exists a homotopy $H : \closedint 0 1 \times \closedint 0 1 \to X$ such that:
:$\forall s \in \closedint 0 1: \map H {s, 0} = \map f s$
:$\forall s \in \closedint 0 1: \map H {s, 1} = \map g s$
Homotopic Paths have Same Endpoints shows that:
:$\map f 0 = \map g 0$
and:
:$\map f 1 = \map g 1$
As $f$ and $g$ are homotopic relative to $\set{ 0 , 1 }$, we have:
:$\forall s' \in \set{ 0 , 1 } , t \in \closedint 0 1 : \map H { s' , t } = \map f { s' } = \map g { s' }$
By {{Defof|Path Homotopy}}, it follows that $H$ is a {{Defof|Path Homotopy}}.
{{qed|lemma}}
Theorem: Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation

Let $\Bbb K = \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a bounded sequence in $\Bbb K$.
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the normed $2$-sequence space.
Let $\mathbf x = \tuple {a_1, a_2, a_3, \ldots} \in \ell^2$.
Suppose $\Lambda : \ell^2 \to \ell^2$ is a diagonal operator such that:
:$\Lambda \tuple {a_1, a_2, a_3, \ldots} = \tuple {\lambda_1 a_1, \lambda_2 a_2, \lambda_3 a_3, \ldots}$
Then $\Lambda \in \map {CL} {\ell^2}$.

Proof:
=== Linearity ===
Let $\sequence {a_n}_{n \mathop \in \N_{> 0} }, \sequence {b_n}_{n \mathop \in \N_{> 0} } \in \ell^2$.
{{begin-eqn}}
{{eqn | l = \map \Lambda {k \sequence {a_n} + \sequence {b_n} }
      | r = \map \Lambda {\sequence {k a_n + b_n} }
      | c = $P$-sequence space is a vector space
}}
{{eqn | r = \sequence {\lambda_n \paren{k a_n + b_n} }
}}
{{eqn | r = \sequence {\lambda_n k a_n + \lambda_n b_n }
}}
{{eqn | r = k \sequence {\lambda_n a_n} + \sequence {\lambda_n b_n}
      | c = $P$-sequence space is a vector space
}}
{{eqn | r = k \map \Lambda {\sequence {a_n} } + \map \Lambda {\sequence {b_n} }
}}
{{end-eqn}}
By linearity of transformations:
:$\Lambda \in \map \LL {\ell^2}$
Theorem: Supremum Operator Norm of Diagonal Operator over 2-Sequence Space

Let $\Bbb K = \set {\R, \C}$.
Let $\sequence {\lambda_n}_{n \mathop \in \N_{> 0} }$ be a bounded sequence in $\Bbb K$.
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the normed $2$-sequence space.
Let $\mathbf x = \tuple {a_1, a_2, a_3, \ldots} \in \ell^2$.
Suppose $\Lambda : \ell^2 \to \ell^2$ be a diagonal operator such that:
:$\Lambda \tuple {a_1, a_2, a_3, \ldots} = \tuple {\lambda_1 a_1, \lambda_2 a_2, \lambda_3 a_3, \ldots}$
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Then $\ds \Lambda \in \map {CL} {\ell^2}$ and $\ds \norm \Lambda = \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$

Proof:
From Diagonal Operator over 2-Sequence Space is Continuous Linear Transformation:
:$\ds \norm {\Lambda \mathbf x }_2^2 \le \paren {\sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} }^2 \norm {\mathbf x}_2^2$
Therefore:
:$\ds \norm {\Lambda \mathbf x}_2 \le \sup_{n \mathop \in \N_{> 0} } \size{\lambda_n} \norm {\mathbf x}_2$
Suppose $\norm {\mathbf x}_2 \le 1$.
Then:
:$\ds \norm {\Lambda \mathbf x}_2 \le \sup \norm {\Lambda \mathbf x}_2 \le \sup_{n \mathop \in \N_{> 0} } \size{\lambda_n}$
By definition of supremum operator norm:
:$\norm \Lambda = \sup \set {\norm {\Lambda \mathbf x}_2 : \forall \mathbf x \in \ell^2 : \norm {\mathbf x}_2 \le 1 }$
Hence:
:$\ds \norm {\Lambda} \le \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
Let $\mathbf e_n \in \ell^2$ such that:
:$\mathbf e_n = \tuple {\underbrace{0, \ldots, 0}_{n - 1}, 1, 0, \ldots}$
Then:
{{begin-eqn}}
{{eqn | ll = \forall n \in \N_{> 0} :
      | l = \norm \Lambda
      | r = \norm \Lambda \cdot 1
}}
{{eqn | r = \norm \Lambda \norm {\mathbf e_n}_2
}}
{{eqn | o = \ge
      | r = \norm {\Lambda \mathbf e_n}_2
      | c = Supremum Operator Norm as Universal Upper Bound
}}
{{eqn | r = \norm {\tuple {\underbrace{0, \ldots, 0}_{n - 1}, \lambda_n \cdot 1, 0 \ldots} }_2
}}
{{eqn | r = \size {\lambda_n}
}}
{{end-eqn}}
Take the supremum of both sides over $n \in N$.
On the left-hand dize nothing happens because it is an $n$-independent real number.
On the left-hand dize we have $\ds \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
Hence:
:$\ds \norm \Lambda \ge \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
Combining the results yields:
:$\ds \norm \Lambda = \sup_{n \mathop \in \N_{> 0} } \size {\lambda_n}$
{{qed}}
Theorem: Reparameterization of Directed Smooth Curve with Given Domain

Let $\gamma: \closedint a b \to \C$ be a smooth path in the complex plane.
Let $C$ be a directed smooth curve with parameterization $\gamma$.
Let $\closedint { a_0 }{ b_0 }$ be a closed real interval, where $a_0 < b_0$.
Then there exists a smooth path
:$\gamma_0 : \closedint { a_0 }{ b_0 } \to \C$
that is a reparameterization of $C$.

Proof:
Define $\phi : \closedint a b \to \closedint { a_0 }{ b_0 }$ by:
:$\map{ \phi }{ t } = \dfrac{ b_0 - a_0 }{ b - a } \paren{ t - a } + a_0$
Power Rule for Derivatives shows that
:$\map{ \phi' }{ t } = \dfrac{ b_0 - a_0 }{ b - a }$
Real Function with Strictly Positive Derivative is Strictly Increasing shows that $\phi$ is strictly increasing, as $\map{ \phi' }{ t } > 0$ for all $t \in \closedint a b$.
Strictly Monotone Real Function is Bijective shows that $\phi$ is bijective.
Set $\gamma_0 = \gamma \circ \phi : \closedint a b \to \C$.
By definition of directed smooth curve, it follows that $\gamma_0$ is a reparameterization of $C$.
{{qed}}
Theorem: Interior of Simple Closed Contour is Well-Defined

Let $C$ be a simple closed contour in the complex plane.
Let $f : \closedint 0 1 \to \R^2$ be a Jordan curve.
Let $\phi : \R^2 \to \C$ be defined by:
:$\map \phi {x, y} = x + i y$
Let $\Img C = \map \phi {\Img f}$, where $\Img C$ denotes the image of $C$, and $\Img f$ denotes the image of $f$.
Then the interior of $C$:
:$\Int C = \map \phi {\Int f}$
is well-defined.
Here, $\Int f$ denotes the interior of $f$.

Proof:
Let $C$ be defined as a concatenation of a (finite) sequence of directed smooth curves $\sequence {C_1, \ldots, C_n}$.
We show existence of a Jordan curve $f : \closedint 0 1 \to \R^2$ that fulfills the criteria $\map \phi {\Img f} = \Img C$.
Reparameterization of Directed Smooth Curve with Given Domain shows that $C_k$ can be reparameterized by a smooth path:
:$\gamma_k: \closedint {\dfrac {k - 1} n} {\dfrac k n} \to \C$ for all $k \in \set {1, \ldots, n}$
Define $\gamma: \closedint 0 1 \to \C$ by:
:$\map \gamma t = \map {\gamma_k} t$ for all $t \in \closedint {\dfrac {k - 1} n} {\dfrac k n}$
Reparameterization of Directed Smooth Curve Preserves Image shows that $\Img \gamma = \Img C$.
Pasting Lemma for Pair of Continuous Mappings on Closed Sets shows that $\gamma$ is continuous.
By definition of simple contour, it follows that:
:$\map \gamma {t_1} \ne \map \gamma {t_2}$ for all $t_1 \in \hointr 0 1, t_2 \in \hointr 0 1$ with $t_1 \ne t_2$
By definition of closed contour, it follows that:
:$\map \gamma 0 = \map \gamma 1$
Complex Plane is Homeomorphic to Real Plane shows that $\phi$ is a homeomorphism.
Composite of Continuous Mappings between Metric Spaces is Continuous shows that $\phi^{-1} \circ \gamma : \closedint 0 1 \to \R^2$ is continuous.
As $\phi^{-1}$ is a homeomorphism, it follows by the definition of Jordan curve that $\phi^{-1} \circ \gamma$ is a Jordan curve.
Set $f = \phi^{-1} \circ \gamma$.
We now have $\Img C = \map \phi {\Img f}$.
{{qed|lemma}}
We show uniqueness of the definition of $\Int C$.
Let $g: \closedint 0 1 \to \R^2$ be a Jordan curve such that $\Img g = \Img f$.
Jordan Curve Theorem shows that the definition of the interior of a Jordan curve only depends on the image of the Jordan curve.
Then:
:$\Int g = \Int f$
{{qed}}
Category:Complex Contour Integrals
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Theorem: Jordan Curve Characterization of Simply Connected Set

Let $D \subseteq \R^2$ be an open path-connected subset of the Euclidean plane.
Then $D$ is simply connected, {{Iff}} the following condition holds:
:For all Jordan curves $\gamma : \closedint 0 1 \to \R^2$ with $\Img \gamma \subseteq D$, we have:
:$\Int \gamma \subseteq D$
Here $\Img \gamma$ denotes the image of $\gamma$, and  $\Int \gamma$ denotes the interior of $f$.

Proof:
=== Sufficient condition ===
Suppose that $D$ is simply connected.
Let $\gamma : \closedint 0 1 \to \R^2$ be a Jordan curve with $\Img \gamma \subseteq D$.
By definition of simple connectedness, there exists a constant loop $c : \closedint 0 1 \to D$ such that $\gamma$ and $c$ are path-homotopic in $D$.
Let $H : \closedint 0 1 \times \closedint 0 1 \to D$ be the path homotopy between $\gamma$ and $c$.
From Interior of Jordan Curve is Subset of Image of Null-Homotopy, it follows that:
:$\Int \gamma \subseteq \Img H$
As:
:$\Img H \subseteq D$
it follows that:
:$\Int \gamma \subseteq D$
{{qed|lemma}}
Theorem: Coprime Integers cannot Both be Zero

Let $a$ and $b$ be integers.
Let $a$ and $b$ be coprime.
Then it cannot be the case that $a = b = 0$.

Proof:
Let $a$ and $b$ be coprime.
Then by definition:
:$\gcd \set {a, b} = 1$
{{AimForCont}} $a = b = 0$.
Then $\gcd \set {a, b}$ is undefined.
But it is not possible both:
:for $\gcd \set {a, b}$ to be undefined
:for $\gcd \set {a, b} = 1$.
Hence by Proof by Contradiction it follows that it cannot be the case that $a = b = 0$.
{{qed}}
Category:Coprime Integers
Category:Zero
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Theorem: Complex Plane is Homeomorphic to Real Plane

Let $\C$ denote the complex plane.
Let $\R^2$ denote the Euclidean plane.
Then the function $\phi : \R^2 \to \C$ defined by:
:$\map \phi {x, y} = x + i y$
is a homeomorphism between $\R^2$ and $\C$.

Proof:
Define $\phi^{-1} : \C \to \R^2$ by:
:$\map {\phi^{-1} } z = \tuple {\map \Re z, \map \Im z}$
Then $\phi^{-1}$ is the inverse of $\phi$, as:
:$\map \phi {\map { \phi^{-1} } z} = \map \Re z + i \map \Im z = z$
:$\map {\phi^{-1} } {\map \phi {x, y} } = \tuple {\map \Re {x + i y}, \map \Im {x + i y} } = \tuple {x, y}$
By definition of bijection, it follows that $\phi$ is bijective.
We show continuity of $\phi$.
Complex Plane is Metric Space shows that for $\map \phi {x_1, y_1} = x_1 + i y_1, \map \phi {x_2, y_2} = x_2 + i y_2$, we have:
:$\cmod {\map \phi {x_2, y_2} - \map \phi {x_1, y_1} } = \sqrt {\paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2}$
which is equal to the distance between $\tuple {x_1, x_2}$ and $\tuple {y_1, y_2}$ in the Euclidean metric on the Euclidean plane.
Let $\epsilon \in \R_{>0}$.
With $\delta = \epsilon$, we have:
:$\sqrt {\paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2} < \delta \implies \cmod {\map \phi {x_2, y_2} - \map \phi {x_1 , y_1} } < \epsilon$
which shows the continuity of $\phi$.
The continuity of $\phi^{-1}$ is proven the same way.
{{qed}}
Category:Complex Analysis
Category:Examples of Homeomorphisms
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