nyuuzyou/wikis huggingface dataset:
Upper Section with no Minimal Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $U \subseteq S$.
Then:
:$U$ is an upper section in $S$ with no minimal element
:
:$\ds U = \bigcup \set {u^\succ: u \in U}$
where $u^\succ$ is the strict upper closure of $u$.
== Proof ==
=== Forward implication ===
Let $U$ be an upper section in $S$ with no minimal element.
Then by the definition of upper section:
:$\ds \bigcup \set {u^\succ: u \in U} \subseteq U$
Let $x \in U$.
Since $U$ has no minimal element, $x$ is not minimal.
Thus there is a $u \in U$ such that $u \prec x$.
Then $x \in u^\succ$, so:
:$\ds x \in \bigcup \set {u^\succ: u \in U }$
Since this holds for all $x \in U$:
:$\ds U \subseteq \bigcup \set {u^\succ: u \in U}$
Thus the theorem holds by definition of set equality.
=== Reverse implication ===
Let:
:$\ds U = \bigcup \set {u^\succ: u \in U}$
Then:
:$\forall u \in U: u^\succ \subseteq U$
so $U$ is an upper section.
Furthermore:
:$\forall x \in U: \exists u \in U: x \in u^\succ$
But then:
:$u \prec x$
so $x$ is not minimal.
Since this holds for all $x \in U$, $U$ has no minimal element.
== Also see ==
* Lower Section with no Maximal Element
Category:Upper Sections
Upper Section with no Smallest Element is Open in GO-Space

== Theorem ==
Let $\struct {S, \preceq, \tau}$ be a generalized ordered space.
Let $U$ be an upper section in $S$ with no smallest element.
Then $U$ is open in $\struct {S, \preceq, \tau}$.
== Proof ==
By Minimal Element in Toset is Unique and Smallest, $U$ has no minimal element.
By Upper Section with no Minimal Element:
:$U = \bigcup \set {u^\succ: u \in U}$
where $u^\succ$ is the strict upper closure of $u$.
By Open Ray is Open in GO-Space and the fact that a union of open sets is open, $U$ is open.
== Also see ==
* Lower Section with no Greatest Element is Open in GO-Space
Category:Generalized Ordered Spaces
Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $f$ be a bounded real function defined on $\closedint a b$.
Let $P$ and $Q$ be finite subdivisions of $\closedint a b$.
Let $\map L P$ be the lower Darboux sum of $f$ on $\closedint a b$ with respect to $P$.
Let $\map U Q$ be the upper Darboux sum of $f$ on $\closedint a b$ with respect to $Q$.
Then $\map L P \le \map U Q$.
== Proof ==
Let $P' = P \cup Q$.
We observe:
:$P'$ is either equal to $P$ or finer than $P$
:$P'$ is either equal to $Q$ or finer than $Q$
We find:
:$\map L P \le \map L {P'}$ by Lower Sum of Refinement
:$\map L {P'} \le \map U {P'}$ by Upper Darboux Sum Never Smaller than Lower Darboux Sum
:$\map U {P'} \le \map U Q$ by Upper Sum of Refinement
By combining these inequalities, we conclude:
:$\map L P \le \map U Q$
== Sources ==
* : $2.5$: The Riemann Integral
Category:Real Analysis
Upper Sum of Refinement

== Theorem ==
Let $\closedint a b$ be a closed interval.
Let $P$ be a finite subdivision of $\closedint a b$. 
Let $Q$ be a refinement of $P$. 
Then: 
:$\map U {f, P} \le \map U {f, Q}$
where $\map U {f, P}$ and $\map U {f, Q}$ denote the upper Darboux sum of $f$ with respect to $P$ and $Q$ respectively.
== Proof ==
Write: 
:$P = \set {x_0, x_1, \ldots, x_k}$
and:
:$Q = \set {y_0, y_1, \ldots, y_l}$
where:
:$a = x_0 < x_1 < \ldots < x_k = b$
and:
:$a = y_0 < y_1 < \ldots < y_l = b$
Since $P \subseteq Q$, we have $k \le l$ from Cardinality of Subset of Finite Set.
Set: 
:$M_i = \sup \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for each $1 \le i \le k$.
Also set: 
:${\tilde M}_j = \sup \set {\map f x : x \in \closedint {y_{j - 1} } {y_j} }$
for each $1 \le j \le l$. 
Consider a pair of elements $\tuple {x_{i - 1}, x_i}$ in $P$. 
Since $P \subseteq Q$, there exists $a_i, b_i$ such that: 
:$\tuple {x_{i - 1}, x_i} = \tuple {y_{a_i}, y_{b_i} }$
We can see that: 
:$a_1 = 0$
and:
:$b_k = l$ 
We also clearly have: 
:$b_{i - 1} = a_i$ for each $1 \le i \le k$. 
Note that: 
:$\closedint {y_{j - 1} } {y_j} \subseteq \closedint {x_{i - 1} } {x_i}$
for all $a_i + 1 \le j \le b_i$. 
So:
:$\set {\map f x : x \in \closedint {y_{j - 1} } {y_j} } \subseteq \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for all $a_i + 1 \le j \le b_i$. 
So, from Supremum of Subset, we have:
:$\sup \set {\map f x : x \in \closedint {y_{j - 1} } {y_j} } \le \sup \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for all $a_i + 1 \le j \le b_i$.
That is: 
:${\tilde M}_j \le M_i$
for each $\tuple {i, j}$ with $a_i + 1 \le j \le b_i$
We can then write: 
:$\ds x_i - x_{i - 1} = \sum_{j \mathop = a_i + 1}^{b_i} \paren {y_j - y_{j - 1} }$
for each $1 \le i \le k$, giving:
== Sources ==
* : $2.5$: The Riemann Integral
Category:Real Analysis
Upper Triangular Matrix/Examples

== Examples of Upper Triangular Matrices ==
=== Upper Triangular Matrix with fewer Rows than Columns ===
=== Upper Triangular Matrix with more Rows than Columns ===
=== Square Upper Triangular Matrix ===
=== Example of Square Upper Triangular Matrix ===
=== Example of Non-Square Upper Triangular Matrix ===
=== Upper Triangular Matrix not in Echelon Form ===
=== Arbitrary Example ===
Category:Examples of Triangular Matrices
Upper Triangular Matrix/Examples/Non-Square Example

== Example of Upper Triangular Matrix ==
This is an arbitrary example of an upper triangular matrix which is specifically not square:
:$\begin {pmatrix} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 0 & 0 & 0 & 10 \\ 0 & 0 & 0 & 0 \end {pmatrix}$
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.3$: $m \times n$ matrices
Category:Examples of Triangular Matrices
Upper Triangular Matrix/Examples/Not in Echelon Form

== Example of Upper Triangular Matrix ==
This is an arbitrary example of an upper triangular square matrix which is specifically not in echelon form (non-unity variant):
:$\begin {pmatrix} 1 & 2 & 3 & 4 \\ 0 & 0 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 0 & 0 & 0 & 10 \end {pmatrix}$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.8$
Category:Examples of Triangular Matrices
Upper Triangular Matrix/Examples/Square Example

== Example of Upper Triangular Matrix ==
This is an arbitrary example of an upper triangular square matrix:
:$\begin {pmatrix} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 0 & 0 & 0 & 10 \end {pmatrix}$
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.3$: $m \times n$ matrices
Category:Examples of Triangular Matrices
Upper Triangular Matrix/Examples/Square Matrix

== Example of Upper Triangular Matrix ==
An upper triangular square matrix of order $n$:
:$\mathbf U = \begin{bmatrix}
a_{11} & a_{12} & a_{13} & \cdots & a_{1, n - 1} & a_{1n} \\
 0 & a_{22} & a_{23} & \cdots & a_{2, n - 1} & a_{2n} \\
 0 & 0 & a_{33} & \cdots & a_{3, n - 1} & a_{3n} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
 0 & 0 & 0 & \cdots & a_{n - 1, n - 1} & a_{n - 1, n} \\
 0 & 0 & 0 & \cdots & 0 & a_{nn} \\
\end{bmatrix}$
Category:Examples of Triangular Matrices
Upper Triangular Matrix/Examples/m greater than n

== Example of Upper Triangular Matrix ==
An upper triangular matrix of order $m \times n$ such that $m > n$:
:$\mathbf U = \begin{bmatrix}
a_{11} & a_{12} & a_{13} & \cdots & a_{1, n - 1} & a_{1n} \\
 0 & a_{22} & a_{23} & \cdots & a_{2, n - 1} & a_{2n} \\
 0 & 0 & a_{33} & \cdots & a_{3, n - 1} & a_{3n} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
 0 & 0 & 0 & \cdots & a_{n - 1, n - 1} & a_{n - 1, n} \\
 0 & 0 & 0 & \cdots & 0 & a_{nn} \\
 0 & 0 & 0 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
 0 & 0 & 0 & \cdots & 0 & 0 \\
 0 & 0 & 0 & \cdots & 0 & 0 \\
\end{bmatrix}$
Category:Examples of Triangular Matrices
Upper Triangular Matrix/Examples/m less than n

== Example of Upper Triangular Matrix ==
An upper triangular matrix of order $m \times n$ such that $m < n$:
:$\mathbf U = \begin{bmatrix}
a_{11} & a_{12} & a_{13} & \cdots & a_{1, m - 1} & a_{1m} & \cdots & a_{1, n - 1} & a_{1n} \\
 0 & a_{22} & a_{23} & \cdots & a_{2, m - 1} & a_{2m} & \cdots & a_{2, n - 1} & a_{2n} \\
 0 & 0 & a_{33} & \cdots & a_{3, m - 1} & a_{3m} & \cdots & a_{3, n - 1} & a_{3n} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\
 0 & 0 & 0 & \cdots & a_{m - 1, m - 1} & a_{m - 1, m} & \cdots & a_{m - 1, n - 1} & a_{m - 1, n} \\
 0 & 0 & 0 & \cdots & 0 & a_{mm} & \cdots & a_{m, n - 1} & a_{mn} \\
\end{bmatrix}$
Category:Examples of Triangular Matrices
Upper and Lower Bound of Fibonacci Number

== Theorem ==
For all $n \in \N_{> 0}$:
:$\phi^{n - 2} \le F_n \le \phi^{n - 1}$
where:
:$F_n$ is the $n$th Fibonacci number
:$\phi$ is the golden section: $\phi = \dfrac {1 + \sqrt 5} 2$
== Proof ==
From Fibonacci Number greater than Golden Section to Power less Two:
:$F_n \ge \phi^{n - 2}$
From Fibonacci Number less than Golden Section to Power less One:
:$F_n \le \phi^{n - 1}$
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers
Category:Fibonacci Numbers
Category:Golden Mean
Upper and Lower Closures are Convex

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $a \in S$.
Then $a^\succeq$, $a^\succ$, $a^\preceq$, and $a^\prec$ are convex in $S$.
== Proof ==
The cases for upper and lower closures are dual, so we need only prove the case for upper closures.
Suppose, then, that $C = a^\succeq$ or $C = a^\succ$.
Suppose that $x, y, z \in S$, $x \prec y \prec z$, and $x, z \in C$.
Then $a \preceq x \prec y$, so $a \prec y$ by Extended Transitivity.
Therefore $y \in a^\succ \subseteq C$.
Thus $C$ is convex.
Category:Lower Closures
Category:Upper Closures
Category:Convex Sets (Order Theory)
Upper and Lower Closures of Open Set in GO-Space are Open

== Theorem ==
Let $\struct {X, \preceq, \tau}$ be a Generalized Ordered Space/Definition 1.
Let $A$ be open in $X$.
Then the upper and lower closures of $A$ are open.
== Proof ==
We will show that the upper closure $U$ of $A$ is open.
The lower closure can be proven open by the same method.
By the definition of upper closure:
:$U = \set {u \in X: \exists a \in A: a \preceq u}$
But then:
where $a^\preceq$ denotes the upper closure of $a$.
By Open Ray is Open in GO-Space/Definition 1, each $a^\succeq$ is open.
Thus $U$ is a union of open sets.
Thus $U$ is open by the definition of a topology.
Category:Topology
Category:Total Orderings
Upward Löwenheim-Skolem Theorem

== Theorem ==
Let $T$ be an $\LL$-theory with an infinite model.
Then for each infinite cardinal $\kappa \ge \card \LL$, there exists a model of $T$ with cardinality $\kappa$.
== Proof ==
The idea is:
:to extend the language by adding $\kappa$ many new constants
and:
:to extend the theory by adding sentences asserting that these constants are distinct.
It is shown that this new theory is finitely satisfiable using an infinite model of $T$.
Compactness then implies that the new theory has a model.
Some care needs to be taken to ensure that we construct a model of exactly size $\kappa$.
Let $\LL^*$ be the language formed by adding new constants:
:$\set {c_\alpha: \alpha < \kappa}$ to $\LL$.
Let $T^*$ be the $\LL^*$-theory formed by adding the sentences:
:$\set {c_\alpha \ne c_\beta: \alpha, \beta < \kappa, \ \alpha \ne \beta}$
to $T$.
We show that $T^*$ is finitely satisfiable:
Let $\Delta$ be a finite subset of $T^*$.
Then $\Delta$ contains:
:finitely many sentences from $T$
along with:
:finitely many sentences of the form $c_\alpha \ne c_\beta$ for the new constant symbols.
Since $T$ has an infinite model, it must have a model $\MM$ of cardinality at most:
:$\card \LL + \aleph_0$
This model already satisfies everything in $T$.
So, since we can find arbitrarily many distinct elements in it, it can also be used as a model of $\Delta$ by interpreting the finitely many new constant symbols in $\Delta$ as distinct elements of $\MM$.
Since $T^*$ is finitely satisfiable, it follows by the Compactness Theorem that $T^*$ itself is satisfiable.
Since $T^*$ ensures the existence of $\kappa$ many distinct elements, this means it has models of size at least $\kappa$.
It can be proved separately or observed from the ultraproduct proof of the compactness theorem that $T^*$ then has a model $\MM^*$ of exactly size $\kappa$.
Since $T^*$ contains $T$, $\MM^*$ is a model of $T$ of size $\kappa$.
== Also see ==
* Overflow Theorem
== Sources ==
* : $\mathrm{II}.7$ First-Order Logic Semantics: Theorem $\mathrm{II.7.17}$
Category:Model Theory for Predicate Logic
Urysohn's Lemma Converse

== Lemma ==
Let $T = \struct {S, \tau}$ be a topological space.
Let there exist an Urysohn function for any two $A, B \subseteq S$ which are closed sets in $T$ such that $A \cap B = \O$.
Then $T = \struct {S, \tau}$ is a $T_4$ space.
== Proof ==
Let $A$ and $B$ be arbitrary closed sets of $T$ and let $f$ be an Urysohn function for $A$ and $B$.
Let $C = \hointr 0 {\dfrac 1 4}$ and $D = \hointl {\dfrac 3 4} 1$.
Since:
:$C = \openint {-1} {\dfrac 1 4} \cap \closedint 0 1$
:$D = \openint {\dfrac 3 4} 2 \cap \closedint 0 1$
each is open in $\closedint 0 1$ under the subspace topology by definition.
Hence, $\map {f^{-1} } C$ and $\map {f^{-1} } D$ are open in $T$.
Furthermore, by definition of Urysohn function, $A \subset \map {f^{-1} } C$ and $B \subset \map {f^{-1} } D$.
Also, from Preimage of Intersection under Mapping:
:$C \cap D = \O \implies \map {f^{-1} } C \cap \map {f^{-1} } D = \O$
Therefore, $T$ is a $T_4$ space.
== Also see ==
* Urysohn's Lemma
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Category:Closed Sets
Category:T4 Spaces
Category:Urysohn Functions
Urysohn's Metrization Theorem

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is regular and second-countable.
Then $T$ is metrizable.
== Proof ==
From Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube:
:$T$ is homeomorphic to a subspace of the Hilbert cube $\struct{I^\omega, d_2}$ 
where $d_2$ is the metric: $\ds \forall x = \sequence {x_i}, y = \sequence {y_i} \in I^\omega: \map {d_2} {x, y} := \paren {\sum_{k \mathop \in \N_{>0} } \paren {x_k - y_k}^2}^{\frac 1 2}$
By definition of metrizable topology:
:$\struct{I^\omega, \tau_{d_2}}$ is metrizable
where $\tau_{d_2}$ is the topology induced by $d_2$.
From Subspace of Metrizable Space is Metrizable Space:
:$T$ is homeomorphic to a metrizable space
It follows that $T$ is metrizable by definition.
== Also see ==
* Metrizable Space is not necessarily Second-Countable, indicating that the converse does not hold.
* Metrization of Regular Second Countable Space, for necessary and sufficient conditions for the metrization of regular second-countable spaces.
== Historical Note ==
== Sources ==
* : Chapter $4$: Embedding and Metrization, $\S$ Metrization, Theorem $16$
* : Chapter $7$: Metrizable Spaces: $\S23$: Metrization: Definition $23.1$
* : Chapter $4$: Countability and Separation Axioms: $\S34$: The Urysohn Metrization Theorem: Theorem $34.1$
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Metrizability
Category:Metrization Theorems
Category:Metrizable Topologies
Category:Regular Spaces
Category:Second-Countable Spaces
Urysohn Space is Completely Hausdorff Space

== Theorem ==
Let $\struct {S, \tau}$ be an Urysohn space.
Then $\struct {S, \tau}$ is also a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
== Proof ==
Let $T = \struct {S, \tau}$ be an Urysohn space.
Then for any distinct points $x, y \in S$ (i.e. $x \ne y$), there exists an Urysohn function for $\set x$ and $\set y$.
Thus:
:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ 
which is precisely the definition of a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties
* : Problems: $\S 2: \ 12$
Category:Completely Hausdorff Spaces
Category:Urysohn Spaces
Utility Theory/Examples/Arbitrary Example 1

== Examples of Utility Theory ==
Suppose the organizers of a social event are to decide whether to hold it indoors or out of doors.
Suppose they are able to estimate, based on past experience, the likely attendance.
Numbers attending would then be a useful utility measure as a generator of income, based on every attendee paying the same admission charge.
Suppose the probability of rain is $0.1$ and of no rain is $0.9$.
Then the likely attendances are as given in the following contingency table:
:$\begin{array}{l|cc|c}
& \text {Rain} & \text {No rain} \\
& p = 0.1 & p = 0.9 \\
\hline
\text {Held outdoors} & 100 & 650 \\
\text {Held indoors} & 250 & 400 \\
\end{array}$
If held outdoors, the expected utility is given by:
:$\expect U = 100 \times 0.1 + 650 \times 0.9 = 595$
while if held indoors, the expected utility is given by:
:$\expect U = 250 \times 0.1 + 400 \times 0.9 = 385$
Hence to maximize expected utility, the event should be held outdoors.
== Sources ==
* 
Category:Examples of Utility Theory
V Operator on Hilbert Space is Unitary

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ be the Hilbert space direct sum of $\HH$ with itself.
Define $V : \HH \times \HH \to \HH \times \HH$ by:
:$\map V {x, y} = \tuple {-y, x}$
for each $\tuple {x, y} \in \HH \times \HH$.
Then $V : \struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} } \to \struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ is unitary.
== Proof ==
Let $\tuple {x, y}, \tuple {u, v} \in \HH \times \HH$. 
We have:
so $V$ is linear.
Further, for $x, y \in \HH$ we have:
:$V \tuple {y, -x} = \tuple {x, y}$
So $V$ is surjective.
We also have:
We conclude that $V : \struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} } \to \struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ is unitary.
== Sources ==
* : $13.7$: Graphs
Category:V Operators on Hilbert Spaces
Vague Convergence of Uniformly Bounded Sequence of Finite Borel Measures implies Convergence of Integrals of Bounded Continuous Functions Vanishing at Infinity

== Theorem ==
Let $M > 0$.
Let $\mu$ be a finite Borel measure on $\R$ with $\map \mu \R \le M$. 
Let $\sequence {\mu_n}_{n \mathop \in \N}$ be a sequence of finite Borel measures converging vaguely to $\mu$ such that:
:$\map {\mu_n} \R \le M$ for each $n \in \N$.
Let $f : \R \to \R$ be continuous such that:
:$\ds \lim_{\size x \mathop \to \infty} \map f x = 0$
Then:
:$\ds \int f \rd \mu_n \to \int f \rd \mu$
== Proof ==
Let $\epsilon > 0$.
Since:
:$\ds \lim_{\size x \mathop \to \infty} \map f x = 0$
there exists $K > 0$ such that:
:$\ds \size {\map f x} < \frac \epsilon 2$ for $\size x \ge K$.
Define $f_1 : \R \to \R$ by:
:$\ds \map {f_1} x = \begin {cases} \map f x & : x \in \closedint {-K} K \\ \paren {x - K + 1} \map f K & : x \in \closedint K {K + 1} \\ \paren {x + K + 1} \map f {-K} & : x \in \closedint {-K - 1} {-K} \\ 0 & : \text {otherwise} \end {cases}$
Note that since $\map {f_1} x = 0$ for $x \in \R \setminus \closedint {-K - 1} {K + 1}$, $f_1$ has compact support.
Further, we can see that $f_1$ is continuous. 
Hence, from the definition of vague convergence, we have:
:$\ds \int f_1 \rd \mu_n \to \int f_1 \rd \mu$
as $n \to \infty$. 
Define $f_2 : \R \to \R$ by $\map {f_2} x = \map f x - \map {f_1} x$ for each $x \in \R$.
We show that $\size {\map {f_2} x} < \epsilon$ for all $x \in \R$. 
For $x \in \closedint {-K} K$, we have:
:$\map {f_1} x = \map f x$
and hence:
:$\map {f_2} x = 0$
For $x \in \R \setminus \closedint {-K - 1} {K + 1}$, we have:
:$\map {f_1} x = 0$
and hence:
:$\map {f_2} x = \map f x$
Since $\size x \ge K$ in this case, we have:
:$\size {\map {f_2} x} < \dfrac \epsilon 2 < \epsilon$
Now, for $x \in \closedint K {K + 1}$, we have:
Finally, for $x \in \closedint {-K - 1} {-K}$, we have:
We now have:
Since:
:$\ds \int f_1 \rd \mu_n \to \int f_1 \rd \mu$ as $n \to \infty$
there exists $N \in \N$ such that:
:$\ds \size {\int f_1 \rd \mu - \int f_1 \rd \mu_n} < \epsilon$
for $n \ge N$. 
From Triangle Inequality and Triangle Inequality for Integrals, we have:
:$\ds \size {\int f_2 \rd \mu - \int f_2 \rd \mu_n} \le \int \size {f_2} \rd \mu + \int \size {f_2} \rd \mu_n$
From Integral of Integrable Function is Monotone and Integral of Characteristic Function, we have:
:$\ds \int \size {f_2} \rd \mu + \int \size {f_2} \rd \mu_n \le \epsilon \map \mu \R + \epsilon \map {\mu_n} \R \le 2 M \epsilon$ 
Hence we have:
:$\ds \size {\int f \rd \mu - \int f \rd \mu_n} \le \paren {2 M + 1} \epsilon$
for $n \ge N$.
Since $\paren {2 M + 1} \epsilon \to 0$ as $\epsilon \to 0^+$, and $\epsilon > 0$ was arbitrary, we have:
:$\ds \size {\int f \rd \mu - \int f \rd \mu_n} \to 0$
as desired.
== Sources ==
* : $\text A.8$: Vague convergence of measures
Category:Vague Convergence of Borel Measures
Vajda's Identity/Formulation 1

== Theorem ==
Let $F_n$ be the $n$th Fibonacci number.
Then:
:$F_{n + i} F_{n + j} - F_n F_{n + i + j} = \paren {-1}^n F_i F_j$
== Proof ==
From Honsberger's Identity:
Therefore:
Category:Vajda's Identity
Vajda's Identity/Formulation 2

== Theorem ==
Let $F_n$ be the $n$th Fibonacci number.
Then:
:$F_{n + k} F_{m - k} - F_n F_m = \left({-1}\right)^n F_{m - n - k} F_k$
== Proof ==
We have:
Now substitute:
where:
: $\phi$ denotes the golden mean
: $\hat \phi = 1 - \phi$
first into $(2)$:
and then into $(1)$:
$(1) = (2)$ and hence the result.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $17$
Category:Vajda's Identity
Valid Argument/Examples/Bats have Lungs

== Example of Valid Argument ==
The following is a valid argument with true premises:
:All bats are mammals.
:All mammals have lungs.
:Therefore all bats have lungs.
== Proof ==
This is a categorical syllogism of the first figure, of the form $AAA$.
From Valid Patterns of Categorical Syllogism, this is seen to be a valid argument.
The premises are true.
It follows that the argument is sound.
== Sources ==
* : $1$ Introduction: Logic and Language: $1.3$: Truth and Validity
Category:Examples of Valid Arguments
Valid Argument/Examples/Trout have Wings

== Example of Valid Argument ==
The following is a valid argument with false premises:
:All trout are mammals.
:All mammals have wings.
:Therefore all trout have wings.
== Proof ==
This is a categorical syllogism of the first figure, of the form $AAA$.
From Valid Patterns of Categorical Syllogism, this is seen to be a valid argument.
The premises, however, are false.
It follows that the argument, while valid, is unsound.
== Sources ==
* : $1$ Introduction: Logic and Language: $1.3$: Truth and Validity
Category:Examples of Valid Arguments