avewright/proofwiki-math huggingface dataset:
Theorem: Ring by Idempotent

Let $\struct {A, +, \circ}$ be a commutative ring.
Let $e$ be an idempotent element of $A$.
Then the ideal $I := \ideal e$ generated by $e$ is a commutative ring with unity $\struct {I, +, \circ}$ with unity $e$.

Proof:
Because $\struct {I, +}$ is an ideal of $\struct {A, +, \circ}$, it follows that $\struct {I, +, \circ}$ is a ring (not necessarily unital).
{{explain|Why is $\struct {I, +}$ an ideal?
By definition it is the ideal generated by $e$. --Wandynsky (talk) 13:04, 30 July 2021 (UTC)
I don't see this anywhere in the Definition:Generator of Ideal of Ring or Definition:Generated Ideal of Ring.}}
By definition of generated ideal, $I$ is the intersection of all ideals of $A$ containing $e$.
Thus $e \mathop \in I$.
{{explain|No not "clearly", we need to link to a result. I'm sure it's there, somewhere under Cosets.
Ah i produced some confusion with the notation. Here $eA$ is defined as $\ideal e$. I don't want to use any cosets.
There should be some theorem like $(e) {{=}} eA$, where the RHS are the actual cosets. But I can't find it. --Wandynsky (talk) 13:26, 30 July 2021 (UTC)
There is certainly a page which gives $e \in I$, for a start. As for a page containing $\ideal e {{=}} e A$, there is no such page.
Thank you. Now the only things left are refactoring Definition:Generator of Ideal, proving it is an ideal (above) and proving the general form of an element of $I$ (below). --Wandynsky (talk) 15:59, 30 July 2021 (UTC)}}
Let $e a \in I$ for some arbitrary $a \in A$.
{{explain|theorem on the form of a general element of $I$ --Wandynsky (talk) 15:50, 30 July 2021 (UTC)}}
Then
{{begin-eqn}}
{{eqn | l = e \paren {e a}
      | r = e^2 a
      | c = 
}}
{{eqn | r = e a
      | c = 
}}
{{end-eqn}}
It follows that $e$ is a unity of $\struct {I, +, \circ}$.
Hence $\struct {I, +, \circ}$ is a commutative ring with unity with unity $e$.
{{qed}}
Category:Ring Theory
oblef72j5iknbj4bf5yhrk1mljgrwb8
Theorem: Ring Homomorphism by Idempotent

Let $A$ be a commutative ring.
Let $e \in A$ be an idempotent element.
Let $\ideal e$ be the ideal of $A$ generated by $e$.
Then the mapping:
:$f: A \to \ideal e: a \mapsto e a$ 
is a surjective ring homomorphism with kernel the ideal $\ideal {1 - e}$ generated by $1 - e$.

Proof:
Let $e a \in \ideal e$ for some arbitrary $a \in A$.
{{explain|Technically, one needs to prove that a general element of $\ideal e$ has this form. --Wandynsky (talk) 15:38, 30 July 2021 (UTC)
Go to it.}}
Then
{{begin-eqn}}
{{eqn | l = \map f {e a}
      | r = e e a
      | c = Definition of $f$
}}
{{eqn | r = ea
      | c = $e$ is idempotent
}}
{{end-eqn}}
So $f$ is surjective.
Let $x, y \in A$.
Then 
{{begin-eqn}}
{{eqn | l = \map f {x + y}
      | r = e \paren {x + y}
      | c = Definition of $f$
}}
{{eqn | r = e x + e y
      | c = {{Ring-axiom|D}}
}}
{{eqn | r = \map f x + \map f y
      | c = Definition of $f$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map f {x y}
      | r = e x y
      | c = Definition of $f$
}}
{{eqn | r = e^2 x y
      | c = $e$ is idempotent
}}
{{eqn | r = e x e y
      | c = $A$ is commutative
}}
{{eqn | r = \map f x \map f y
      | c = Definition of $f$
}}
{{end-eqn}}
So $f$ is a ring homomorphism.
We have
{{begin-eqn}}
{{eqn | l = \map f {1 - e}
      | r = e \paren {1 - e}
      | c = Definition of $f$
}}
{{eqn | l = 
      | r = e^2 - e
      | c = {{Ring-axiom|D}}
}}
{{eqn | l = 
      | r = 0
      | c = $e$ is idempotent
}}
{{end-eqn}}
Thus $1 - e \in \map \ker f$ and $\ideal {1 - e} \subseteq \map \ker f$.
{{explain|Technically, one needs a result of the form: 'set contained in ideal' implies 'ideal generated by set contained in ideal' --Wandynsky (talk) 15:38, 30 July 2021 (UTC)
Go to it.}}
Suppose $a \in \map \ker f$.
Then $e a = \map f a = 0$.
{{begin-eqn}}
{{eqn | l = a
      | r = e a + \paren {1 - e} a
      | c = {{Ring-axiom|D}}
}}
{{eqn | l = 
      | r = \paren {1 - e} a
      | c = $e a = 0$
}}
{{end-eqn}}
Thus:
:$\map \ker f \subseteq \ideal {1 - e}$
so:
:$\map \ker f = \ideal {1 - e}$
{{qed}}
Theorem: Inner Product is Sesquilinear

Let $\mathbb F$ be a subfield of $\C$. 
Let $V$ be a inner product space over $V$ with inner product $\innerprod \cdot \cdot$. 
Define the $f : V \times V \to \mathbb F$ by: 
:$\map f {x, y} = \innerprod x y$
for each $x, y \in V$. 
Then $f$ is sesquilinear.

Proof:
Let $\alpha \in \mathbb F$. 
Let $x_1, x_2, y \in V$. 
By the definition of the inner product, $f$ is linear in its first argument.
So, we have: 
:$\innerprod {\alpha x_1 + x_2} y = \alpha \innerprod {x_1} y + \innerprod {x_2} y$
From the definition of the inner product, we also have that $f$ is conjugate symmetric, so: 
{{begin-eqn}}
{{eqn	| l = \innerprod y {\alpha x_1 + x_2}
	| r = \overline {\innerprod {\alpha x_1 + x_2} y}
}}
{{eqn	| r = \overline {\alpha \innerprod {x_1} y + \innerprod {x_2} y}
}}
{{eqn	| r = \overline {\alpha \innerprod {x_1} y} + \overline {\innerprod {x_2} y}
	| c = Sum of Complex Conjugates
}}
{{eqn	| r = \overline \alpha \overline {\innerprod {x_1} y} + \overline {\innerprod {x_2} y}
	| c = Product of Complex Conjugates
}}
{{eqn	| r = \overline \alpha \innerprod y {x_1} + \innerprod y {x_2}
	| c = using the conjugate symmetry of the inner product
}}
{{end-eqn}}
So we have: 
:$\innerprod {\alpha x_1 + x_2} y = \alpha \innerprod {x_1} y + \innerprod {x_2} y$
and:
:$\innerprod y {\alpha x_1 + x_2} = \overline \alpha \innerprod y {x_1} + \innerprod y {x_2}$
for $\alpha \in \mathbb F$ and $x_1, x_2, y \in V$. 
So $f$ is sesquilinear.
{{qed}}
Category:Inner Product Spaces
4y814i9slpnyvgbwatumvsz5rf1gmxv
Theorem: Unital Ring Homomorphism by Idempotent

Let $A$ be a commutative ring with unity.
Let $e \in A$ be an idempotent element.
Let $\ideal e$ be the ideal of $A$ generated by $e$.
Then the mapping:
:$f: A \to \ideal e: a \mapsto e a$ 
is a surjective unital ring homomorphism from $\struct {A, +, \circ}$ to $\struct {\ideal e, +, \circ}$ with kernel the ideal $\ideal {1 - e}$ generated by $1 - e$.

Proof:
By Ring Homomorphism by Idempotent $f$ is a surjective ring homomorphism with kernel $\ideal {1 - e}$.
By Ring by Idempotent $A$ is a commutative ring with unity whose unity is $e$.
Then
{{begin-eqn}}
{{eqn | l = \map f 1
      | r = e \cdot 1
      | c = Definition of $f$
}}
{{eqn | r = e
      | c = $1$ is the unity of $A$.
}}
{{end-eqn}}
Thus $f$ is a unital ring homomorphism.
{{qed}}
Category:Ring Theory
Category:Commutative Rings
Category:Ring Homomorphisms
6w10zbhlmoxphyv6uztpkwfskpir8kc
Theorem: Equivalence of Definitions of Generated Normal Subgroup

{{TFAE|def = Generated Normal Subgroup}}
Let $G$ be a group.
Let $S \subseteq G$ be a subset.

Proof:
Let $H$ be the smallest normal subgroup containing $S$.
Let $\mathbb S$ be the set of normal subgroups containing $S$.
To show the equivalence of the two definitions, we need to show that $\ds H = \bigcap \mathbb S$.
Since $H$ is a normal subgroup containing $S$:
:$H \in \mathbb S$
By Intersection is Subset:
:$\ds \bigcap \mathbb S \subseteq H$
On the other hand, by Intersection of Normal Subgroups is Normal:
:$\ds \bigcap \mathbb S$ is a normal subgroup containing $S$.
Since $H$ be the smallest normal subgroup containing $S$:
:$\ds H \subseteq \bigcap \mathbb S$
By definition of set equality:
:$\ds H = \bigcap \mathbb S$
Hence the result.
{{qed}}
Category:Generated Normal Subgroups
0j0h8emlipxplaqioage29l1xjsocww
Theorem: Grothendieck Universe is Closed under Mappings

Let $\mathbb U$ be a Grothendieck universe.
Let $u, v \in \mathbb U$.
Let $f: u \to v$ be a mapping realized as a relation consisting of ordered pairs in Kuratowski formalization.
Then $f \in \mathbb U$.

Proof:
Let $u \times v$ be the finite cartesian product of $u$ and $v$ realized as a set of ordered pairs in Kuratowski formalization.
By definition of mapping, we have $f \subseteq u \times v$.
Then:
{{begin-eqn}}
{{eqn | l = u, v
      | o = \in
      | r = \mathbb U
}}
{{eqn | ll= \leadsto
      | l = u \times v
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe is Closed under Binary Cartesian Product
}}
{{eqn | ll= \leadsto
      | l = f
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe is Closed under Subset and $f \subseteq u \times v$
}}
{{end-eqn}}
{{qed}}
Category:Grothendieck Universes
97jqacy0lmpvpw5ahpgqfypzsios0k1
Theorem: Grothendieck Universe is Closed under Subset

Let $\mathbb U$ be a Grothendieck universe.
Let $u \in \mathbb U$.
Let $v \subseteq u$ be a subset of $u$.
Then $v \in \mathbb U$.

Proof:
{{begin-eqn}}
{{eqn | l = v
      | o = \in
      | r = \powerset u
      | c = {{Defof|Power Set}}
}}
{{eqn | l = u
      | o = \in
      | r = \mathbb U
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = \powerset u
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe: Axiom $(3)$
}}
{{eqn | ll= \leadsto
      | l = v
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe: Axiom $(1)$
}}
{{end-eqn}}
{{qed}}
Category:Grothendieck Universes
f74l7s0ckv3ftnc555bvimch11wu572
Theorem: Grothendieck Universe is Closed under Binary Cartesian Product

Let $\mathbb U$ be a Grothendieck universe.
Let $u, v \in \mathbb U$.
Let $u \times v$ be the binary cartesian product of $u$ and $v$ realized as a set of ordered pairs in Kuratowski formalization.
Then $u \times v \in \mathbb U$.

Proof:
From Binary Cartesian Product in Kuratowski Formalization contained in Power Set of Power Set of Union:
:$u \times v \subseteq \powerset {\powerset {u \cup v} }$ 
Then:
{{begin-eqn}}
{{eqn | l = u \cup v 
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe is Closed under Binary Union
}}
{{eqn | ll= \leadsto
      | l = \powerset {u \cup v}
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe: Axiom (3)
}}
{{eqn | ll= \leadsto
      | l = \powerset {\powerset {u \cup v} }
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe: Axiom (3)
}}
{{eqn | ll= \leadsto
      | l = u \times v
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe is Closed under Subset
}}
{{end-eqn}}
{{qed}}
Category:Grothendieck Universes
4nl4n0lkdpuey3a13u16cqju8okhjmy
Theorem: Grothendieck Universe is Closed under Binary Union

Let $\mathbb U$ be a Grothendieck universe.
Let $u, v \in \mathbb U$.
Then $u \cup v \in \mathbb U$.

Proof:
If $\mathbb U = \O$, the claim is true.
Assume $\mathbb U \ne \O$.
By Nonempty Grothendieck Universe contains Von Neumann Natural Numbers, every von Neumann natural number is an element of $\mathbb U$.
In particular:
:$2 = \set {\O, \set \O} \in \mathbb U$
Using $2$ as an indexing set, we remember that $0 = \O$ and $1 = \set \O$, and define:
{{begin-eqn}}
{{eqn | l = w_0
      | o = :=
      | r = u
}}
{{eqn | l = w_1
      | o = :=
      | r = v
}}
{{end-eqn}}
This sets up the structures needed to exploit Grothendieck Universe: Axiom $(4)$ below.
Then:
{{begin-eqn}}
{{eqn | l = u, v
      | o = \in
      | r = \mathbb U
}}
{{eqn | ll= \leadsto
      | l = \set {u, v}
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe: Axiom $(2)$
}}
{{eqn | ll= \leadsto
      | l = \bigcup_{i \mathop \in 2} w_i = u \cup v
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe: Axiom $(4)$
}}
{{end-eqn}}
{{qed}}
Category:Grothendieck Universes
0vfi08fyykhx3p8m5h4lfv2oq7uktwl
Theorem: Faithful Functor Reflects Monomorphisms

Let $\mathbf C$ and $\mathbf D$ be categories.
Let $F: \mathbf C \to \mathbf D$ be a faithful functor.
Let $x$ and $y$ be objects in $\mathbf C$.
Let $f: x \to y$ be a morphism in $\mathbf C$.
Let $\map F f : \map F x \to \map F y$ be a monomorphism in $\mathbf D$.
Then $f$ is a monomorphism in $\mathbf C$.

Proof:
Let $z$ be an object in $\mathbf C$.
Let $g: z \to x$ and $h: z \to x$ be morphisms in $\mathbf C$ such that $f \circ g = f \circ h$.
{{begin-eqn}}
{{eqn | l = f \circ g
      | r = f \circ h
}}
{{eqn | ll= \leadsto
      | l = \map F f \circ \map F g
      | r = \map F f \circ \map F h
      | c = {{Defof|Functor}}
}}
{{eqn | ll= \leadsto
      | l = \map F g
      | r = \map F h
      | c = $\map F f$ is a monomorphism in $\mathbf D$
}}
{{eqn | ll= \leadsto
      | l = g
      | r = h
      | c = {{Defof|Faithful Functor}}
}}
{{end-eqn}}
Hence $f$ is a monomorphism in $\mathbf C$.
{{qed}}
Category:Functors
Category:Monomorphisms (Category Theory)
h53b6sb8ghn0dh5dyt9q1q4tvx0g2kp
Theorem: Subring Module is Module/Special Case/Unitary Module

Let $\struct {R, +, \circ}$ be a ring with unity such that $1_R$ is that unity.
Let $1_R \in S$.
Then $\struct {R, +, \circ_S}_S$ is a unitary $S$-module.

Proof:
From Subring Module is Module: Special Case, we have that $\struct {R, +, \circ_S}_S$ is an $S$-module.
Then {{hypothesis}} $1_R$ is the unity of $\struct {S, +, \circ_S}$.
Thus $\struct {S, +, \circ_S}$ is also a ring with unity.
It follows from Ring with Unity is Module over Itself that $\struct {R, +, \circ_S}_S$ is a unitary module.
{{qed}}
Theorem: Integral Ideal iff Set of Integer Multiples

Let $J$ be a non-empty subset of the set of integers $\Z$.
Then:
:$J$ is an integral ideal
{{iff}}:
:$\exists m \in \Z: J = m \Z$.

Proof:
=== Sufficient Condition ===
{{:Integral Ideal is Set of Integer Multiples}}
{{qed|lemma}}
Theorem: Principal Ideal Domain is Bézout Domain

Let $D$ be a principal ideal domain.
Then $D$ is a Bézout domain.

Proof:
Let $J_1, J_2$ be ideals of $D$.
From Sum of Ideals is Ideal, $J_1 + J_2$ is an ideal of $D$.
By definition of principal ideal domain, $J_1 + J_2$ is principal.
Hence $D$ is a Bézout domain.
{{qed}}
Category:Bézout Domains
Category:Principal Ideal Domains
t17929ufe1vmkt57g3c8426k5448zdn
Theorem: Ring with Unity is Module over Itself

Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.
Then $\struct {R, +, \circ}_R$ is a unitary $R$-module.

Proof:
From Ring is Module over Itself we have that $\struct {R, +, \circ}_R$ is an $R$-module. 
We have {{hypothesis}} that $\struct {R, +, \circ}$ has a unity $1_R$. 
For $\struct {R, +, \circ}_R$ to be unitary, it must satisfy the additional axiom:
{{begin-axiom}}
{{axiom | n = 4
        | q = \forall x \in R
        | m = 1_R \circ x = x
}}
{{end-axiom}}
The axiom follows from the definition of a unity.
Category:Module Theory
Category:Ring is Module over Itself
h69dce260a72q8kuqm0d6dq6b80fg23
Theorem: Subring Module is Module/Unitary

Let $\struct {R, +, \times}$ be a ring with unity.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Let $1_R \in S$.
Then $\struct{G, +_G, \circ_S}_S$ is also unitary.

Proof:
From Subring Module is Module, we have that $\struct {G, +_G, \circ_S}_S$ is an $S$-module.
It remains to be demonstrated that $\struct{G, +_G, \circ_S}_S$ is unitary.
To show this, we must prove that:
:$\forall x \in G: 1_R \circ_S x = x$
Since $1_R \in S$ by assumption, the product $1_R \circ_S x$ is defined. 
We now have:
{{begin-eqn}}
{{eqn | l = 1_R \circ_S x
      | r = 1_R \circ x
}}
{{eqn | r = x
      | c = {{Module-axiom|4}} on $\struct {G, +_G, \circ}_R$
}}
{{end-eqn}}
and the proof is complete.
{{qed}}
Theorem: Nonempty Grothendieck Universe contains Von Neumann Natural Numbers

Let $\mathbb U$ be a non-empty Grothendieck universe.
Let $\N$ denote the set of von Neumann natural numbers.
Then $\N$ is a subset of $\mathbb U$.

Proof:
We prove the claim by induction.
Theorem: Empty Set is Element of Nonempty Grothendieck Universe

Let $\mathbb U$ be a non-empty Grothendieck universe.
Then $\O \in \mathbb U$.

Proof:
Let $A \in \mathbb U$ be an arbitrary element.
{{begin-eqn}}
{{eqn | l = \O
      | o = \subseteq
      | r = A
      | c = Empty Set is Subset of All Sets
}}
{{eqn | ll = \leadsto
      | l = \O
      | o = \in
      | r = \mathbb U
      | c = Grothendieck Universe is Closed under Subset
}}
{{end-eqn}}
{{qed}}
Category:Grothendieck Universes
8zk918omq9am4dhs8m9yg1mght821nx
Theorem: Adjoint of Identity Transformation

Let $\tuple {\HH, \innerprod \cdot \cdot_\HH}$ be a Hilbert space.
Let $I_\HH$ be the identity transformation on $\HH$.
Then: 
:${I_\HH}^* = I_\HH$
where ${I_\HH}^*$ denotes the adjoint of $I_\HH$.

Proof:
From Identity Mapping on Normed Vector Space is Bounded Linear Operator:
:$I_\HH$ is a bounded linear transformation.
So, from the existence part of Existence and Uniqueness of Adjoint:
:$I_\HH$ has an adjoint ${I_\HH}^*$.
That is:
:$\innerprod {I_\HH h} g_\HH = \innerprod h { {I_\HH}^* g}_\HH$
for all $h, g \in \HH$.
Note that by the definition of the identity transformation, we also have: 
:$\innerprod {I_\HH h} g_\HH = \innerprod h {I_\HH g}_\HH$
for all $h, g \in \HH$.
So, from the uniqueness part of Existence and Uniqueness of Adjoint:
:$I_\HH = {I_\HH}^*$ 
{{qed}}
Category:Adjoints
Category:Identity Mappings
ao51ozv6fbh9knthasjmiqbud8c2n2a
Theorem: Characterization of Invertible Bounded Linear Transformations

<onlyinclude>
Let $\struct {U, \norm \cdot_U}$ and $\struct {V, \norm \cdot_V}$ be normed vector spaces.
Let $A : V \to U$ be a linear transformation with inverse $A^{-1} : U \to V$.
Then $A^{-1}$ is a bounded linear transformation {{iff}}:
:there exists a real number $c > 0$ such that $\norm {A x}_U \ge c \norm x_V$ for all $x \in V$.
That is, $A$ is invertible as a bounded linear transformation {{iff}}:
:there exists a real number $c > 0$ such that $\norm {A x}_U \ge c \norm x_V$ for all $x \in V$.
<onlyinclude>

Proof:
From Inverse of Linear Transformation is Linear Transformation, we have: 
:$A^{-1}$ is a linear transformation.
So we are interested in determining when $A^{-1}$ is bounded.
Theorem: Sum of Bounded Linear Transformations is Bounded Linear Transformation

Let $\mathbb F \in \set {\R, \C}$. 
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\mathbb F$. 
Let $A, B : \HH \to \KK$ be bounded linear transformations.
Let $\norm \cdot$ be the norm on the space of bounded linear transformations.
Then: 
:$A + B$ is a bounded linear transformation
with: 
:$\norm {A + B} \le \norm A + \norm B$

Proof:
From Addition of Linear Transformations, we have that: 
:$A + B$ is a linear transformation.
It remains to show that $A + B$ is bounded.
Let $\norm \cdot_\HH$ be the inner product norm on $\HH$. 
Let $\norm \cdot_\KK$ be the inner product norm on $\KK$.
Since $A$ is a bounded linear transformation, from Fundamental Property of Norm on Bounded Linear Transformation, we have: 
:$\norm {A x}_\KK \le \norm A \norm x_\HH$
for all $x \in \HH$. 
Similarly, since $B$ is a bounded linear transformation we have: 
:$\norm {B x}_\KK \le \norm B \norm x_\HH$ 
for all $x \in \HH$. 
Let $x \in \HH$. 
Then, we have: 
{{begin-eqn}}
{{eqn	| l = \norm {\paren {A + B} x}_\KK
	| r = \norm {A x + B x}_\KK
}}
{{eqn	| o = \le
	| r = \norm {A x}_\KK + \norm {B x}_\KK
	| c = {{Defof|Norm on Vector Space}}
}}
{{eqn	| o = \le
	| r = \norm A \norm x_\HH + \norm B \norm x_\HH
}}
{{eqn	| r = \paren {\norm A + \norm B} \norm x_\HH
}}
{{end-eqn}}
So, taking $c = \norm A +\norm B$, we have: 
:$\norm {\paren {A + B} x}_\KK \le c \norm x_\HH$
for all $x \in \HH$.
So:
:$A + B$ is a bounded linear transformation.
Note that:
:$\norm A + \norm B \in \set {c > 0: \forall h \in \HH: \norm {\paren {A + B} h}_\KK \le c \norm h_\HH}$
while, by the definition of the norm, we have: 
:$\norm {A + B} = \inf \set {c > 0: \forall h \in \HH: \norm {\paren {A + B} h}_\KK \le c \norm h_\HH}$
So, by the definition of infimum:
:$\norm {A + B} \le \norm A + \norm B$
{{qed}}
Category:Linear Transformations on Hilbert Spaces
reps7f1oo0qk03kgwa1yhjp7q1t5zt2
Theorem: Suprema and Infima of Combined Bounded Functions/Bounded Below

Let both $f$ and $g$ be bounded below on $S \subseteq \R$.
Then:
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$
where $\ds \map \inf {\map f x}$ is the infimum of $\map f x$.

Proof:
First we show that:
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$
Let $T = \set {\map f x: x \in S}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\inf_{x \mathop \in S} } {\map f x + c}
      | r = \map {\inf_{y \mathop \in T} } {y + c}
      | c = 
}}
{{eqn | r = c + \map {\sup_{y \mathop \in T} } y
      | c = Infimum Plus Constant
}}
{{eqn | r = c + \map {\inf_{x \mathop \in S} } {\map f x}
      | c = 
}}
{{end-eqn}}
Next we show that:
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$
Let:
:$\ds H = \map {\inf_{x \mathop \in S} } {\map f x}$
:$\ds K = \map {\inf_{x \mathop \in S} } {\map g x}$
Then:
:$\forall x \in S: \map f x + g \map x \ge H + K$
Hence $H + K$ is a lower bound for $\set {\map f x + \map g x: x \in S}$.
The result follows.
{{qed}}
Theorem: Suprema and Infima of Combined Bounded Functions/Bounded Above

Let both $f$ and $g$ be bounded above on $S \subseteq \R$.
Then:
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$
where $\ds \map \sup {\map f x}$ is the supremum of $\map f x$.

Proof:
First we show that:
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$
Let $T = \set {\map f x: x \in S}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\sup_{x \mathop \in S} } {\map f x + c}
      | r = \map {\sup_{y \mathop \in T} } {y + c}
      | c = 
}}
{{eqn | r = c + \map {\sup_{y \mathop \in T} } y
      | c = Supremum Plus Constant
}}
{{eqn | r = c + \map {\sup_{x \mathop \in S} } {\map f x}
      | c = 
}}
{{end-eqn}}
Next we show that $\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$:
Let:
:$\ds H = \map {\sup_{x \mathop \in S} } {\map f x}$
:$\ds K = \map {\sup_{x \mathop \in S} } {\map g x}$
Then:
:$\forall x \in S: \map f x + \map g x \le H + K$
Hence $H + K$ is an upper bound for $\set {\map f x + \map g x: x \in S}$.
The result follows.
{{qed}}
Theorem: Kernel of Linear Transformation contained in Kernel of different Linear Transformation implies Transformations are Proportional

Let $V$ be a complex vector space.
Let $\map \LL {V, \C}$ be the space of all linear transformations from $V$ to complex numbers $\C$.
Let $\ell, L \in \map \LL {V, \C}$ be such that:
:$\ker \ell \subseteq \ker L$
where $\ker$ denotes the kernel.
Then:
:$\exists c \in \C : L = c \ell$

Proof:
Suppose $\ell = \mathbf 0$.
Then:
:$\ker \ell = V$
That is, the kernel of $\ell$ is the entire vector space $V$.
Moreover:
:$\ker \ell \subseteq \ker L \implies \ker L = V$
Therefore:
:$L = \mathbf 0$
and we can set $c = 0$ to have:
:$L = \mathbf 0 = 0 \cdot \ell$
Suppose $\ell \ne \mathbf 0$.
By Linear Transformation Maps Zero Vector to Zero Vector:
:$\exists v_0 \in V : v_0 \ne \mathbf 0 : \map \ell {v_0} \ne 0$
Let $v \in V$.
Let:
:$w = v - c_v v_0$
where $c_v \in \C$.
Then:
{{begin-eqn}}
{{eqn | l = \map L v
      | r = \map L {c_v v_0 + w}
}}
{{eqn | r = c_v \map L {v_0} + \map L w
      | c = {{Defof|Linear Transformation}}
}}
{{end-eqn}}
Suppose $w \in \ker \ell$.
Then:
:$w \in \ker L$
and it follows that:
:$\map \ell w = \map L w = 0$
Now we will seek for $w$ such that $\map \ell w = 0$.
{{begin-eqn}}
{{eqn | l = \map \ell w
      | r = \map \ell v - c_v \map \ell {v_0}
}}
{{eqn | r = 0
}}
{{eqn | ll = \leadsto
      | l = c_v
      | r = \frac {\map \ell v }{\map \ell {v_0} }
}}
{{end-eqn}}
Hence, if $\ds c_v = \frac {\map \ell v} {\map \ell {v_0} }$ then:
{{begin-eqn}}
{{eqn | l = \map L v
      | r = \frac {\map \ell v }{\map \ell {v_0} } \map L {v_0}
}}
{{eqn | r = \frac {\map L {v_0} }{\map \ell {v_0} } \map \ell v
}}
{{end-eqn}}
Therefore:
:$\ds c = \frac {\map L {v_0} }{\map \ell {v_0} }$
{{qed}}