avewright/proofwiki-math huggingface dataset:
Theorem: Exponentiation Functor is Functor

Let $\mathbf C$ be a Cartesian closed metacategory.
Let $A$ be an object of $\mathbf C$.
Let $\left({-}\right)^A: \mathbf C \to \mathbf C$ be the exponentiation functor.
Then $\left({-}\right)^A$ is a functor.

Proof:
Let $B$ be an object of $\mathbf C$.
Let $\epsilon_B: B^A \times A \to B$ be the evaluation morphism at $B$.
Then, since:
{{begin-eqn}}
{{eqn | l = \operatorname{id}_B \epsilon_B
      | r = \epsilon_B
}}
{{eqn | r = \epsilon_B \operatorname{id}_{B^A \times A}
}}
{{eqn | r = \map {\epsilon_B} {\operatorname{id}_{B^A} \times \operatorname{id}_A}
}}
{{end-eqn}}
or equivalently:
::<nowiki>$\begin{xy}\xymatrix@L+2mu@+1em{
B^A \times A \ar[r]^*+{\epsilon_B} & B \\
B^A \times A
  \ar[u]^*+{\operatorname{id}_{B^A} \times \operatorname{id}_A}
  \ar[ur]_*+{\operatorname{id}_B \epsilon_B}
}\end{xy}$</nowiki>
it follows:
:$\paren {\operatorname{id}_B}^A = \operatorname{id}_{B^A}$
Next, let $f: B \to C, g: C \to D$ be composable morphisms.
Then:
{{begin-eqn}}
{{eqn | l = g f \epsilon_B
      | r = g \map {\epsilon_C} {f^A \times \operatorname{id}_A}
      | c = Definition of $f^A$
}}
{{eqn | r = \map {\map {\epsilon_D} {g^A \times \operatorname{id}_A} } {f^A \times \operatorname{id}_A}
      | c = Definition of $g^A$
}}
{{eqn | r = \map {\epsilon_D} {g^A f^A \times \operatorname{id}_A}
}}
{{end-eqn}}
That is:
::<nowiki>$\begin{xy}\xymatrix@L+2mu@+1em{
D^A\times A \ar[rrr]^*+{\epsilon_D} &&& D \\
C^A\times A \ar[u]^*+{g^A \times \operatorname{id}_A} \ar[rr]^*+{\epsilon_C} && C \ar[ur]_*+{g} & \\
B^A\times A \ar[u]^*+{f^A \times \operatorname{id}_A} \ar[r]^*+{\epsilon_B} & B \ar[ur]_*+{f} && \\
}\end{xy}$</nowiki>
so that:
::<nowiki>$\begin{xy}\xymatrix@L+2mu@+1em{
D^A \times A \ar[r]^*+{\epsilon_D} & D \\
B^A \times A
  \ar[u]^*+{g^A f^A \times \operatorname{id}_A}
  \ar[ur]_*+{g f \epsilon_B}
}\end{xy}$</nowiki>
Thus, $g^A f^A$ is the morphism satisfying the universal mapping property for $\paren {g f}^A$.
That is to say:
:$\paren {g f}^A = g^A f^A$
Hence, $\paren -^A$ is a functor.
{{qed}}
Theorem: Euler's Cosine Identity

:$\cos z = \dfrac {e^{i z} + e^{-i z} } 2$

Proof:
Recall the definition of the cosine function:
{{begin-eqn}}
{{eqn | l = \cos z
      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}
      | c = 
}}
{{eqn | r = 1 - \frac {z^2} {2!} + \frac {z^4} {4!} - \frac {z^6} {6!} + \cdots + \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!} + \cdots
      | c = 
}}
{{end-eqn}}
Recall the definition of the exponential as a power series:
{{begin-eqn}}
{{eqn | l = e^z
      | r = \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}
      | c = 
}}
{{eqn | r = 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots
      | c = 
}}
{{end-eqn}}
Then, starting from the {{RHS}}:
{{begin-eqn}}
{{eqn | l = \frac {e^{i z} + e^{-i z} } 2
      | r = \frac 1 2 \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i z}^n} {n!} + \sum_{n \mathop = 0}^\infty \frac {\paren {-i z}^n} {n!} }
}}
{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^n + \paren {-i z}^n} {n!} }
      | c = Cosine Function is Absolutely Convergent
}}
{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^{2 n} + \paren {-i z}^{2 n} } {\paren {2 n}!} + \frac {\paren {i z}^{2 n + 1} + \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!} }
      | c = split into even and odd $n$
}}
{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n} + \paren {-i z}^{2 n} } {\paren {2 n}!}
      | c = $\paren {-i z}^{2 n + 1} = -\paren {i z}^{2 n + 1}$
}}
{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {2 \paren {i z}^{2 n} } {\paren {2 n}!}
      | c = $\left({ -1 }\right)^{2n} = 1$
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n} } {\paren {2 n}!}
      | c = cancel $2$
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}
      | c = $i^{2 n} = \paren {-1}^n$
}}
{{eqn | r = \cos z
}}
{{end-eqn}}
{{qed}}
Theorem: Euler's Cosine Identity

:$\cos z = \dfrac {e^{i z} + e^{-i z} } 2$

Proof:
Recall Euler's Formula:
:$e^{i z} = \cos z + i \sin z$
Then, starting from the {{RHS}}:
{{begin-eqn}}
{{eqn | l = \frac {e^{i z} + e^{-i z} } 2
      | r = \frac {\cos z + i \sin z + \map \cos {-z} + i \map \sin {-z} } 2
}}
{{eqn | r = \frac {\cos z + \map \cos {-z} } 2
      | c = Sine Function is Odd
}}
{{eqn | r = \frac {2 \cos z} 2
      | c = Cosine Function is Even
}}
{{eqn | r = \cos z
}}
{{end-eqn}}
{{qed}}
Theorem: Euler's Cosine Identity

:$\cos z = \dfrac {e^{i z} + e^{-i z} } 2$

Proof:
{{begin-eqn}}
{{eqn | n = 1
      | l = e^{i z}
      | r = \cos z + i \sin z
      | c = Euler's Formula
}}
{{eqn | n = 2
      | l = e^{-i z}
      | r = \cos z - i \sin z
      | c = {{Corollary|Euler's Formula}}
}}
{{eqn | ll= \leadsto
      | l = e^{i z} + e^{-i z}
      | r = \paren {\cos z + i \sin z} + \paren {\cos z - i \sin z}
      | c = $(1) + (2)$
}}
{{eqn | r = 2 \cos z
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = \frac {e^{i z} + e^{-i z} } 2
      | r = \cos z
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Euler's Cosine Identity

:$\cos z = \dfrac {e^{i z} + e^{-i z} } 2$

Proof:
Recall the definition of the real cosine function:
{{begin-eqn}}
{{eqn | l = \cos x
      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n!} }
      | c = 
}}
{{eqn | r = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \frac {x^6} {6!} + \cdots + \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} + \cdots
      | c = 
}}
{{end-eqn}}
Recall the definition of the exponential as a power series:
{{begin-eqn}}
{{eqn | l = e^x
      | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}
      | c = 
}}
{{eqn | r = 1 + x + \frac {x^2} 2 + \frac {x^3} 6 + \cdots + \frac {x^n} {n!} + \cdots
      | c = 
}}
{{end-eqn}}
Then, starting from the {{RHS}}:
{{begin-eqn}}
{{eqn | l = \frac {e^{i x} + e^{-i x} } 2
      | r = \frac 1 2 \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i x}^n} {n!} + \sum_{n \mathop = 0}^\infty \frac {\paren {-i x}^n} {n!} }
}}
{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^n + \paren {-i x}^n} {n!} }
      | c = Cosine Function is Absolutely Convergent
}}
{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i x}^{2 n} + \paren {-i x}^{2 n} } {\paren {2 n}!} + \frac {\paren {i x}^{2 n + 1} + \paren {-i x}^{2 n + 1} } {\paren {2 n + 1}!} }
      | c = split into even and odd $n$
}}
{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n} + \paren {-i x}^{2 n} } {\paren {2 n}!}
      | c = as $\paren {-i x}^{2 n + 1} = -\paren {i x}^{2 n + 1}$
}}
{{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {2 \paren {i x}^{2 n} } {\paren {2 n}!}
      | c = as $\paren {-1}^{2 n} = 1$
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {i x}^{2 n} } {\paren {2 n}!}
      | c = cancel $2$
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}
      | c = as $i^{2 n} = \paren {-1}^n$
}}
{{eqn | r = \cos x
}}
{{end-eqn}}
{{qed}}
Theorem: Euler's Cosine Identity

:$\cos z = \dfrac {e^{i z} + e^{-i z} } 2$

Proof:
Recall Euler's Formula:
:$e^{i x} = \cos x + i \sin x$
Then, starting from the {{RHS}}:
{{begin-eqn}}
{{eqn | l = \frac {e^{i x} + e^{-i x} } 2
      | r = \frac {\cos x + i \sin x + \map \cos {-x} + i \map \sin {-x} } 2
}}
{{eqn | r = \frac {\cos x + \map \cos {-x} } 2
      | c = Sine Function is Odd
}}
{{eqn | r = \frac {2 \cos x} 2
      | c = Cosine Function is Even
}}
{{eqn | r = \cos x
}}
{{end-eqn}}
{{qed}}
Theorem: Euler's Cosine Identity

:$\cos z = \dfrac {e^{i z} + e^{-i z} } 2$

Proof:
{{begin-eqn}}
{{eqn | n = 1
      | l = e^{i x}
      | r = \cos x + i \sin x
      | c = Euler's Formula
}}
{{eqn | n = 2
      | l = e^{-i x}
      | r = \cos x - i \sin x
      | c = Euler's Formula: Corollary
}}
{{eqn | ll= \leadsto
      | l = e^{i x} + e^{-i x}
      | r = \paren {\cos x + i \sin x} + \paren {\cos x - i \sin x}
      | c = $(1) + (2)$
}}
{{eqn | r = 2 \cos x
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = \frac {e^{i x} + e^{-i x} } 2
      | r = \cos x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Euler's Cosine Identity

:$\cos z = \dfrac {e^{i z} + e^{-i z} } 2$

Proof:
Consider the differential equation:
:$(1): \quad D^2_x \map f x = -\map f x$
subject to the initial conditions:
:$(2): \quad \map f 0 = 1$
:$(3): \quad D_x \map f 0 = 0$
=== Step 1 ===
We will prove that $y = \cos x$ is a particular solution of $(1)$.
{{begin-eqn}}
{{eqn | l = y
      | r = \cos x
      | c = 
}}
{{eqn | l = D^2_x y
      | r = D^2_x \cos x
      | c = taking second derivative of both sides
}}
{{eqn | r = \map {D_x} {-\sin x}
      | c = Derivative of Cosine Function
}}
{{eqn | r = -\map {D_x} {\sin x}
      | c = Derivative of Constant Multiple
}}
{{eqn | r = -\cos x
      | c = Derivative of Sine Function
}}
{{eqn | r = -y
      | c = 
}}
{{end-eqn}}
Thus $y = \cos x$ fulfils $(1)$.
Then from Cosine of Zero is One:
:$\cos 0 = 1$
Thus $y = \cos x$ fulfils $(2)$.
Then:
{{begin-eqn}}
{{eqn | l = D_x \cos 0
      | r = -\sin 0
      | c = Derivative of Cosine Function
}}
{{eqn | r = 0
      | c = Sine of Zero is Zero
}}
{{end-eqn}}
Thus $y = \cos x$ fulfils $(3)$.
So $y = \cos x$ is a particular solution of $(1)$.
{{qed|lemma}}
=== Step 2 ===
We will prove that $z = \dfrac {e^{i x} + e^{-i x} } 2$ is a particular solution of $(1)$.
{{begin-eqn}}
{{eqn | l = z
      | r = \frac {e^{i x} + e^{-i x} } 2
      | c = 
}}
{{eqn | l = D^2_x z
      | r = \map {D^2_x} {\frac {e^{i x} + e^{-i x} } 2}
      | c = taking second derivative of both sides
}}
{{eqn | r = \frac 1 2 \paren {D^2_x e^{i x} + D^2_x e^{-i x} }
      | c = Linear Combination of Derivatives
}}
{{eqn | r = \frac 1 2 \paren {i D_x e^{i x} - i D_x e^{-i x} }
      | c = Derivative of Exponential Function
}}
{{eqn | r = \frac 1 2 \paren {i^2 e^{i x} - i \paren {-i} e^{-i x} }
      | c = Derivative of Exponential Function
}}
{{eqn | r = \frac 1 2 \paren {- e^{i x} - e^{-i x} }
      | c = $i^2 = -1$
}}
{{eqn | r = -\frac {e^{i x} + e^{-i x} }2
      | c = 
}}
{{eqn | r = -z
      | c = 
}}
{{end-eqn}}
Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(1)$.
Then:
{{begin-eqn}}
{{eqn | l = \frac {e^{i \times 0} + e^{-i \times 0} } 2
      | r = \frac {1 + 1} 2
      | c = Exponential of Zero
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(2)$.
Then:
{{begin-eqn}}
{{eqn | l = \intlimits {D_x \frac {e^{i x} + e^{-i x} } 2} {x \mathop = 0} {}
      | r = \intlimits {\frac {i e^{i x} - i e^{-i x} } 2} {x \mathop = 0} {}
      | c = Derivative of Exponential Function
}}
{{eqn | r = \frac {i - i} 2
      | c = Exponential of Zero
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
Thus $z = \dfrac {e^{i x} + e^{-i x} } 2$ fulfils $(3)$.
So $z = \dfrac {e^{i x} + e^{-i x} } 2$ is a particular solution of $(1)$.
{{qed|lemma}}
We have shown that $y$ and $z$ are both particular solutions of $(1)$.
But a particular solution to a differential equation is unique.
{{MissingLinks|This result definitely needs backing up with an appropriate link}}
Therefore $y = z$.
That is:
:$\cos x = \dfrac {e^{i x} + e^{-i x} } 2$
{{qed}}
Theorem: Reciprocal of i

:$\dfrac 1 i = -i$

Proof:
{{begin-eqn}}
{{eqn | l = i^2
      | r = -1
      | c = {{Defof|Imaginary Unit}}
}}
{{eqn | ll= \leadsto
      | l = \frac {i^2} i
      | r = \frac {-1} i
}}
{{eqn | ll= \leadsto
      | l = i
      | r = \frac {-1} i
}}
{{eqn | ll= \leadsto
      | l = -i
      | r = \frac 1 i
}}
{{end-eqn}}
{{qed}}
Category:Imaginary Unit
Category:Examples of Reciprocals
m05ti2i4f7dvmp10bxd839pzlkqcc74
Theorem: Cosine of Difference

:$\map \cos {a - b} = \cos a \cos b + \sin a \sin b$

Proof:
{{begin-eqn}}
{{eqn | l = \map \cos {a - b}
      | r = \cos a \map \cos {-b} - \sin a \map \sin {-b}
      | c = Cosine of Sum
}}
{{eqn | r = \cos a \cos b - \sin a \map \sin {-b}
      | c = Cosine Function is Even
}}
{{eqn | r = \cos a \cos b + \sin a \sin b
      | c = Sine Function is Odd
}}
{{end-eqn}}
{{qed}}
Theorem: Rational Numbers are not Connected

The set of rational numbers $\Q$ is not a connected topological space.

Proof:
Let $\alpha \in \R$ be an irrational number.
By definition, $\alpha \notin \Q$.
Consider the sets:
:$S := \Q \cap \openint \gets \alpha$
:$T := \Q \cap \openint \alpha \to$
Let $x \in S$.
Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball of $x$ in $\Q$.
Then:
:$\forall x \in S: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq S$
by setting $\epsilon = \alpha - x$.
Similarly:
:$\forall x \in T: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq T$
by setting $\epsilon = x - \alpha$.
Thus $S$ and $T$ are open sets of $\Q$.
Then:
:$S \cup T = \Q$
:$S \cap T = \O$
:$S, T \ne \O$
So $S$ and $T$ fulfil the conditions for $S \mid T$ to be a separation of $\Q$.
Hence the result by definition of connected topological space.
{{qed}}
Theorem: Closure of Connected Set is Connected

Let $T$ be a topological space.
Let $H$ be a connected set of $T$.
Let $H^-$ denote the closure of $H$ in $T$. 
Then $H^-$ is connected in $T$.

Proof:
By Set is Subset of Itself, the result follows by setting $K = H^-$ in Set between Connected Set and Closure is Connected.
{{qed}}
Theorem: Closed Topologist's Sine Curve is Connected

Let $T$ be the closed topologist's sine curve.
Then $T$ is connected.

Proof:
{{Recall|Closed Topologist's Sine Curve}}
{{:Definition:Closed Topologist's Sine Curve}}
Because the open interval $\openint 0 \infty$ is connected, then so is $G$ by Continuous Image of Connected Space is Connected.
It is enough, from Set between Connected Set and Closure is Connected, to show that $J \subseteq \map \cl G$.
Let $p \in J$, say, $\tuple {0, y}$ where $-1 \le y \le 1$.
We need to show that:
:$\forall \epsilon > 0: \map {N_\epsilon} p \cap G \ne \O$
where $\map {N_\epsilon} p$ is the $\epsilon$-neighborhood of $p$.
Let us choose $n \in \N: \dfrac 1 {2 n \pi} < \epsilon$.
From Sine of Half-Integer Multiple of Pi:
:$\map \sin {\dfrac {\paren {4 n + 1} \pi} 2} = 1$
and:
:$\map \sin {\dfrac {\paren {4 n + 3} \pi} 2} = -1$
So by the Intermediate Value Theorem, $\map \sin {\dfrac 1 x}$ takes every value between $-1$ and $1$ in the closed interval $\closedint {\dfrac 2 {\paren {4 n + 3} \pi} } {\dfrac 2 {\paren {4 n + 1} \pi} }$.
In particular, $\map \sin {\dfrac 1 {x_0} } = y$ for some $x_0$ in this interval.
The distance between the points $\tuple {0, y}$ and $\tuple {x_0, \map \sin {\dfrac 1 {x_0} } } = \tuple {x_0, y}$ is $x_0 < \epsilon$.
So:
:$\tuple {x_0, \map \sin {\dfrac 1 {x_0} } } \in \map {N_\epsilon} p \cap G$
as required.
{{qed}}
Theorem: Separated Sets are Clopen in Union

Let $T = \left({S, \tau}\right)$ be a topological space.
Let $A$ and $B$ be separated sets in $T$.
Let $H = A \cup B$ be given the subspace topology.
Then $A$ and $B$ are each both open and closed in $H$.

Proof:
By hypothesis, $A$ and $B$ are separated:
:$A \cap B^- = A^- \cap B = \O$
Then:
{{begin-eqn}}
{{eqn | l = H \cap B^-
      | r = \paren {A \cup B} \cap B^-
}}
{{eqn | r = \paren {A \cap B^-} \cup \paren {B \cap B^-}
      | c = Intersection Absorbs Union
}}
{{eqn | r = \O \cup B
      | c = Set is Subset of its Topological Closure and Intersection with Subset is Subset
}}
{{eqn | r = B
      | c = Union with Empty Set
}}
{{end-eqn}}
Since the intersection of a closed set with a subspace is closed in the subspace, $B$ is closed in $H$.
{{explain|Link to the above result}}
Since $A = H \setminus B$ and $B$ is closed in $H$, $A$ is open in $H$.
{{explain|$A {{=}} H \setminus B$: it may be trivial but all statements are linked to or explained.}}
By the same argument with the roles of $A$ and $B$ reversed, $A$ is closed in $H$ and $B$ is open in $H$.
Hence the result.
{{qed}}
Category:Separated Sets
m1f02h1s1lcidd77jo0wfdkarkqzuml
Theorem: Compatibility of Atlases is Equivalence Relation

Let $M$ be a topological space.
Let $d$ and $k$ be natural numbers.
Let $\AA$ denote the set of all $d$-dimensional atlases of class $\CC^k$ on $M$.
Define a relation $\sim$ on $\AA$ by putting, for any two $\CC^k$-atlases $\FF$ and $\GG$:
:$\FF \sim \GG$ {{iff}} $\FF$ and $\GG$ are $C^k$-compatible.
Then $\sim$ is an equivalence relation on $\AA$.

Proof:
It is to be shown that $\sim$ is reflexive, symmetric and transitive.
Theorem: Addition of Real and Imaginary Parts

Let $z_0, z_1 \in \C$ be two complex numbers.
Then:
:$\map \Re {z_0 + z_1} = \map \Re {z_0} + \map \Re {z_1}$
and:
:$\map \Im {z_0 + z_1} = \map \Im {z_0} + \map \Im {z_1}$
Here, $\map \Re {z_0}$ denotes the real part of $z_0$, and $\map \Im {z_0}$ denotes the imaginary part of $z_0$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = z_0 + z_1
      | r = \paren {\map \Re {z_0} + i \, \map \Im {z_0} } + \paren {\map \Re {z_1} + i \, \map \Im {z_1} }
      | c = {{Defof|Complex Number}}
}}
{{eqn | r = \paren {\map \Re {z_0} + \map \Re {z_1} } + i \paren {\map \Im {z_0} + \map \Im {z_1} }
      | c = {{Defof|Complex Addition}}
}}
{{eqn | ll= \leadsto
      | l = \map \Re {z_0 + z_1}
      | r = \map \Re {z_0} + \map \Re {z_1}
      | c = {{Defof|Real Part}}
}}
{{eqn | l = \map \Im {z_0 + z_1}
      | r = \map \Im {z_0} + \map \Im {z_1}
      | c = {{Defof|Imaginary Part}}
}}
{{end-eqn}}
{{qed}}
Category:Complex Addition
ao3h6u72s4zm8bjt7jh7d7re2gufozy
Theorem: Multiplication of Real and Imaginary Parts

Let $w, z \in \C$ be complex numbers.
$(1)$ If $w$ is wholly real, then:
:$\map \Re {w z} = w \, \map \Re z$
and:
:$\map \Im {w z} = w \, \map \Im z$
$(2)$ If $w$ is wholly imaginary, then:
:$\map \Re {w z} = -\map \Im w \, \map \Im z$
and:
:$\map \Im {w z} = \map \Im w \, \map \Re z$
Here, $\map \Re z$ denotes the real part of $z$, and $\map \Im z$ denotes the imaginary part of $z$.

Proof:
Assume that $w$ is wholly real.
Then:
{{begin-eqn}}
{{eqn | l = w z
      | r = \map \Re w \, \map \Re z - \map \Im w \, \map \Im z + i \paren {\map \Re w \, \map \Im z + \map \Im w \, \map \Re z}
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = w \, \map \Re z + i w \, \map \Im z 
      | c = as $\map \Re w = w$ and $\map \Im w = 0$
}}
{{end-eqn}}
This equation shows that $\map \Re {w z} = w \, \map \Re z$, and $\map \Im {w z} = w \, \map \Im z$.
This proves $(1)$.
Now, assume that $w$ is wholly imaginary.
Then:
{{begin-eqn}}
{{eqn | l = w z
      | r = \map \Re w \, \map \Re z - \map \Im w \, \map \Im z + i \paren {\map \Re w \, \map \Im z + \map \Im w \, \map \Re z}
}}
{{eqn | r = -\map \Im w \, \map \Im z) + i \, \map \Im w \, \map \Re z
      | c = as $\map \Re w = 0$
}}
{{end-eqn}}
This equation shows that $\map \Re {w z} = -\map \Im w \, \map \Im z$, and $\map \Im {w c} = \map \Im w \, \map \Re z$.
This proves $(2)$.
{{qed}}
Category:Complex Multiplication
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Theorem: Rational Numbers are Totally Disconnected

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is a totally disconnected space.

Proof:
Let $x, y \in \Q$ such that $x \ne y$.
From Between two Rational Numbers exists Irrational Number, there exists $\alpha \in \R \setminus \Q$ such that $x < \alpha < y$.
From Rational Numbers are not Connected, it follows that $x$ and $y$ belong to different components of $\Q$.
As $x$ and $y$ are arbitrary, it follows that no rational number is in the same component as any other rational number.
That is, the components of $\Q$ are singetons.
Hence the result, by definition of a totally disconnected space.
{{qed}}
Theorem: Rational Numbers are Totally Disconnected

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ is a totally disconnected space.

Proof:
Follows from:
: Rational Number Space is Totally Separated
: Totally Separated Space is Totally Disconnected
{{qed}}
Theorem: Discrete Space is Totally Disconnected

Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Then $T$ is totally disconnected.

Proof:
Follows from:
* Discrete Space is Extremally Disconnected Hausdorff
* Extremally Disconnected Hausdorff Space is Totally Separated
* Totally Separated Space is Totally Disconnected
{{qed}}
Theorem: Rational Numbers are not Discrete Space

The set of rational numbers $\Q$ does not form a discrete space.

Proof:
Follows from:
:Rational Number Space is not Extremally Disconnected
:Discrete Space is Extremally Disconnected Hausdorff.
{{qed}}
Theorem: Rational Number Space is not Extremally Disconnected

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then $\struct {\Q, \tau_d}$ does not form an extremally disconnected space.

Proof:
{{Recall|Extremally Disconnected Space|extremally disconnected space}}
{{Definition:Extremally Disconnected Space/Definition 1}}
Hence the existence will be demonstrated of an open set in $\struct {\Q, \tau_d}$ whose closure is not open.
We have that $\openint 0 1$ is open in $\struct {\R, \tau_d}$.
Thus $\openint 0 1 \cap \Q$ is open in $\struct {\Q, \tau_d}$.
We show that $0$ and $1$ are limit points of $\openint 0 1 \cap \Q$.
For any $\epsilon \in \R_{>0}$:
:$\openint {-\epsilon} \epsilon \cap \Q \cap \openint 0 1 = \openint 0 \epsilon \cap \Q$
From Between two Real Numbers exists Rational Number:
:$\openint 0 \epsilon \cap \Q \ne \O$
Similarly:
:$\openint {1 - \epsilon} 1 \cap \Q \ne \O$
Thus $0$ and $1$ are limit points of $\openint 0 1 \cap \Q$.
Hence:
:$0, 1 \in \paren {\openint 0 1 \cap \Q}'$
Thus:
:$\closedint 0 1 \cap \Q \subseteq \paren {\openint 0 1 \cap \Q}^-$
where $\paren {\openint 0 1 \cap \Q}^-$ denotes the closure of $\openint 0 1 \cap \Q$.
Now we show that for any $x \notin \closedint 0 1$, $x$ is not a limit point of $\openint 0 1 \cap \Q$.
Suppose $x < 0$.
Then $\dfrac x 2 < 0$.
Thus:
:$\openint {x + \dfrac x 2} {x - \dfrac x 2} \cap \Q \cap \openint 0 1 = \O$
Similarly for $y > 1$:
:$\openint {y - \dfrac {y - 1} 2} {y + \dfrac {y - 1} 2} \cap \Q \cap \openint 0 1 = \O$
Thus:
:$\map \complement {\closedint 0 1 \cap \Q} \subseteq \map \complement {\paren {\openint 0 1 \cap \Q}^-}$
Hence we have:
:$\closedint 0 1 \cap \Q = \paren {\openint 0 1 \cap \Q}^-$
From Between two Real Numbers exists Rational Number:
:$\openint {-\epsilon} \epsilon \cap \Q \cap \map \complement {\closedint 0 1} = \openint {-\epsilon} 0 \cap \Q \ne \O$
A neighborhood of $0$ must intersect $\map \complement {\closedint 0 1}$.
So $\closedint 0 1 \cap \Q$ is not open in $\struct {\Q, \tau_d}$.
Hence $\openint 0 1$ is that open set in $\struct {\Q, \tau}$ whose closure is not open of which we were to demonstrate the existence.
Thus the result follows from definition of extremally disconnected space.
{{qed}}
Category:Rational Number Space
Category:Extremally Disconnected Spaces
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