avewright/proofwiki-math huggingface dataset:
Theorem: Normed Vector Space is Hausdorff

Let $\struct {X, \norm {\, \cdot \,}_X }$ be a normed vector space.
Let $\tau$ be the topology on $X$ that consists of all open sets in $X$.
That is, $U \in \tau$ {{iff}}:
:$\forall x \in U: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.
Then $\struct {X, \tau}$ is a $T_2$ (Hausdorff) space.

Proof:
From Metric Induced by Norm is Metric, it follows that the norm $\norm {\,\cdot\,}$ induces a metric $d$ on $X$.
From the definition of open set in normed vector space, it follows that $\tau$ is the topology induced by the metric $d$.
The result now follows from Metric Space is Hausdorff.
{{qed}}
Category:Normed Vector Spaces
Category:Hausdorff Spaces
51kjlmrvcq7x9swa5b660bclxhwxkzu
Theorem: Normed Vector Space is Hausdorff Topological Vector Space

Let $\struct { K, +_K, \circ_K }$ be a valued field with norm  $\norm {\,\cdot\,}_K$.
Let $\struct {X, \norm {\, \cdot \,}_X }$ be a normed vector space over $K$.
Let $\tau$ be the topology on $X$ that consists of all open sets in $X$.
That is, $U \in \tau$ {{iff}}:
:$\forall x \in U: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.
Then $\struct { X, \tau }$ is a Hausdorff topological vector space.

Proof:
By its definition, a valued field is a normed division ring.
Let $\tau_K$ be the topology on $K$ induced by the metric induced by the norm $\norm {\,\cdot\,}_K$
From Normed Division Ring Operations are Continuous:Corollary, it follows that $\struct {K, \tau_K}$ is a topological field.
Let $d_X : X \times X \to \R_{\ge 0}$ be the metric induced by the inner product norm $\norm {\,\cdot\,}_X$ on $X$.
By definition of open set in normed vector space, it follows that $\tau$ is the topology on $X$ induced by the metric $d_X$.
To show that $\tau$ is a topology on $X$, we examine each of the open set axioms in turn:
Theorem: Hilbert Space is Hausdorff Topological Vector Space

Let $\struct {H, \innerprod \cdot \cdot}$ be a Hilbert space over $\mathbb F \in \set {\R, \C}$.
Let $d: H \times H \to \R_{\ge 0}$ be the metric induced by the inner product norm $\norm {\,\cdot\,}$ on $H$.
Let $\tau$ be the the topology on $H$ induced by the metric $d$.
Then $\struct {H, \tau}$ is a Hausdorff topological vector space.

Proof:
If $\mathbb F = \R$, then Real Numbers form Field shows that $\mathbb F$ is a field.
From Euclidean Space is Normed Vector Space, it follows for $r \in \mathbb F$ that:
:$\size r = \sqrt {r^2}$
is a norm on $\mathbb F$.
If $\mathbb F = \C$, then Complex Numbers form Field shows that $\mathbb F$ is a field.
From Complex Modulus is Norm, it follows for $z \in \mathbb F$ that:
:$\cmod z$
is a norm on $\mathbb F$.
In both cases, it follows from definition of valued field that $\mathbb F$ is a normed division ring.
By definition of normed vector space, $\struct {H, \norm {\,\cdot\,} }$ is a normed vector space.
The result follows from Normed Vector Space is Hausdorff Topological Vector Space.
{{qed}}
Category:Hilbert Spaces
Category:Hausdorff Topological Vector Spaces
1y21oen13t7vqibcq1ydglf0ugwnivx
Theorem: Furstenberg Topology is Topology

Let $\struct {\Z, \tau}$ be the topological space formed by the Furstenberg topology on the set of integers $\Z$.
Then $\tau$ is indeed a topology on $\Z$.

Proof:
Recall the definition of the Furstenberg topology:
{{:Definition:Furstenberg Topology}}{{qed|lemma}}
In view of Union from Synthetic Basis is Topology it suffices to show that $\BB$ is a synthetic basis on $\Z$.
Recall the definition of synthetic basis:
{{Definition:Synthetic Basis/Definition 1}}
Theorem: Equivalence of Definitions of Hilbert Space

Let $H$ be a vector space over $\mathbb F \in \set {\R, \C}$.
{{TFAE|def = Hilbert Space|view = Hilbert space}}
=== Definition 1===
{{:Definition:Hilbert Space/Definition 1}}
=== Definition 2===
{{:Definition:Hilbert Space/Definition 2}}

Proof:
=== Definition 1 implies Definition 2 ===
By definition of complete metric space, every Cauchy sequence in $H$ is convergent.
By definition of Banach space, it follows that $\struct { H, \norm {\,\cdot\,}_H }$ is a Banach space.
{{qed|lemma}}
Theorem: Open Ball is Simply Connected

Let $\struct {V, \norm {\,\cdot\,} }$ be a normed vector space over $\R$ or $\C$.
Let $d: V \times V \to \R_{\ge 0}$ be the metric induced by the norm $\norm {\,\cdot\,}$ on $V$.
Let $\tau$ be the the topology on $V$ induced by the metric $d$.
Let $v \in V$ and $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} v$ be the open $\epsilon$-ball of $v$.
Let $\tau_B$ be the subspace topology on $\map {B_\epsilon} v$ induced by $\tau$.
Then $\struct{ \map {B_\epsilon} v, \tau_B }$ is simply connected.

Proof:
Normed Vector Space is Hausdorff Topological Vector Space shows that $\struct {V, \tau}$ is a topological vector space.
The result now follows from Open Ball is Convex Set and Convex Set is Simply Connected.
{{qed}}
Category:Normed Vector Spaces
Category:Open Balls
Category:Simply Connected Spaces
hzf6v9nvwl00p0jon7rg5s360l3us3x
Theorem: Basis Element of Furstenberg Topology is Clopen

Let $\tau$ be the Furstenberg topology on the set of integers $\Z$.
Let $a, b \in \Z$ such that $a \ne 0$.
Then $a \Z + b$ is clopen in $\struct {\Z, \tau}$.

Proof:
$a \Z + b \in \tau$ by {{Defof|Furstenberg Topology}}.
It remains to show:
:$\Z \setminus \paren {a \Z + b} \in \tau$
As $a \Z = \paren {-a} \Z$, we may assume $a > 0$.
If $a = 1$, then $\Z \setminus \Z = \O \in \tau$.
Thus we assume that $a \ge 2$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \Z \setminus \paren {a \Z + b}
}}
{{eqn | ll= \leadstoandfrom
      | l = x - b
      | o = \not \in
      | r = a \Z
}}
{{eqn | ll= \leadstoandfrom
      | l = x - b
      | o = \not \equiv
      | r = 0
      | rr = \pmod a
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall k \in \set {1, \ldots , a-1}
      | l = x - b
      | o = \equiv
      | r = k
      | rr = \pmod a
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall k \in \set {1, \ldots , a-1}
      | l = x - b
      | o = \in
      | r = a \Z + k
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | o = \in
      | r = \bigcup _{k \mathop \in \set {1, \ldots , a-1} } a \Z + k
}}
{{end-eqn}}
That is:
:$\ds \Z \setminus \paren {a \Z + b} = \bigcup _{k \mathop \in \set {1, \ldots , a-1} } a \Z + k$
where $a \Z + k \in \tau$ for all $k \in \set {1, \ldots , a-1}$.
Therefore $\Z \setminus \paren {a \Z + b} \in \tau$.
{{qed}}
Category:Furstenberg Topology
dwwt6jf0dujth8dbxh4t87icc4mtv4u
Theorem: Cosine of 2 is Strictly Negative

:$\cos 2 < 0$

Proof:
Recall the definition of the cosine:
{{Definition:Real Cosine Function}}
Thus:
{{begin-eqn}}
{{eqn | l = \cos 2
      | r = 1 - \paren {\frac {2^2} {2!} - \frac {2^4} {4!} } - \paren {\frac {2^6} {6!} - \frac {2^8} {8!} } - \cdots 
}}
{{eqn | r = 1 - \sum_{n \mathop = 1}^\infty A_n
}}
{{end-eqn}}
where:
{{begin-eqn}}
{{eqn | l = A_n
      | o = :=
      | r = \dfrac {2^{4 n - 2} } {\paren {4 n - 2}!} - \dfrac {2^{4 n} } {\paren {4 n}!}
}}
{{eqn | r = \dfrac {4 n \paren {4n - 1} 2^{4 n - 2} - 2^{4 n} } {\paren {4 n}!}
}}
{{eqn | r = \dfrac {n \paren {4n - 2} 2^{4 n} } {\paren {4 n}!}
}}
{{eqn | o = \ge
      | r = 0
      | c = for all $n \ge 1$
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \cos 2
      | o = \le
      | r = 1 - A_1
}}
{{eqn | r = 1 - \paren {\frac {2^2} {2!} - \frac {2^4} {4!} }
}}
{{eqn | r = - \dfrac 1 3
}}
{{eqn | o = <
      | r = 0
}}
{{end-eqn}}
{{qed}}
Category:Cosine Function
7hd7czp51ci0o80jtz66789ild9sten
Theorem: Non-Empty Open Set of Furstenberg Topology is Infinite

Let $\struct {\Z, \tau}$ be the topological space formed by the Furstenberg topology on the set of integers $\Z$.
Let $U \in \tau$ such that $U \ne \O$.
Then $U$ is an infinite set.

Proof:
{{ProofWanted}}
Category:Furstenberg Topology
bsxfvjbqv2zdugidjfr0ebjr5tsn8a2
Theorem: Finite Hausdorff Measure Implies Zero Higher Dimensional Measure

Let $n \in \N_{>0}$.
Let $F \subseteq \R^n$ be a subset of the real Euclidean space.
Let $s \in \R_{\ge 0}$.
Let $\map {\HH^s} \cdot$ denote the $s$-dimensional Hausdorff measure.
Then:
:$\map {\HH^s} F < +\infty \implies \forall t \in \R_{>s} : \map {\HH^t} F = 0$

Proof:
For each $\delta$-cover $\sequence {U_i}$ of $F$:
{{begin-eqn}}
{{eqn | l = \sum \size {U_i}^t
      | r = \sum \size {U_i}^s \size {U_i}^{t - s}
}}
{{eqn | o = \le
      | r = \delta^{t - s} \sum \size {U_i}^s
}}
{{end-eqn}}
Thus:
:$\map {\HH^t_\delta} F \le \delta^{t - s} \map {\HH^s_\delta} F$
Therefore:
{{begin-eqn}}
{{eqn | l = \map {\HH^t} F
      | r = \lim_{\delta \to 0^+} \map {\HH^t_\delta} F
}}
{{eqn | o = \le
      | r = \lim_{\delta \to 0^+} \delta^{t - s} \map {\HH^s_\delta} F
}}
{{eqn | r = \lim_{\delta \to 0^+} \delta^{t - s} \cdot \lim_{\delta \to 0^+} \map {\HH^s_\delta} F
}}
{{eqn | r = 0 \cdot \map {\HH^s} F
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{finish|details}}
Category:Hausdorff Measures
Category:Fractals
f5fcc8j5xvyc170xe3rv8fs49mvxpcv
Theorem: Strictly Positive Hausdorff Measure implies Infinite Lower Dimensional Measure

Let $n \in \N_{>0}$.
Let $F \subseteq \R^n$ be a subset of the real Euclidean space.
Let $\map {\HH^s} \cdot$ denote the $s$-dimensional Hausdorff measure.
Let $s \in \R_{\ge 0}$.
Then:
:$\map {\HH^s} F > 0 \implies \forall t \in \hointr 0 s : \map {\HH^t} F = +\infty$

Proof:
Let:
:$\exists t \in \hointr 0 s : \map {\HH^t} F < +\infty$
Then by Finite Hausdorff Measure Implies Zero Higher Dimensional Measure:
:$\map {\HH^s} F = 0$
Hence the result by Proof by Contraposition.
{{qed}}
Category:Measure Theory
Category:Fractals
pmp0ltjih7enimrhedgf40avfbbjz1b
Theorem: Hausdorff-Besicovitch Dimension is Well-Defined

Let $F \subseteq \R^n$ be a subset of the $n$-dimensional Euclidean space.
Let $\map {\HH^s} F$ be the $s$-dimensional Hausdorff measure on $\R^n$ of $F$ for each $s \in \R_{\ge 0}$.
Then:
{{begin-eqn}}
{{eqn | o =
      | r = \inf \set {s \in \R_{\ge 0} : \map {\HH^s} F = 0}
}}
{{eqn | r = \sup \set {s \in \R_{\ge 0} : \map {\HH^s} F = +\infty}
}}
{{eqn | o = \in
      | r = \closedint 0 n
}}
{{end-eqn}}
In particular, the '''Hausdorff-Besicovitch dimension''' is well-defined.

Proof:
{{ProofWanted}}
Category:Hausdorff-Besicovitch Dimension
br8947i6jj969phdfqr0ryvavqdl3wl
Theorem: Higher Dimensional Hausdorff Measure than Euclidean Space is Zero

Let $\R^n$ be the $n$-dimensional Euclidean space.
Let $s \in \R_{>n}$.
Then:
:$\map {\HH ^s} {\R^n} = 0$
where $\map {\HH^s} \cdot$ denotes the $s$-dimensional Hausdorff measure on $\R^n$

Proof:
Let $s \in \R_{>n}$.
Consider the $n$-cube:
:$Q := {\closedint 0 1}^n$
Then:
:$\ds \R^n = \bigcup _{x \mathop \in \Z^n} Q + x$
As $\map {\HH ^s} \cdot$ is countably subadditive and translation invariant, it suffices to show:
:$\map {\HH ^s} Q = 0$
Let $N \in \N_{>0}$.
Let $\CC_N$ be the set of $n$-cubes:
:$\ds \closedint {\dfrac {i_1 - 1} {2^N} } {\dfrac {i_1} {2^N} } \times \closedint {\dfrac {i_2 - 1} {2^N} } {\dfrac {i_2} {2^N} } \times \cdots \times \closedint {\dfrac {i_n - 1} {2^N} } {\dfrac {i_n} {2^N} }$
where $i_1,\ldots ,i_n \in \set {1, 2, 3, \ldots, 2^N}$.
By Diameter of N-Cube:
:$\forall Z \in \CC_N : \size Z = \dfrac {\sqrt n} {2^N}$
where $\size Z$ denotes the diameter of $Z$.
Thus:
{{begin-eqn}}
{{eqn | l = \map {\HH ^s} Q
      | o = \le
      | r = \sum_{Z \mathop \in \CC_N} {\size Z}^s
}}
{{eqn | r = 2^{nN} \paren {\dfrac {\sqrt n} {2^N} }^s
}}
{{eqn | r = {\sqrt n}^s 2^{N \paren {n - s} }
}}
{{eqn | o = \to
      | r = 0
      | c = as $N \to +\infty$
}}
{{end-eqn}}
{{finish|Linking is not yet compete}}
Category:Hausdorff Measures
Category:Fractals
meo5i1y6umtats8yd4b65n4fom1rtsm
Theorem: Orientation of Contour is Well-Defined

Let $C$ be a contour in the complex plane $\C$.
Let $D \subseteq \C$ be a connected domain.
Let $\Img C \subseteq \partial D$, where $\Img C$ denotes the image of $C$, and $\partial D$ denotes the boundary of $D$.
The orientation of $C$ with respect to $D$, if it exists, does not depend on the choice of parameterization of $C$.

Proof:
By definition of contour, $C$ is a concatenation of a finite sequence of directed smooth curves $C_1, \ldots, C_n$.
Let each directed smooth curve $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k}{b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Let $\sigma_k: \closedint {c_k}{d_k}$ be a reparameterization of $C_k$.
By definition of directed smooth curve, there exists a bijective differentiable strictly increasing real function:
:$\phi_k: \closedint c d \to \closedint a b$
such that $\sigma_k = \gamma_k \circ \phi_k$.
Let $\sigma: \closedint c d \to \C$ be the parameterization of $C$ defined by $\sigma_1, \ldots, \sigma_n$.
Set $K_\sigma := \set { t \in \closedint c d : \textrm{ $\sigma$ is not differentiable at $t$ }  }$.
If $t \in \openint c d \setminus K_\sigma$, then there exists $k \in \set {1, \ldots, n}$ such that $t \in \openint {c_k}{d_k}$, and:
{{begin-eqn}}
{{eqn | l = \map {\gamma_k'}{ \map {\phi_k} t } \map {\phi_k'} t
      | r = \map {\sigma_k'} t 
      | c = by Derivative of Complex Composite Function
}}
{{eqn | ll = \leadsto
      | l = \map {\gamma_k'}{ \map {\phi_k} t }
      | r = \dfrac { \map {\sigma_k'} t }{ \map {\phi_k'} t }
}}
{{end-eqn}}
Suppose that $C$ is positively oriented with respect to $D$.
It follows that there exists $r \in \R_{>0}$ such that:
:$\forall \epsilon \in \openint 0 r : \map \gamma t + \epsilon i \map {\gamma_k'}{ \map {\phi_k} t } \in D, \map \gamma t - \epsilon i \map {\gamma_k'}{ \map {\phi_k} t } \notin D$
When we replace the expression for $\map {\gamma_k'}{ \map {\phi_k} t }$, we have:
:$\map \gamma t + \dfrac {\epsilon}{ \map {\phi_k'} t } i \map {\sigma_k'} t \in D,\map \gamma t -\dfrac {\epsilon}{ \map {\phi_k'} t } i \map {\sigma_k'} t \notin D$
where $\dfrac {\epsilon}{ \map {\phi_k'} t } \in \openint 0 {\dfrac r {\map {\phi_k'} t} }$.
This shows that $C$ is positively oriented with respect to $D$, when we use the parameterization $\sigma$ of $C$.
When $C$ is negatively oriented with respect to $D$, the proof is similar.
{{qed}}
Category:Orientation of Complex Contour
q7tjxlomgk2u6rtzpzhwglzrnvk4yzc
Theorem: Reversed Contour Reverses Orientation

Let $C$ be a contour in the complex plane $\C$.
Let $D \subseteq \C$ be a connected domain.
Let $\Img C \subseteq \partial D$, where $\Img C$ denotes the image of $C$, and $\partial D$ denotes the boundary of $D$.
Let $-C$ be the reversed contour of $C$.
If $C$ is positively oriented {{WRT}} $D$, then $-C$ is negatively oriented {{WRT}} $D$.
If $C$ is negatively oriented {{WRT}} $D$, then $-C$ is positively oriented {{WRT}} $D$.

Proof:
By definition of contour, $C$ is a concatenation of a finite sequence of directed smooth curves $C_1, \ldots, C_n$.
Let each directed smooth curve $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Let $\gamma: \closedint a b \to \C$ be the parameterization of $C$ defined by $\gamma_1, \ldots, \gamma_n$.
By definition of reversed contour, $-C$ is a concatenation of $-C_n , -C_{n-1}, \ldots, -C_1$.
By definition of reversed directed smooth curve, $-C_k$ is parameterized by the smooth path $\rho_k: \closedint {a_k} {b_k} \to \C$, defined by $\map {\rho_k} t = \map {\gamma_k} {a_k + b_k - t}$.
Let $\rho: \closedint c d \to \C$ be the parameterization of $-C$ defined by $\rho_n, \rho_{n-1}, \ldots, \rho_1$.
With $\psi_k: \closedint {a_k} {b_k} \to \closedint {a_k} {b_k}$ defined by $\map {\psi_k} t = a_k + b_k - t$, we have $\rho_k = \gamma_k \circ \psi_k$.
Suppose $C$ is positively oriented {{WRT}} $D$.
Set $K_\rho := \set {u \in \closedint c d : \textrm{ $\rho$ is not differentiable at $u$ } }$.
If $u \in \openint c d \setminus K_\rho$, then there exists $k \in \set {1, \ldots, n}$ such that $t \in \openint {a_k} {b_k}$ with $\map \rho u = \map {\rho_k} t = \map {\gamma_k} {a_k + b_k - t}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\rho'} u
      | r = \map {\rho_k'} t 
      | c = {{Defof|Reversed Contour (Complex Plane)}}
}}
{{eqn | r = \map {\gamma_k'} {\map \psi t} \map {\psi'} t
      | c = Derivative of Complex Composite Function
}}
{{eqn | r = - \map {\gamma_k'} {a_k + b_k - t}
      | c = Derivatives of Function of a x + b
}}
{{end-eqn}}
By definition of positive orientation, there exists $r \in \R_{>0}$ such that:
:for all $\epsilon \in \openint 0 r$ : $\map {\gamma_k} {a_k + b_k - t} + \epsilon i \map {\gamma_k'} {a_k + b_k - t} \in D$, and $\map {\gamma_k} {a_k + b_k - t} - \epsilon i \map {\gamma_k'} {a_k + b_k - t} \notin D$
It follows that $-C$ is negatively oriented {{WRT}} $D$, as:
:for all $\epsilon \in \openint 0 r$ : $\map \rho u - \epsilon i \map {\rho'} u \in D$, and $\map \rho u + \epsilon i \map {\rho'} u \notin D$
If $C$ is negatively oriented {{WRT}} $D$, the proof is similar.
{{qed}}
Category:Reversed Contour Reverses Orientation
Category:Orientation of Complex Contour
131z172jsj9jardjzmxhwh00kjdryvx
Theorem: L-2 Space forms Hilbert Space

Let $\struct{ X, \Sigma, \mu }$ be a measure space.
Let $\map {L^2} \mu$ be the $L^2$ space of $\mu$.
Let $\innerprod \cdot \cdot$ be the inner product on $\map {L^2} \mu$.
Then $\map {L^2} \mu$ endowed with $\innerprod \cdot \cdot$ is a Hilbert space.

Proof:
{{ProofWanted|We only need the completeness, rest is already up}}
Theorem: Real Vector Space with Dot Product is Hilbert Space

Let $\R^d$ be a real vector space with $d$ dimensions.
Let $\innerprod \cdot \cdot$ be the dot product on $\R^d$.
Then $\R^d$ endowed with $\innerprod \cdot \cdot$ is a Hilbert space.

Proof:
Let us explore the inner product norm of $\innerprod \cdot \cdot$:
:$\ds \norm x = \sqrt{ \sum_{i \mathop = 1}^d x_i^2 }$
and subsequently the metric induced by $\norm \cdot$:
:$\ds \map d {x, y} = \norm {x - y} = \sqrt{ \sum_{i \mathop = 1}^d \paren{ x_i - y_i }^2 }$
which is seen to be the definition of the Euclidean metric.
By Euclidean Space is Complete Metric Space, it follows that $\R^d$ is complete with respect to $d$.
That is, $\R^d$ is a Hilbert space.
{{qed}}
Theorem: Complex Vector Space with Dot Product is Hilbert Space

Let $\C^d$ be a complex vector space with $d$ dimensions.
Let $\innerprod \cdot \cdot$ be the dot product on $\C^d$.
{{mistake|This dot product is only defined on $\C$, not on $\C^d$}}
Then $\C^d$ endowed with $\innerprod \cdot \cdot$ is a Hilbert space.

Proof:
Let us explore the inner product norm of $\innerprod \cdot \cdot$:
:$\ds \norm z = \sqrt{ \sum_{i \mathop = 1}^d x_i^2 + y_i^2 }$
and subsequently the metric induced by $\norm \cdot$:
:$\ds \map d {z, z'} = \norm {z - z'} = \sqrt{ \sum_{i \mathop = 1}^d \paren{ x_i - x'_i }^2 + \paren{ y_i - y'_i }^2 }$
{{ProofWanted|Elementary results around $\C^d$ seem to be missing}}
Theorem: Sum of Degenerate Linear Transformation is Degenerate

Let $U, V$ be vector spaces over a field $K$.
Let $S: U \to V$ be a degenerate linear transformation.
Let $T: U \to V$ be a degenerate linear transformation.
Then $S + T$ is a degenerate linear transformation.

Proof:
Let $\set {s_1, \ldots, s_m}$ be a generator of $\Img S$.
Let $\set {t_1, \ldots, t_n}$ be a generator of $\Img T$.
Then $\set {s_1, \ldots, s_m, t_1, \ldots, t_n}$ is a generator of $\Img {S + T}$.
By Cardinality of Generator of Vector Space is not Less than Dimension:
:$\map \dim {\Img {S + T}} \le m + n$
{{qed}}
Theorem: Product with Degenerate Linear Transformation is Degenerate

Let $U, V, W$ be vector spaces over a field $K$.
Let $G: U \to V$ be a degenerate linear transformation.
Let $N: W \to U$ be a linear transformation.
Then $G \circ N$ is degenerate.

Proof:
Recall that the dimension of subspace is not greater than its super space.
Thus the claim follows from:
:$\Img {G \circ N} \subseteq \Img G$
{{qed}}
Theorem: Right Product with Degenerate Linear Transformation is Degenerate

Let $U, V, W$ be vector spaces over a field $K$.
Let $G: U \to V$ be a degenerate linear transformation.
Let $M: V \to W$ be a linear transformation.
Then $M \circ G$ is degenerate.

Proof:
Let $\set {s_1, \ldots, s_n}$ be a generator of $\Img G$.
Then $\set {\map M {s_1}, \ldots, \map M {s_n} }$ is a generator of $\Img {M \circ G}$.
By Cardinality of Generator of Vector Space is not Less than Dimension:
:$\map \dim {\Img {M \circ G}} \le n$
{{qed}}
Theorem: Invariance of Pseudoinverse under Addition of Degenerate Transformation

Let $U, V$ be vector spaces over a field $K$.
Let $S: U \to V$ be a linear transformation.
Let $T: V \to U$ be a linear transformation.
Let $S$ and $T$ are pseudoinverse to each other.
Then $S + G_1$ and $T + G_2$ are pseudoinverse to each other, where:
:$G_1: U \to V$ is an arbitrary degenerate linear transformation
:$G_2: V \to U$ is an arbitrary degenerate linear transformation

Proof:
Let:
{{begin-eqn}}
{{eqn | n = 1
      | l = G_3
      | o = :=
      | r = T \circ S - I_U
}}
{{eqn | n = 2
      | l = G_4
      | o = :=
      | r = S \circ T - I_V
}}
{{end-eqn}}
By {{Defof|Pseudoinverse of Linear Transformation}}, $G_3, G_4$ are degenerate.
Then:
{{begin-eqn}}
{{eqn | l = \paren {T + G_2} \circ \paren {S + G_1} - I_U
      | r = T \circ S + T \circ G_1 + G_2 \circ \paren {S + G_1} - I_U
}}
{{eqn | r = \paren {T \circ S - I_U} + T \circ G_1 + G_2 \circ \paren {S + G_1}
}}
{{eqn | r = G_3 +  T \circ G_1 + G_2 \circ \paren {S + G_1}
      | c = by $\paren 1$
}}
{{end-eqn}}
which is degenerate in view of:
:Product with Degenerate Linear Transformation is Degenerate
:Right Product with Degenerate Linear Transformation is Degenerate
:Product with Degenerate Linear Transformation is Degenerate
Similarly:
{{begin-eqn}}
{{eqn | l = \paren {S + G_1} \circ \paren {T + G_2} - I_V
      | r = S \circ T + S \circ G_2 + G_1 \circ \paren {T + G_2} - I_V
}}
{{eqn | r = \paren {S \circ T - I_V} + S \circ G_2 + G_1 \circ \paren {T + G_2}
}}
{{eqn | r = G_4 + S \circ G_2 + G_1 \circ \paren {T + G_2}
      | c = by $\paren 2$
}}
{{end-eqn}}
which is degenerate.
{{qed}}
Theorem: Number of Permutations of All Elements

Let $S$ be a set of $n$ elements.
The number of permutations of $S$ is $n!$

Proof:
We are seeking to calculate the number of $r$-permutations of $S$, that is ${}^n P_r$, where $r = n$.
Hence:
{{begin-eqn}}
{{eqn | l = {}^n P_n
      | r = \dfrac {n!} {\paren {n - n}!}
      | c = Number of Permutations
}}
{{eqn | r = n!
      | c = {{Defof|Factorial}}
}}
{{end-eqn}}
{{qed}}
Theorem: Number of Permutations of All Elements

Let $S$ be a set of $n$ elements.
The number of permutations of $S$ is $n!$

Proof:
We pick the elements of $S$ in an arbitrary order.
There are $n$ elements of $S$, so there are $n$ options for the first element.
Then there are $n - 1$ elements left in $S$ that we have not picked, so there are $n - 1$ options for the second element.
Then there are $n - 2$ elements left, so there are $n - 2$ options for the third element.
And so on, until there are $3$, then $2$, then $1$ remaining elements.
Each mapping is independent of the choices made for all the other mappings.
So, by the Product Rule for Counting, the total number of ordered selections from $S$:
{{begin-eqn}}
{{eqn | l = {}^n P_n
      | r = n \paren {n - 1} \paren {n - 2} \cdots 3 \times 2 \times 1
}}
{{eqn | r = n!
      | c = {{Defof|Factorial}}
}}
{{end-eqn}}
{{qed}}
Theorem: Number of Permutations of All Elements

Let $S$ be a set of $n$ elements.
The number of permutations of $S$ is $n!$

Proof:
From the definition, it can be seen that a bijection $f: S \to S$ is an $n$-permutation.
Hence, from Cardinality of Set of Bijections the number of $n$-permutations on a set of $n$ elements is:
:${}^n P_n = \dfrac {n!} {\paren {n - n}!} = n!$
{{qed}}
Theorem: Real Semi-Inner Product is Complex Semi-Inner Product

Let $V$ be a vector space over a real subfield $\GF$.
Let $\innerprod \cdot \cdot: V \times V \to \GF$ be a real semi-inner product.
Then $\innerprod \cdot \cdot$ is a complex semi-inner product.

Proof:
This follows immediately from:
* {{Defof|Real Semi-Inner Product}}
* {{Defof|Complex Semi-Inner Product}}
* Complex Number equals Conjugate iff Wholly Real
{{qed}}