avewright/proofwiki-math huggingface dataset:
Theorem: Evaluation Linear Transformation is Linear Transformation

Let $R$ be a commutative ring with unity.
Let $G$ be an $R$-module.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the double dual of $G^*$.
Let the mapping $J: G \to G^{**}$ be the evaluation linear transformation from $G$ into $G^{**}$ defined as:
:$\forall x \in G: \map J x = x^\wedge$
where for each $x \in G$, $x^\wedge: G^* \to R$ is defined as:
:$\forall t \in G^*: \map {x^\wedge} t = \map t x$
Then $J$ is a linear transformation.

Proof:
From Underlying Mapping of Evaluation Linear Transformation is Element of Double Dual, we have that:
:$x^\wedge \in G^{**}$
Hence $x^\wedge$ a fortiori is a linear transformation.
It remains to be shown that $J: G \to G^{**}$ is a linear transformation.
That is, that the following conditions are satisfied by $J$:
:$(1): \quad \forall x, y \in G: \map J {x + y} = \map J x + \map J y$
:$(2): \quad \forall x \in G: \forall \lambda \in R: \map J {\lambda \times x} = \lambda \times J x$
Hence:
{{begin-eqn}}
{{eqn | n = 1
      | l = \map J {x + y}
      | r = \paren {x + y}^\wedge
      | c = Definition of $J$
}}
{{eqn | r = x^\wedge + y^\wedge
      | c = {{Defof|Pointwise Addition of Linear Transformations}}
}}
{{eqn | r = \map J x + \map J y
      | c = Definition of $x^\wedge$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | n = 2
      | l = \map J {\lambda \times x}
      | r = \paren {\lambda \times x}^\wedge
      | c = Definition of $x^\wedge$
}}
{{eqn | r = \lambda \times x^\wedge
      | c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \lambda \times \map J x
      | c = Definition of $x^\wedge$
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Evaluation Isomorphism is Isomorphism

Let $R$ be a commutative ring with unity.
Let $G$ be a unitary $R$-module whose dimension is finite.
Then the evaluation linear transformation $J: G \to G^{**}$ is an isomorphism.

Proof:
Let $\sequence {a_n}$ be an ordered basis of $G$.
Then by definition, every $x \in G$ can be written in the form:
:$\ds \sum_{k \mathop = 1}^n \lambda_k a_k$
where $\lambda_k \in R$.
From Unique Representation by Ordered Basis, this representation is unique.
From Existence of Ordered Dual Basis:
:$\sequence {\map J {a_n} }$ is the ordered basis of $G^{**}$ dual to the ordered basis of $G^*$ which itself is dual to $\sequence {a_n}$.
From Unique Linear Transformation Between Modules, there exists a unique linear transformation $\phi: G \to H$ satisfying:
:$\forall k \in \closedint 1 n: \map \phi {a_k} = b_k$
From the preceding it follows that this unique $\phi$ has to be $J$.
As every element of $G^{**}$ is so representable, it follows that $J$ is surjective.
Now let $x, y \in G$ such that $\map J x = \map J y$.
We have that:
:$x = \ds \sum_{k \mathop = 1}^n \lambda_k a_k$
for some unique $\sequence {\lambda_n}$ in $R$
:$y = \ds \sum_{k \mathop = 1}^n \mu_k a_k$
for some unique $\sequence {\mu_n}$ in $R$
Hence from Unique Linear Transformation Between Modules:
:$\map J x = \ds \sum_{k \mathop = 1}^n \lambda_k b_k$
for some unique $\sequence {\lambda_n}$ in $R$
:$\map J y = \ds \sum_{k \mathop = 1}^n \mu_k b_k$
for some unique $\sequence {\mu_n}$ in $R$
But as $\map J x = \map J y$ it must be the case that:
:$\forall k \in \set {1, 2, \ldots, n}: \lambda_k = \mu_k$
and it follows that $x = y$.
Thus $J$ is injective.
It has been shown that $J$ is both injective and surjective.
Hence, by definition, $J$ is bijective.
By definition, a linear transformation is a homomorphism.
Hence, again by definition, $J$ is an isomorphism.
{{qed}}
Theorem: Conditions for Homogeneity/Straight Line

The line $L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$ is homogeneous {{iff}} $\beta = 0$.

Proof:
=== Sufficient Condition ===
Let the line $L = \alpha_1 x_1 + \alpha_2 x_2 = \beta$ be homogeneous.
Then the origin $\tuple {0, 0}$ lies on $L$.
Hence:
{{begin-eqn}}
{{eqn | l = \alpha_1 0 + \alpha_2 0
      | r = \beta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \beta
      | r = 0
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Conditions for Homogeneity/Plane

The plane $P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ is homogeneous {{iff}} $\gamma = 0$.

Proof:
=== Sufficient Condition ===
Let the plane $P = \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be homogeneous.
Then the origin $\tuple {0, 0, 0}$ lies on $P$.
Hence:
{{begin-eqn}}
{{eqn | l = \alpha_1 0 + \alpha_2 0 + \alpha_3 0
      | r = \gamma
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \gamma
      | r = 0
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Zero Matrix is Identity for Hadamard Product

Let $\struct {S, \cdot}$ be a monoid whose identity is $e$.
Let $\map {\MM_S} {m, n}$ be an $m \times n$ matrix space over $S$.
Let $\mathbf e = \sqbrk e_{m n}$ be the zero matrix of $\map {\MM_S} {m, n}$.
Then $\mathbf e$ is the identity element for Hadamard product.

Proof:
Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_S} {m, n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A \circ \mathbf e
      | r = \sqbrk a_{m n} \circ \sqbrk e_{m n}
      | c = Definition of $\mathbf A$ and $\mathbf e$
}}
{{eqn | r = \sqbrk {a \cdot e}_{m n}
      | c = {{Defof|Hadamard Product}}
}}
{{eqn | r = \sqbrk a_{m n}
      | c = {{Defof|Identity Element}}
}}
{{eqn | ll= \leadsto
      | l = \mathbf A \circ \mathbf e
      | r = \mathbf A
      | c = {{Defof|Zero Matrix over General Monoid}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \mathbf e \circ \mathbf A
      | r = \sqbrk e_{m n} \circ \sqbrk a_{m n}
      | c = Definition of $\mathbf A$ and $\mathbf e$
}}
{{eqn | r = \sqbrk {e \cdot a}_{m n}
      | c = {{Defof|Hadamard Product}}
}}
{{eqn | r = \sqbrk e_{m n}
      | c = {{Defof|Identity Element}}
}}
{{eqn | ll= \leadsto
      | l = \mathbf e \circ \mathbf A
      | r = \mathbf A
      | c = {{Defof|Zero Matrix over General Monoid}}
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Ideals is Ideal/General Result

Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.
Then:
:$J = J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.
where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product with respect to $\struct{R, +}$.

Proof:
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.
$\map P 1$ is true, as this just says $J_1$ is an ideal of $R$.
Theorem: Integer is Expressible as Product of Primes

Let $n$ be an integer such that $n > 1$.
Then $n$ can be expressed as the product of one or more primes.

Proof:
{{AimForCont}} this supposition is false.
Let $m$ be the smallest integer which can not be expressed as the product of primes.
As a prime number is trivially a product of primes, $m$ can not itself be prime.
Hence:
:$\exists r, s \in \Z: 1 < r < m, 1 < s < m: m = r s$
As $m$ is our least counterexample, both $r$ and $s$ can be expressed as the product of primes.
Say $r = p_1 p_2 \cdots p_k$ and $s = q_1 q_2 \cdots q_l$, where all of $p_1, \ldots, p_k, q_1, \ldots, q_l$ are prime.
Hence $m = r s = p_1 p_2 \cdots p_k q_1 q_2 \cdots q_l$, which is a product of primes.
Hence there is no such counterexample.
{{qed}}
Theorem: Integer is Expressible as Product of Primes

Let $n$ be an integer such that $n > 1$.
Then $n$ can be expressed as the product of one or more primes.

Proof:
If $n$ is prime, the result is immediate.
Let $n$ be composite.
Then by Composite Number has Two Divisors Less Than It:
:$\exists r, s \in \Z: n = r s, 1 < r < n, 1 < s < n$
This being the case, the set $S_1 = \set {d: d \divides n, 1 < d < n}$ is nonempty, and bounded below by $1$.
By Set of Integers Bounded Below by Integer has Smallest Element, $S_1$ has a smallest element, which we will call $p_1$.
{{AimForCont}} $p_1$ is composite.
By Composite Number has Two Divisors Less Than It, there exist $a, b$ such that $a, b \divides p_1$ and $1 < a < p_1, 1 < b < p_1$.
But by Divisor Relation on Positive Integers is Partial Ordering, it follows that $a, b \divides n$ and hence $a, b \in S_1$.
This contradicts the assertion that $p_1$ is the smallest element of $S_1$.
Thus, $p_1$ is necessarily prime.
We may now write $n = p_1 n_1$, where $n > n_1 > 1$.
If $n_1$ is prime, the proof is complete.
Otherwise, the set $S_2 = \set {d: d \divides n_1, 1 < d < n_1}$ is nonempty, and bounded below by $1$.
By the above argument, the smallest element $p_2$ of $S_2$ is prime.
Thus we may write $n_1 = p_2 n_2$, where $1 < n_2 < n_1$.
This gives us $n = p_1 p_2 n_2$.
If $n_2$ is prime, we are done.
Otherwise, we continue this process.
Since $n > n_1 > n_2 > \cdots > 1$ is a (strictly) decreasing sequence of positive integers, there must be a finite number of $n_i$'s.
That is, we will arrive at some prime number $n_{k - 1}$, which we will call $p_k$.
This results in the prime decomposition $n = p_1 p_2 \cdots p_k$.
{{qed}}
Theorem: Integer is Expressible as Product of Primes

Let $n$ be an integer such that $n > 1$.
Then $n$ can be expressed as the product of one or more primes.

Proof:
The proof proceeds by induction.
For all $n \in \N_{> 1}$, let $\map P n$ be the proposition:
:$n$ can be expressed as a product of prime numbers.
First note that if $n$ is prime, the result is immediate.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$n$ can be expressed as a product of prime numbers.
As $2$ itself is a prime number, and the result is immediate.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $2 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:For all $j \in \N$ such that $2 \le j \le k$, $j$ can be expressed as a product of prime numbers.
from which it is to be shown that:
:$k + 1$ can be expressed as a product of prime numbers.
=== Induction Step ===
This is the induction step:
If $k + 1$ is prime, then the result is immediate.
Otherwise, $k + 1$ is composite and can be expressed as:
:$k + 1 =  r s$
where $2 \le r < k + 1$ and $2 \le s < k + 1$
That is, $2 \le r \le k$ and $2 \le s \le k$.
Thus by the induction hypothesis, both $r$ and $s$ can be expressed as a product of primes.
So $k + 1 = r s$ can also be expressed as a product of primes.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore, for all $n \in \N_{> 1}$:
:$n$ can be expressed as a product of prime numbers.
Theorem: Expression for Integer as Product of Primes is Unique

Let $n$ be an integer such that $n > 1$.
Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear.

Proof:
{{AimForCont}} the supposition false.
That is, suppose there is at least one positive integer that can be expressed in more than one way as a product of primes.
Let the smallest of these be $m$.
Thus:
:$m = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s$
where all of $p_1, \ldots p_r, q_1, \ldots q_s$ are prime.
By definition, $m$ is not itself prime.
Therefore:
:$r, s \ge 2$
Let us arrange that the primes which compose $m$ are in order of size:
:$p_1 \le p_2 \le \dots \le p_r$
and:
:$q_1 \le q_2 \le \dots \le q_s$
Let us arrange that $p_1 \le q_1$.
Suppose $p_1 = q_1$.
Then:
:$\dfrac m {p_1} = p_2 p_3 \cdots p_r = q_2 q_3 \cdots q_s = \dfrac m {q_1}$
But then we have the positive integer $\dfrac m {p_1}$ being expressible in two different ways.
This contradicts the fact that $m$ is the smallest positive integer that can be so expressed.
Therefore:
:$p_1 \ne q_1 \implies p_1 < q_1 \implies p_1 < q_2, q_3, \ldots, q_s$
as we arranged them in order.
From Prime not Divisor implies Coprime:
:$1 < p_1 < q_j: 1 < j < s \implies p_1 \nmid q_j$
But:
:$p_1 \divides m \implies p_1 \divides q_1 q_2 \ldots q_s$
where $\divides$ denotes divisibility.
Thus from Euclid's Lemma for Prime Divisors:
:$\exists j: 1 \le j \le s: p_1 \divides q_j$
But $q_j$ was supposed to be a prime.
This is a contradiction.
Hence, by Proof by Contradiction, the supposition was false.
{{Qed}}
Theorem: Expression for Integer as Product of Primes is Unique

Let $n$ be an integer such that $n > 1$.
Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear.

Proof:
{{AimForCont}} $n$ has two prime factorizations:
:$n = p_1 p_2 \dots p_r = q_1 q_2 \dots q_s$
where $r \le s$ and each $p_i$ and $q_j$ is prime with $p_1 \le p_2 \le \dots \le p_r$ and $q_1 \le q_2 \le \dots \le q_s$. 
Since $p_1 \divides q_1 q_2 \dots q_s$, it follows from Euclid's Lemma for Prime Divisors that $p_1 = q_j$ for some $1 \le j \le s$.
Thus:
:$p_1 \ge q_1$
Similarly, since $q_1 \divides p_1 p_2 \dots p_r$, from Euclid's Lemma for Prime Divisors:
:$q_1 \ge p_1$
Thus, $p_1 = q_1$, so we may cancel these common factors, which gives:
:$p_2 p_3 \cdots p_r = q_2 q_3 \dots q_s$
This process is repeated to show that:
:$p_2 = q_2, p_3 = q_3, \ldots, p_r = q_r$
If $r < s$, we arrive at $1 = q_{r + 1} q_{r + 2} \cdots q_s$ after canceling all common factors.
But by Divisors of One, the only divisors $1$ are $1$ and $-1$.
Hence $q_{r + 1}, q_{r + 2}, \ldots, q_s$ cannot be prime numbers
From that contradiction it follows that $r = s$.
Thus:
:$p_1 = q_1, p_2 = q_2, \ldots, p_r = q_s$
which means the two factorizations are identical.
Therefore, the prime factorizations of $n$ is unique.
{{Qed}}
Theorem: Expression for Integer as Product of Primes is Unique

Let $n$ be an integer such that $n > 1$.
Then the expression for $n$ as the product of one or more primes is unique up to the order in which they appear.

Proof:
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:the prime decomposition for $n$ is unique up to order of presentation.
Note that it has been established in Integer is Expressible as Product of Primes that $n$ does in fact have at least $1$ prime decomposition.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$n = 2$
which is trivially unique.
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:the prime decomposition for all $j$ such that $0 \le j \le k$ is unique up to order of presentation.
from which it is to be shown that:
:the prime decomposition for $k + 1$ is unique up to order of presentation.
=== Induction Step ===
Either $k + 1$ is prime or it is composite.
If $k + 1$ is prime, there is only one way to express it, that is, as the prime $k + 1$ itself.
So suppose $k + 1$ is composite.
{{AimForCont}} $k + 1$ has the two prime decompositions:
$k + 1 = p_1 p_2 \dotsm p_r = q_1 q_2 \dotsm q_s$
where:
:$p_1 \le p_2 \le \dotsb \le p_r$
and:
:$q_1 \le q_2 \le \dotsb \le q_s$
Because $q_1$ is a divisor of $k + 1$:
:$q_1 \divides p_1 \le p_2 \le \dotsb \le p_r$
where $\divides$ indicates divisibility.
Thus by Euclid's Lemma for Prime Divisors:
:$\exists p_i \in \set {p_1, p_2, \ldots, p_r}: q_1 \divides p_i$
But as $q_1$ and $p_i$ are both primes, it follows by definition that $q_1 = p_i$.
In a similar way it can be shown that:
:$\exists q_j \in \set {q_1, q_2, \ldots, q_s}: p_1 = q_j$
So we have:
:$p_1 = q_j \ge q_1$
and:
:$q_1 = p_i \ge p_1$
Thus:
:$p_1 \ge q_1 \ge p_1$
and so:
:$p_1 = q_1$
Thus $\dfrac {k + 1} {p_1}$ is an integer such that:
:$\dfrac {k + 1} {p_1} \le k$
and so:
:$p_2 p_3 \dotsm p_r \dfrac {k + 1} {p_1} = q_2 q_3 \dotsm q_s$
But by the induction hypothesis:
:$p_2 = q_2, p_3 = q_3, \dotsc, p_r = q_s$
where furthermore $r = s$.
Therefore the prime decomposition for $k + 1$ is unique.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction:
:$\forall n \in \Z_{\ge 2}$, the prime decomposition for $n$ is unique up to order of presentation.
{{qed}}
Theorem: Modulo Addition is Closed/Integers

Let $m \in \Z$ be an integer.
Then addition modulo $m$ on the set of integers modulo $m$ is closed:
:$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m +_m \eqclass y m \in \Z_m$.

Proof:
From the definition of addition modulo $m$, we have:
:$\eqclass x m +_m \eqclass y m = \eqclass {x + y} m$
By the Division Theorem:
:$x + y = q m + r$ where $0 \le r < m$
Therefore for all $0 \le r < m$:
:$\eqclass {x + y} m = \eqclass r m$.
Therefore from the definition of integers modulo $m$:
:$\eqclass {x + y} m \in \Z_m$
{{Qed}}
Theorem: Modulo Addition is Closed/Real Numbers

Let $z \in \R$ be a real number.
Then addition modulo $z$ on the set of residue classes modulo $z$ is closed:
:$\forall \eqclass x z, \eqclass y z \in \R_z: \eqclass x z +_z \eqclass y z \in \R_z$.

Proof:
From the definition of addition modulo $z$, we have:
:$\eqclass x z +_z \eqclass y z = \eqclass {x + y} z$
As $x, y \in R$, we have that $x + y \in \R$ as Real Addition is Closed.
Hence by definition of congruence, $\eqclass {x + y} z \in \R_z$.
{{qed}}
Theorem: Field Homomorphism Preserves Subfields

Let $\struct {F_1, +_1, \circ_1}$ and $\struct {F_2, +_2, \circ_2}$ be fields.
Let $\phi: F_1 \to F_2$ be a field homomorphism such that $\phi$ is not the trivial homomorphism.
If $K$ is a subfield of $F_1$, then $\phi \sqbrk K$ is a subfield of $F_2$.

Proof:
First note that if $\phi$ is the trivial homomorphism then $\phi \sqbrk K = 0_{F_2}$ and so is not a field.
Since $K$ is a field, we have that:
:$0_{F_1} \in K$
:$1_{F_1} \in K$
and so by Ring Homomorphism Preserves Zero and Field Homomorphism Preserves Unity:
:$\map \phi {0_{F_1} } = 0_{F_2} \in \phi \sqbrk K$
:$\map \phi {1_{F_1} } = 1_{F_2} \in \phi \sqbrk K$
So $\phi \sqbrk K$ contains at least the zero and unity.
From Group Homomorphism Preserves Subgroups:
:$\struct {\phi \sqbrk K, +_2}$ is a subgroup of $\struct {F_2, +_2}$
:$\struct {\phi \sqbrk K, \circ_2}$ is a subgroup of $\struct {F_2, \circ_2}$.
Thus, as $\struct {F_2, +_2}$ and $\struct {F_2, \circ_2}$ are both groups, the result follows.
{{qed}}
Category:Field Homomorphisms
Category:Subfields
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Theorem: Set is Subset of Union/General Result

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\mathbb S \subseteq \powerset S$.
Then:
:$\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$

Proof:
Let $x \in T$ for some $T \in \mathbb S$.
Then: 
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = T
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | o = \in
      | r = \bigcup \mathbb S
      | c = {{Defof|Set Union}}
}}
{{eqn | ll= \leadsto
      | l = T
      | o = \subseteq
      | r = \bigcup \mathbb S
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
As $T$ was arbitrary, it follows that:
:$\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
{{qed}}
Category:Set Union
Category:Subsets
2ube9fms855oeum7f1i45d1hph2b2hw
Theorem: Intersection Distributes over Union/General Result

Let $S$ and $T$ be sets.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
:$\ds S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$

Proof:
=== Intersection Subset of Union ===
Let $\ds x \in S \cap \bigcup \mathbb T$.
We need to show that $\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$ and then by definition of subset we will have shown that $\ds S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$.
So, we have that $\ds x \in S \cap \bigcup \mathbb T$.
By definition of set intersection, $x \in S$ and $\ds x \in \bigcup \mathbb T$.
From $\ds x \in \bigcup \mathbb T$ we know that:
:$\exists X \in \mathbb T: x \in X$
and so:
:$\ds \exists X \in \mathbb T: x \in S \cap X$
So by definition of set union:
:$\ds x \in \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$
So:
:$\ds S \cap \bigcup \mathbb T \subseteq \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$
{{qed|lemma}}
Theorem: Union Distributes over Intersection/General Result

Let $S$ and $T$ be sets.
Let $\powerset T$ be the power set of $T$.
Let $\mathbb T$ be a subset of $\powerset T$.
Then:
:$\ds S \cup \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$

Proof:
=== Union Subset of Intersection ===
Let $\ds x \in S \cup \bigcap \mathbb T$.
We need to show that:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
and then by definition of subset we will have shown that:
:$\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.
So, we have that $\ds x \in S \cup \bigcap \mathbb T$.
By definition of set union, $x \in S$ or $\ds x \in \bigcap \mathbb T$.
So there are two cases to consider:
$(1): \quad$ Suppose $x \in S$.
Then by definition of set union, $\forall X \in \mathbb T: x \in S \cup X$.
So:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
$(2): \quad$ Suppose $\ds x \in \bigcap \mathbb T$.
Then by definition of set intersection:
:$\forall X \in \mathbb T: x \in X$
So by definition of set union:
:$\forall X \in \mathbb T: x \in S \cup X$
So:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
In both cases we see that:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
so by Proof by Cases, we have that:
:$\ds S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
{{qed|lemma}}
=== Intersection Subset of Union ===
Let $\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$.
We need to show that:
:$\ds x \in S \cup \bigcap \mathbb T$
and then by definition of subset we will have shown that:
:$\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$.
So, we have that:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
By definition of set intersection:
:$(A): \quad \forall X \in \mathbb T: x \in S \cup X$
There are two cases to consider:
:$(1): \quad \forall X \in \mathbb T: x \in X$
Then  by definition of set intersection:
:$\ds x \in \bigcap_{X \mathop \in \mathbb T} X$
and so by definition of set union:
:$\ds x \in S \cup \bigcap \mathbb T$
:$(2): \quad \exists X \in \mathbb T: x \notin X$
From $(A)$ we have that $x \in S \cup X$.
But as $x \notin X$ it follows that $x \in S$.
Then by definition of set union:
:$\ds x \in S \cup \bigcap \mathbb T$
In both cases we see that:
:$\ds x \in S \cup \bigcap \mathbb T$
so by Proof by Cases, we have that:
:$\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$
{{qed|lemma}}
So we have that:
:$S \cup \ds \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
and
:$\ds \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$
and so by definition of set equality:
:$S \cup \ds \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$
{{qed}}
Theorem: Scalar Multiple of Simple Function is Simple Function

Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function, and let $\lambda \in \R$.
Then the pointwise scalar multiple $\lambda f: X \to \R$ of $f$ is also a simple function.

Proof:
Let $\Img f$ denote the image of $f$.
Let $\Img {\lambda f}$ denote the image of $\lambda f$.
Consider the surjection $l_\lambda: \Img f \to \Img {\lambda f}$ defined by:
:$\map {l_\lambda} {\map f x} := \lambda \map f x$
By Measurable Function is Simple Function iff Finite Image Set, $\card {\Img f}$ is finite.
Hence Cardinality of Codomain of Surjection yields that $\size {\Img {\lambda f} }$ is finite as well.
The result follows from a second application of Measurable Function is Simple Function iff Finite Image Set.
{{qed}}
Category:Simple Functions
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