avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Arcsecant of x over a/Formulation 2

:$\ds \int \arcsec \frac x a \rd x = x \arcsec \frac x a - a \ln \size {x + \sqrt {x^2 - a^2} } + C$
for $x^2 > 1$.
$\arcsec \dfrac x a$ is undefined on the real numbers for $x^2 < 1$.

Proof:
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \arcsec \frac x a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \dfrac a {\size x \sqrt {x^2 - a^2} }
      | c = Derivative of $\arcsec \dfrac x a$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = x
      | c = Primitive of Constant
}}
{{end-eqn}}
We then have:
{{begin-eqn}}
{{eqn | l = \int \arcsec \frac x a \rd x
      | r = x \arcsec \frac x a - \int x \paren {\dfrac a {\size x \sqrt {x^2 - a^2} } } \rd x + C
      | c = Integration by Parts
}}
{{eqn | n = 1
      | r = x \arcsec \frac x a - \int \frac x {\size x} \paren {\frac a {\sqrt {x^2 - a^2} } } \rd x + C
      | c = rearrangement
}}
{{end-eqn}}
Let $x > 1$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac x {\size x} \paren {\frac a {\sqrt {x^2 - a^2} } } \rd x
      | r = \int \frac a { {\sqrt {x^2 - a^2} } } \rd x
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = a \int \frac 1 { {\sqrt {x^2 - a^2} } } \rd x
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = a \size {\ln \size {x + \sqrt {x^2 - a^2} } } + C
      | c = Integral of One Over Square Root of Binomial
}}
{{eqn | r = a \ln \size {x + \sqrt {x^2 - a^2} } + C
      | c = as argument of logarithm is positive
}}
{{end-eqn}}
Similarly, let $x < -1$.
Then: 
{{begin-eqn}}
{{eqn | l = \int \frac x {\size x} \paren {\frac a {\sqrt {x^2 - a^2} } } \rd x
      | r = \int \paren {-1} \frac a { {\sqrt{x^2 - a^2} } } \rd x
      | c = {{Defof|Absolute Value}}
}}
{{eqn | r = -a \int \frac 1 { {\sqrt{x^2 - a^2} } } \rd x
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -a \size {\ln \size {x + \sqrt {x^2 - a^2} } } + C
      | c = Integral of One Over Square Root of Binomial
}}
{{eqn | r = -a \paren {-\ln \size {x + \sqrt {x^2 - a^2} } } + C
      | c = as argument of logarithm is negative
}}
{{eqn | r = a \ln \size {x + \sqrt {x^2 - a^2} } + C
      | c = as argument of logarithm is negative
}}
{{end-eqn}}
{{qed}}
Theorem: Cross Product of Vector with Itself is Zero

Let $\mathbf x$ be a vector in a vector space of $3$ dimensions:
:$\mathbf x = x_i \mathbf i + x_j \mathbf j + x_k \mathbf k$
Then:
:$\mathbf x \times \mathbf x = \mathbf 0$
where $\times$ denotes vector cross product.

Proof:
{{begin-eqn}}
{{eqn | l = \mathbf x \times \mathbf x
      | r = \begin {vmatrix} \mathbf i & \mathbf j & \mathbf k \\ x_i & x_j & x_k \\ x_i & x_j & x_k \end {vmatrix}
      | c = {{Defof|Vector Cross Product}}
}}
{{eqn | r = \mathbf 0
      | c = Square Matrix with Duplicate Rows has Zero Determinant
}}
{{end-eqn}}
{{qed}}
Theorem: Cross Product of Vector with Itself is Zero

Let $\mathbf x$ be a vector in a vector space of $3$ dimensions:
:$\mathbf x = x_i \mathbf i + x_j \mathbf j + x_k \mathbf k$
Then:
:$\mathbf x \times \mathbf x = \mathbf 0$
where $\times$ denotes vector cross product.

Proof:
By definition, a vector is parallel to itself.
The result follows from Cross Product of Parallel Vectors.
{{qed}}
Theorem: Reciprocal of Riemann Zeta Function

For $\map \Re z > 1$:
:$\ds \frac 1 {\map \zeta z} = \sum_{k \mathop = 1}^\infty \frac{\mu \left({k}\right)} {k^z}$
where:
:$\zeta$ is the Riemann zeta function
:$\mu$ is the Möbius function.

Proof:
By Sum of Reciprocals of Powers as Euler Product:
{{begin-eqn}}
{{eqn | l = \frac 1 {\map \zeta z}
      | r = \prod_{\text {$p$ prime} } \paren {1 - p^{-z} }
      | c = 
}}
{{eqn | r = \paren {1 - \frac 1 {2^z} } \paren {1 - \frac 1 {3^z} } \paren {1 - \frac 1 {5^z} } \paren {1 - \frac 1 {7^z} } \paren {1 - \frac 1 {11^z} } \cdots
      | c = 
}}
{{end-eqn}}
The expansion of this product will be:
:$\ds 1 + \sum_{\text {$n$ prime} } \paren {\frac{-1} {n^z} } + \sum_{n \mathop = p_1 p_2} \paren {\frac {-1} {p_1^z} \frac {-1} {p_2^z} } + \sum_{n \mathop = p_1 p_2 p_3} \paren {\frac {-1} {p_1^z} \frac {-1} {p_2^z} \frac {-1}{p_3^z} } + \cdots$
which is precisely:
:$\ds \sum_{n \mathop = 1}^\infty \frac {\map \mu n} {n^z}$
as desired.
{{handwaving|get this from Dirichlet Series of Inverse of Arithmetic Function instead}}
{{qed}}
Category:Riemann Zeta Function
gjanjr8bqs20shngst4bjqip8y3mxjz
Theorem: Condition for Agreement of Family of Mappings

Let $\family {A_i}_{i \mathop \in I}, \family {B_i}_{i \mathop \in I}$ be families of non empty sets.
Let $\family {f_i}_{i \mathop \in I}$ be a family of mappings such that:
:$\forall i \in I: f_i \in \map \FF {A_i, B_i}$
{{explain|Clarify: what is $\map \FF {A_i, B_i}$? From the context it can be understood as being the set of all mappings $f_i: A_i \to B_i$ but this is just a guess. The domain and range of each of the $f_i$ could help with being explicitly defined.<br/>Note also that I have taken the liberty of exchanging the difficult-to-read $\mathscr F$ with the more eye-friendly $\FF$.}}
We have that:
:$\ds \bigcup_{i \mathop \in I} f_i \in \map \FF {\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}$
{{iff}}:
:$\ds \forall i, j \in I: \Dom {f_i} \cap \Dom {f_j} \ne \O \implies \paren {\forall a \in \paren {\Dom {f_i} \cap \Dom {f_j} }, \tuple {a, b} \in f_i \implies \tuple {a, b} \in f_j}$

Proof:
=== Sufficient Condition ===
Let:
:$\ds \bigcup_{i \mathop \in I} f_i \in \map \FF {\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}$
Let $i, j \in I$ be such that:
:$\Dom {f_i} \cap \Dom {f_j} \ne \O$
Let $a \in \paren {\Dom {f_i} \cap \Dom {f_j} }$.
Let $\ds b \in \bigcup_{i \mathop \in I} B_i$ be such that:
:$\tuple {a, b} \in f_i$
{{AimForCont}}:
:$\tuple {a, b} \notin f_j$
As $a \in \paren {\Dom {f_i} \cap \Dom {f_j} }$:
:$\ds \exists c \in \bigcup_{i \mathop \in I} B_i: \tuple {a, c} \in f_j$ 
As $\tuple {a, b} \in f_i$:
:$\ds \tuple {a, b} \in \bigcup_{i \mathop \in I} f_i$
Thus:
:$\ds \tuple {a, b}, \tuple {a, c} \in \bigcup_{i \mathop \in I} f_i$
such that $b \ne c$ and $\ds \bigcup_{i \mathop \in I} f_i$ is a mapping.
This is a contradiction.
Thus the supposition that the fact $\tuple {a, b} \notin f_j$ was false.
So:
:$\tuple {a, b} \in f_j$
{{qed|lemma}}
Theorem: Supremum of Set of Real Numbers is at least Supremum of Subset

Let $S$ be a set of real numbers.
Let $S$ have a supremum.
Let $T$ be a non-empty subset of $S$.
Then $\sup T$ exists and:
:$\sup T \le \sup S$

Proof:
The number $\sup S$ is an upper bound for $S$.
Therefore, $\sup S$ is an upper bound for $T$ as $T$ is a non-empty subset of $S$.
Accordingly, $T$ has a supremum by the Continuum Property.
The number $\sup S$ is an upper bound for $T$.
Therefore, $\sup S$ is greater than or equal to $\sup T$ as $\sup T$ is the least upper bound of $T$.
{{qed}}
Theorem: Supremum of Set of Real Numbers is at least Supremum of Subset

Let $S$ be a set of real numbers.
Let $S$ have a supremum.
Let $T$ be a non-empty subset of $S$.
Then $\sup T$ exists and:
:$\sup T \le \sup S$

Proof:
By the Continuum Property, $T$ admits a supremum.
It follows from Supremum of Subset that $\sup T \le \sup S$.
{{qed}}
Theorem: Supremum of Set of Real Numbers is at least Supremum of Subset

Let $S$ be a set of real numbers.
Let $S$ have a supremum.
Let $T$ be a non-empty subset of $S$.
Then $\sup T$ exists and:
:$\sup T \le \sup S$

Proof:
$S$ is bounded above as $S$ has a supremum.
Therefore, $T$ is bounded above as $T$ is a subset of $S$.
Accordingly, $T$ admits a supremum by the Continuum Property as $T$ is non-empty.
We know that $\sup T$ and $\sup S$ exist.
Therefore by Suprema of two Real Sets:
:$\forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon \iff \sup T \le \sup S$
We have:
{{begin-eqn}}
{{eqn | q = \forall \epsilon \in \R_{>0}
      | l = 0
      | o = <
      | r = \epsilon
}}
{{eqn | ll= \leadsto
      | q = \forall \epsilon \in \R_{>0}: \forall t \in T
      | l = t
      | o = < 
      | r = t + \epsilon
}}
{{eqn | ll= \leadsto
      | q = \forall \epsilon \in \R_{>0}: \forall t \in T
      | l = t
      | o = <
      | r = s + \epsilon
}}
{{eqn | lo= \land
      | l = s
      | r = t
}}
{{eqn | ll= \leadsto
      | q = \forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S
      | l = t
      | o = <
      | r = s + \epsilon
      | c = as $T \subseteq S$
}}
{{eqn | ll= \leadsto
      | l = \sup T
      | o = \le
      | r = \sup S
}}
{{end-eqn}}
{{qed}}
Theorem: Supremum of Set of Real Numbers is at least Supremum of Subset

Let $S$ be a set of real numbers.
Let $S$ have a supremum.
Let $T$ be a non-empty subset of $S$.
Then $\sup T$ exists and:
:$\sup T \le \sup S$

Proof:
By definition $\sup S$ is an upper bound for $S$.
Thus:
:$\forall x \in S: x \le \sup S$
As $T \subseteq S$ we have by definition of subset that:
:$\forall x \in T: x \in S$
Hence:
:$\forall x \in T: x \le \sup S$
So by definition $\sup S$ is an upper bound for $T$.
So $\sup S$ is at least as big as the smallest upper bound for $T$
Thus by definition of supremum:
:$\sup T \le \sup S$
{{qed}}
Theorem: Supremum of Subset of Union Equals Supremum of Union

Let $S$ be a non-empty real set.
Let $S$ have a supremum.
Let $\set {S_i: i \in \set {1, 2, \ldots, n} }$, $n \in \N_{>0}$, be a set of non-empty subsets of $S$.
Let $\bigcup S_i = S$.
Then there exists a $j$ in $\set {1, 2, \ldots, n}$ such that:
:$\sup S_j = \sup S$

Proof:
If $S$ equals $S_j$ for a $j$ in $\set {1, 2, \ldots, n}$, it is trivially true that $\sup S = \sup S_j$.
Now assume that $S$ is unequal to $S_i$ for every $i$ in $\left\{{1, 2, \ldots, n}\right\}$.
By Supremum of Set of Real Numbers is at least Supremum of Subset, $\sup S \ge \sup S_i$ for every $i$ in $\set{1, 2, \ldots, n}$.
There are two alternatives; either:
:$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$
or:
:$\sup S = \sup S_j$ for at least one $j$ in $\set {1, 2, \ldots, n}$.
Suppose that:
:$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$
Let $\epsilon = \sup S - \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}$.
We note that $\epsilon > 0$.
By Supremum of Subset of Real Numbers is Arbitrarily Close, $S$ has an element $x$ that satisfies:
:$x > \sup S - \epsilon$
We have:
{{begin-eqn}}
{{eqn | o = >
      | l = x
      | r = \sup S - \epsilon
}}
{{eqn | r = \sup S - \paren {\sup S - \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n} }
      | c = definition of $\epsilon$
}}
{{eqn | r = \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}
}}
{{end-eqn}}
Therefore:
:$x > \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}$
This means that $x > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$.
However, $x$ must be an element of $S_j$ for some $j$ in $\set {1, 2, \ldots, n}$ as $x \in S$ and $S = \bigcup S_i$.
Accordingly, it is not true that $\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$.
We just concluded that the alternative:
:$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$
is not true.
Therefore, the other alternative:
:$\sup S = \sup S_j$ for a $j$ in $\set {1, 2, \ldots, n}$
is true.
{{qed}}
Category:Real Analysis
2p2kel5xg9ahcfw856r2xlxijud240n
Theorem: Condition for Ideal to be Total Ring

Let $\struct {A, +, \circ}$ be a commutative ring with unity.
Let $I$ be an ideal of $A$ such that the quotient ring $A / I$ is a field.
Let $J$ be an ideal of $A$ such that $I \subsetneq J$.
Then:
:$A = J$

Proof:
Let $A$ be a commutative ring with unity.
Let $I$ be an ideal of $A$ such that the quotient ring $A / I$ is a field.
Let $J$ be an ideal of $A$ such that $I \subsetneq J$.
From Ideal is Subring:
:$J \subseteq A$
It remains to be proved that that $A \subseteq J$.
Let $a \in A$.
As $I \subsetneq J$, it follows from definition of proper subset that:
:$\exists j \in J: j \notin I$
Consider the coset $j + I \in  A / I$.
As $A / I$ is a field:
:$\exists C \in A / I: \paren {j + I} \circ C = 1 + I$
Let $j' \in A$ be such that $C = j' + I$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {j + I} \circ \paren {j' + I}
      | r = 1 + I
      | c = 
}}
{{eqn | ll= \leadsto
      | l = j \circ j' + I
      | r = 1 + I
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a \circ \paren {j \circ j'} + I
      | r = a + I
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists i, i' \in I
      | l = a \circ \paren {j \circ j'} + i
      | r = a + i'
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a
      | r = a \circ j \circ j' + i - i'
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a \circ j
      | o = \in
      | r = J
      | c = {{Defof|Ideal of Ring}}
}}
{{eqn | ll= \leadsto
      | l = a \circ j \circ j'
      | o = \in
      | r = J
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a \cdot j \cdot j' + i - i'
      | o = \in
      | r = J
      | c = {{Defof|Ideal of Ring}} and $I \subsetneq J$
}}
{{eqn | ll= \leadsto
      | l = a
      | o = \in
      | r = J
      | c = {{Defof|Ideal of Ring}} and $I \subsetneq J$
}}
{{end-eqn}}
By definition of subset:
:$A \subseteq J$
Thus $A = J$ follows by definition of set equality.
{{qed}}
{{improve|Might be able to get there quicker by using Ideals of Field on the quotient ring $A / I$.<br>Or, using Ideal is Unit Ideal iff Includes Unity}}
Category:Ideal Theory
jea6ealtwqcnldn7a6mou2866tiza15
Theorem: Area between Two Non-Intersecting Chords

Let $AB$ and $CD$ be two chords of a circle whose center is at $O$ and whose radius is $r$.
:400px
:400px
Let $\alpha$ and $\theta$ be respectively the measures in radians of the angles $\angle COD$ and $\angle AOB$.
Then the area $\AA$ between the two chords is given by:
:$\AA = \dfrac {r^2} 2 \paren {\theta - \sin \theta - \alpha + \sin \alpha}$
if $O$ is not included in the area, and:
:$\AA = r^2 \paren {\pi - \dfrac 1 2 \paren {\theta - \sin \theta + \alpha - \sin \alpha} }$
if $O$ is included in the area.

Proof:
Let $\SS_\alpha$ be the area of the segment whose base subtends $\alpha$.
Let $\SS_\theta$ be the area of the segment whose base subtends $\theta$.
Theorem: Supremum of Sum equals Sum of Suprema

Let $A$ and $B$ be non-empty sets of real numbers.
Let $A + B$ be $\set {x + y: x \in A, y \in B}$.
Let either $A$ and $B$ have suprema or $A + B$ have a supremum.
Then all $\sup A$, $\sup B$, and $\sup \paren {A + B}$ exist and:
:$\sup \paren {A + B} = \sup A + \sup B$

Proof:
Assume first that $A$ and $B$ have suprema.
We have:
:$x \le \sup A$ for an arbitrary $x$ in $A$
:$y \le \sup B$ for an arbitrary $y$ in $B$
Adding these inequalities, we get:
:$x + y \le \sup A + \sup B$
The number $x + y$ is an arbitrary element of $A + B$ as $x$ and $y$ are arbitrary elements of $A$ and $B$ respectively.
Therefore, $\sup A + \sup B$ is an upper bound for $A + B$.
$A + B$ is non-empty as $A$ and $B$ are non-empty.
Accordingly, $A + B$ has a supremum by the Continuum Property.
Next, assume that $\sup \paren {A + B}$ has a supremum.
We need to prove that $A$ and $B$ have suprema.
Let $y$ be a point in $B$.
We have:
{{begin-eqn}}
{{eqn | l = x + y
      | o = \le
      | r = \sup \paren {A + B}
      | c = for every $x$ in $A$ as $x + y$ is a point in $A + B$
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | o = \le
      | r = \sup \paren {A + B} - y
}}
{{end-eqn}}
Therefore, $A$ has an upper bound as $x$ is an arbitrary point in $A$.
Also, we have that $A$ is non-empty.
Accordingly, $A$ has a supremum by the Continuum Property.
A similar argument gives that $B$ has a supremum.
So, we have shown that all $\sup A$, $\sup B$, and $\sup \paren {A + B}$ exist.
We proceed to show that $\sup \paren {A + B} = \sup A + \sup B$.
We have $\sup \paren {A + B} \le \sup A + \sup B$ as $\sup A + \sup B$ is an upper bound for $A + B$.
Accordingly, either:
:$\sup \paren {A + B} < \sup A + \sup B$
or:
:$\sup \paren {A + B} = \sup A + \sup B$.
{{AimForCont}}:
:$\sup \paren {A + B} < \sup A + \sup B$.
Let $\epsilon = \sup A + \sup B - \sup \paren {A + B}$.
We note that $\epsilon > 0$.
Since $\sup A$ is the least upper bound of $A$, there is an element $x$ in $A$ such that:
:$x > \sup A - \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close
Since $\sup B$ is the least upper bound of $B$, there is an element $y$ in $B$ such that:
:$y > \sup B - \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close
Adding these inequalities, we get:
{{begin-eqn}}
{{eqn | l = x + y
      | o = >
      | r = \sup A - \frac \epsilon 2 + \sup B - \frac \epsilon 2
}}
{{eqn | ll= \leadstoandfrom
      | l = x + y
      | o = >
      | r = \sup A + \sup B - \epsilon
}}
{{eqn | ll= \leadstoandfrom
      | l = x + y
      | o = >
      | r = \sup A + \sup B - \paren {\sup A + \sup B - \sup \paren {A + B} }
      | c = definition of $\epsilon$
}}
{{eqn | ll= \leadstoandfrom
      | l = x + y
      | o = >
      | r = \sup \paren {A + B}
}}
{{end-eqn}}
which is impossible since the number $x + y$ is an element of $A + B$ as $x \in A$ and $y \in B$.
We have found that:
:$\sup \paren {A + B} < \sup A + \sup B$ is not true.
Therefore:
:$\sup \paren {A + B} = \sup A + \sup B$ as $\sup \paren {A + B} \le \sup A + \sup B$.
{{qed}}
Category:Real Analysis
3x3xkkrn1xt0c3nfljsduatkyw9cli3
Theorem: Edge is Bridge iff in All Spanning Trees

Let $G$ be a simple graph.
Let $e$ be an edge of $G$.
Then $e$ is a bridge in $G$ {{iff}} $e$ belongs to every spanning tree for $G$.

Proof:
=== Necessary Condition ===
Let $e$ be a bridge.
That is, suppose the edge deletion $G - e$ is disconnected.
Let $T$ be an arbitrary spanning tree for $G$.
By definition $T$ is a connected subgraph of $G$.
If $T$ did not contain $e$, then it would also be a subgraph of $G - e$.
This contradicts the fact that $G - e$ is disconnected.
Therefore $e$ is in $T$.
{{qed|lemma}}
Theorem: Edge is Minimum Weight Bridge iff in All Minimum Spanning Trees

Let $G$ be an undirected network.
Let every edge of $G$ have a unique weight.
Let $e$ be an edge of $G$.
Then $e$ is a bridge of minimum weight in $G$ {{iff}} $e$ belongs to every minimum spanning tree of $G$.

Proof:
=== Necessary Condition ===
{{AimForCont}} $e$ is a bridge of minimum weight that does not belong to some minimum spanning tree $Q$.
Let $e$ be added to $Q$ to make $Q'$.
Then $e$ forms part of a unique cycle $C$ in $Q$.
Thus there exists an edge $f \in C$ such that $\map w Q < \map w {Q + e - f}$.
This contradicts the minimality of $Q$.
{{qed|lemma}}
{{handwaving|The above argument (cleaned up considerably from the original, which was appallingly inadequate) needs to be made rigorous.}}
Theorem: Maximum Weight Edge in all Minimum Spanning Trees is Bridge

Let $G$ be an undirected network.
Let every edge of $G$ have a unique weight.
Let $e$ be an edge of $G$ that belongs to every minimum spanning tree of $G$.
Let $e$ have maximum weight in $G$.
Then $e$ is a bridge in $G$.

Proof:
{{proof wanted}}
Category:Network Theory
qsmxtomolnooyfurqaijieuu9q39qc0
Theorem: Scaling Property of Dirac Delta Function

Let $\map \delta t$ be the Dirac delta function.
Let $a$ be a non zero constant real number.
Then:
:$\map \delta {a t} = \dfrac {\map \delta t} {\size a}$

Proof:
The equation can be rearranged as:
:$\size a \map \delta {a t} = \map \delta t$
We will check the definition of Dirac delta function in turn.
Definition of Dirac delta function:
{{begin-eqn}}
{{eqn | n = 1
      | l = \map \delta t
      | r = \begin{cases} +\infty & : t = 0 \\ 0 & : \text{otherwise} \end{cases}
      | c = 
}}
{{eqn | n = 2
      | l = \int_{-\infty}^{+\infty} \map \delta t \rd t
      | r = 1
      | c = 
}}
{{end-eqn}}
$(1):$
{{begin-eqn}}
{{eqn | l = \size a \map \delta {a t}
      | r = \begin{cases} \paren {\size a} \paren {+\infty} & : a t = 0 \\ \paren {\size a} 0 & : \text{otherwise} \end{cases}
      | c = {{Defof|Dirac Delta Function}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \size a \map \delta {a t}
      | r = \begin{cases} +\infty & : t = 0 \\ 0 & : \text{otherwise} \end{cases}
      | c = simplifying
}}
{{end-eqn}}
$(2):$
The proof of this part will be split into two parts, one for positive $a$ and one for negative $a$.
For $a > 0$:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^{+\infty} \size a \map \delta {a t} \rd t
      | r = \int_{-\infty}^{+\infty} \size a \map \delta t \dfrac {\rd t} a
      | c = Substitute $t \mapsto \dfrac t a$
}}
{{eqn | r = \dfrac {\size a} a \int_{-\infty}^{+\infty} \map \delta t \rd t
      | c = Simplifying
}}
{{eqn | r = \dfrac a a \int_{-\infty}^{+\infty} \map \delta t \rd t
      | c = $a > 0$
}}
{{eqn | r = 1
      | c = {{Defof|Dirac Delta Function}}
}}
{{end-eqn}}
{{qed|lemma}}
For $a < 0$:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^{+\infty} \size a \map \delta {a t} \rd t
      | r = \int_{+\infty}^{-\infty} \size a \map \delta t \dfrac {\rd t} a
      | c = Substitute $t \mapsto \dfrac t a$
}}
{{eqn | r = \dfrac {\size a} a \int_{+\infty}^{-\infty} \map \delta t \rd t
      | c = Simplifying
}}
{{eqn | r = \dfrac {-\size a} a \int_{-\infty}^{+\infty} \map \delta t \rd t
      | c = Reversal of Limits of Definite Integral
}}
{{eqn | r = \dfrac a a \int_{-\infty}^{+\infty} \map \delta t \rd t
      | c = $a < 0$
}}
{{eqn | r = 1
      | c = {{Defof|Dirac Delta Function}}
}}
{{end-eqn}}
{{qed|lemma}}
Therefore, by definition, $\size a \map \delta {a t} = \map \delta t$.
The result follows after rearrangement.
{{qed}}
Category:Dirac Delta Function
1za4mu99dxa1pjuhn9gr7h64lxw4hbc
Theorem: Supremum of Function is less than Supremum of Greater Function

Let $f$ and $g$ be real functions.
Let $S$ be a subset of $\Dom f \cap \Dom g$.
Let $\map f x \le \map g x$ for every $x \in S$.
Let $\ds \sup_{x \mathop \in S} \map g x$ exist.
Then $\ds \sup_{x \mathop \in S} \map f x$ exists and:
:$\ds \sup_{x \mathop \in S} \map f x \le \sup_{x \mathop \in S} \map g x$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \sup g
      | r = \map \sup {f + \paren {g - f} }
}}
{{eqn | r = \sup f + \sup \left({g - f}\right)
      | c = Supremum of Sum equals Sum of Suprema
}}
{{end-eqn}}
Supremum of Sum equals Sum of Suprema also gives that $\sup f$ and $\sup \paren {g - f}$ exist.
We have:
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = \map g x
      | o = \ge
      | r = \map f x
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall x \in S
      | l = \map g x - \map f x
      | o = \ge
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall x \in S
      | l = \map \sup {g - f}
      | o = \ge
      | r = \map g x - \map f x \ge 0
      | c = as $\map \sup {g - f}$ is an upper bound for $\set {\map g x - \map f x: x \in S}$
}}
{{eqn | ll= \leadsto
      | l = \map \sup {g - f}
      | o = \ge
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \sup f + \map \sup {g - f}
      | o = \ge
      | r = \sup f
}}
{{eqn | ll= \leadsto
      | l = \sup g
      | o = \ge
      | r = \sup f
      | c = as $\sup g = \sup f + \map \sup {g - f}$
}}
{{end-eqn}}
{{qed}}
Category:Real Analysis
Category:Suprema
abcrpkv3cgbf2fawmpv51ch6badcotp
Theorem: Characterization of Boundary by Open Sets

Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $T$.
Let $x$ be a point of $T$.
Then $x \in \partial A$ {{iff}}:
:for every open set $U$ of $T$:
::if $x \in U$
::then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$
where:
:$\relcomp S A = S \setminus A$ denotes the complement of $A$ in $S$
:$\partial A$ denotes the boundary of $A$.

Proof:
=== Sufficient Condition ===
Let $x \in \partial A$.
Then by Boundary is Intersection of Closure with Closure of Complement:
:$x \in \paren {\relcomp S A}^-$ and $x \in A^-$
where $A^-$ denotes the closure of $A$.
Hence by Condition for Point being in Closure, for every open set $U$ of $T$:
:$x \in U \implies A \cap U \ne \O$
and:
:$x \in U \implies \relcomp S A \cap U \ne \O$