nyuuzyou/wikis huggingface dataset:
Moment of Inertia of Uniform Solid Right Circular Cone about Axis

== Theorem ==
Let $\CC$ be a solid right circular cone of uniform density of mass $M$ whose base is of radius $a$.
Let $\AA$ be the straight line coinciding with the axis of $\CC$.
Then the moment of inertia $\II$ of $\CC$ about $\AA$ is given by:
:$\II = \dfrac {3 M a^2 } {10}$
== Proof ==
== Sources ==
* : Appendix: Table $5$ Moments of inertia Values for certain uniform bodies of mass $M$ about certain axes.
* : Appendix: Table $5$ Moments of inertia Values for certain uniform bodies of mass $M$ about certain axes.
Category:Moments of Inertia
Category:Right Circular Cones
Moment of Inertia of Uniform Solid Sphere about Diameter

== Theorem ==
Let $\SS$ be a solid sphere of uniform density of mass $M$ and radius $a$.
Let $\AA$ be the straight line coincident with an arbitrary diameter of $\SS$.
Then the moment of inertia $\II$ of $\SS$ about $\AA$ is given by:
:$\II = \dfrac {2 M a^2} 5$
== Proof ==
== Sources ==
* : Appendix: Table $5$ Moments of inertia Values for certain uniform bodies of mass $M$ about certain axes.
* : Appendix: Table $5$ Moments of inertia Values for certain uniform bodies of mass $M$ about certain axes.
Category:Moments of Inertia
Category:Spheres
Moment of Inertia of Uniform Spherical Shell about Diameter

== Theorem ==
Let $\SS$ be a uniform lamina of mass $M$ in the shape of the surface of a sphere of radius $a$.
Let $\AA$ be the straight line coincident with an arbitrary diameter of $\SS$.
Then the moment of inertia $\II$ of $\SS$ about $\AA$ is given by:
:$\II = \dfrac {2 M a^2} 3$
== Proof ==
== Sources ==
* : Appendix: Table $5$ Moments of inertia Values for certain uniform bodies of mass $M$ about certain axes.
* : Appendix: Table $5$ Moments of inertia Values for certain uniform bodies of mass $M$ about certain axes.
Category:Moments of Inertia
Category:Uniform Laminae
Category:Spheres
Monoid/Examples/Real Continuous Functions on Unit Interval

== Example of Monoid ==
Let $S$ be the set of all continuous real functions whose domains and codomains are the closed unit interval $\closedint 0 1$.
Let $\circ: S \times S$ be the mapping composition operation defined on the $S$ as:
:$\forall f, g \in S: \forall x \in \closedint 0 1: \map {f \circ g} x := \map f {\map g x}$
Then $\paren {S, \circ}$ is a monoid whose identity is the identity mapping $I: \closedint 0 1 \to \closedint 0 1$:
:$\forall x \in \closedint 0 1: \map I x = x$
== Proof ==
Let $f, g \in S$ arbitrary.
By Composite of Continuous Mappings is Continuous, $f \circ g \in S$. 
So is satisfied.
Let $f, g, h \in S$ arbitrary.
By Composition of Mappings is Associative and Composite of Continuous Mappings is Continuous:
:$\paren{ f \circ g} \circ h \in \implies f \circ \paren{ g \circ h} \in S$. 
So is satisfied.
By Euclidean Metric on Real Number Line is Metric, the Euclidean metric on the real number line $\R$ is a metric.
By Closed Real Interval is Compact and Compact Subspace of Real Numbers is Closed and Bounded, the Closed unit interval is a metric subspace of the real numbers.
By Subspace of Metric Space is Metric Space the closed unit interval forms a metric space with the metric $d_{ [0,1] }$, where
$d_{[0,1]}: [0,1] \times [0,1] \to \R$ is the restriction $d \restriction_{[0,1] \times [0,1]}$ of $d$ to $[0,1]$.
That is, the metric space is $\paren{ [0, 1], d_{[0,1]}}$.
By Identity Mapping is Continuous/Metric Space, the identity mapping, whose domain and closed unit interval is the closed real interval, is continuous with respect to the metric $d_{[0,1]}$.
So is satisfied.
Hence $\struct {S, \circ}$ is a monoid.
== Sources ==
* 
* 
Category:Examples of Monoids
Monoid/Examples/x+y+xy on Reals

== Example of Monoid ==
Let $\circ: \R \times \R$ be the operation defined on the real numbers $\R$ as:
:$\forall x, y \in \R: x \circ y := x + y + x y$
Then $\struct {\R, \circ}$ is a monoid whose identity is $0$.
== Proof ==
We have that:
:$\forall x, y \in \R: x \circ y \in \R$
and so $\struct {\R, \circ}$ is closed.
Now let $x, y, z \in \R$.
We have:
and:
As can be seen by inspection:
:$x \circ \paren {y \circ z} = \paren {x \circ y} \circ z$
and so $\circ$ is associative.
Then we have:
The result follows by definition of monoid.
== Sources ==
* : Chapter $4$. Groups: Exercise $11$
Category:Examples of Monoids
Monoid Category is Category

== Theorem ==
Let $\left({S, \circ}\right)$ be a monoid with identity $e_S$.
Let $\mathbf S$ be the associated monoid category.
Then $\mathbf S$ is a category.
== Proof ==
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
Suppose that $a, b \in \mathbf S_1$ are morphisms.
Since $\left({S, \circ}\right)$ is closed, $a \circ b$ is also a morphism.
Axiom $(C2)$ follows directly from the axioms for the identity $e_S$ in a monoid.
Axiom $(C3)$ holds because $\circ$ is associative.
This is because $\left({S, \circ}\right)$ is a monoid.
Hence $\mathbf S$ is a metacategory.
As $\mathbf S$ has only one object, and $\mathbf S_1 = S$, it follows that $\mathbf S$ is also a category.
Category:Monoid Categories
Monoid Ring of Commutative Monoid over Commutative Ring is Commutative

== Theorem ==
Let $R$ be a commutative ring.
Let $G$ be a commutative monoid.
Let $\sqbrk R G$ be the monoid ring of $G$ over $R$.
Then $\sqbrk R G$ is commutative.
== Proof ==
Category:Commutative Monoids
Category:Commutative Rings
Monoid is not Empty

== Theorem ==
A monoid cannot be empty.
== Proof ==
Let $\left({S, \circ}\right)$ be a the monoid.
By definition:
:Identity: $\exists e_S \in S: \forall a \in S: a \circ e_S = a = e_S \circ a$
So a monoid must at least have an identity.
Therefore $e_S \in S$ and so $S$ is not the empty set.
== Sources ==
* : $\S 3.1$: Monoids
Category:Monoids
Monoid of Self-Inverse Elements is Abelian Group

== Theorem ==
Let $\struct {S, \circ}$ be a monoid such that:
:$\forall x \in S: x \circ x = e$
where $e$ is the identity element of $\struct {S, \circ}$.
Then $\struct {S, \circ}$ is an abelian group.
== Proof ==
From Equivalence of Definitions of Self-Inverse, $x \circ x = e \implies x = x^{-1}$.
From Inverse in Monoid is Unique, it follows that every element of $\struct {S, \circ}$ has a unique inverse.
So by definition, $\struct {S, \circ}$ is a group.
From All Elements Self-Inverse then Abelian, it follows that $\struct {S, \circ}$ is an abelian group.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.10$
Category:Monoids
Category:Abelian Groups
Monomials form Basis of Polynomial Ring/One Variable

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $R \sqbrk X$ be a polynomial ring over $R$ in the variable $X$.
Then the monomials of $R \sqbrk X$ are a basis of $R \sqbrk X$ as a module over $R$.
== Proof ==
Follows from:
* Polynomial is Linear Combination of Monomials
* Monomials of Polynomial Ring are Linearly Independent
Category:Polynomial Rings
Category:Monomials
Monomials of Polynomial Ring are Linearly Independent/One Variable

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $R \sqbrk X$ be a polynomial ring in one variable $X$ over $R$.
Then the set of monomials $\set {X^k : k \in \N}$ is linearly independent.
== Proof ==
We consider polynomial ring over sequences in $R$:
Let $\struct {R^{\left({\N}\right)}, \oplus, \odot}$ be the underlying ring of sequences of finite support.
Recall:
:$X = \sequence {0, 1, 0, 0, \ldots}$
Observe:
More generally, for all $r \in R$ and $k \in \N$:
Now let $n \in \N$ and $r_0,\ldots,r_n \in R$ be such that:
:$r_0 + r_1 X + r_2 X^2 + \cdots + r_n X^n = 0$
That is:
Therefore:
:$r_0 = \cdots = r_n = 0$
== Also see ==
* Equality of Monomials of Polynomial Ring in One Variable
Category:Monomials
Monomorphism Image is Isomorphic to Domain

== Theorem ==
The image of a monomorphism is isomorphic to its domain.
That is, if $\phi: S_1 \to S_2$ is a monomorphism, then:
:$\phi: S_1 \to \Img \phi$
is an isomorphism.
== Proof ==
Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be closed algebraic structures.
Let $\phi$ be a monomorphism from $\struct {S_1, \circ_1}$ to $\struct {S_2, \circ_2}$.
Let $T = \Img \phi$ be the image of $\phi$.
By Morphism Property Preserves Closure, $\struct {T, \circ_2}$ is closed.
As $\phi$ is a monomorphism, it is by definition an injective homomorphism.
From Restriction of Injection is Injection, $\phi: S_1 \to \Img \phi$ is an injection.
From Restriction of Mapping to Image is Surjection, $\phi: S_1 \to \Img \phi$ is a surjection.
Thus $\phi \to \Img \phi$ is by definition a bijection.
Thus $\phi: S_1 \to \Img \phi$ is a bijective homomorphism.
Hence, by definition, $\phi: S_1 \to \Img \phi$ is an isomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms
* : $\S 47.6$ Homomorphisms and their elementary properties
Category:Monomorphisms (Abstract Algebra)
Monomorphism from Rational Numbers to Totally Ordered Field

== Theorem ==
Let $\struct {F, +, \circ, \le}$ be a totally ordered field.
There is one and only one (ring) monomorphism from the totally ordered field $\Q$ onto $F$.
Its image is the prime subfield of $F$.
== Proof ==
Follows from:
:Characteristic of Ordered Integral Domain is Zero
:Order Embedding between Quotient Fields is Unique.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.11$
Category:Totally Ordered Fields
Category:Ring Monomorphisms
Monomorphism from Total Ordering

== Theorem ==
Let the following conditions hold:
:$(1): \quad$ Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.
:$(2): \quad$ Let $\phi: S \to T$ be a mapping.
:$(3): \quad$ Let $\preceq$ be a total ordering on $S$.
Then $\phi \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ is a (structure) monomorphism :
:$(1): \quad \phi$ is strictly increasing from $\struct {S, \preceq}$ into $\struct {T, \preccurlyeq}$
:$(2): \quad \phi$ is a homomorphism from $\struct {S, \circ}$ into $\struct {T, *}$.
== Proof ==
This follows:
:$(1): \quad$ As a direct consequence of Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing
:$(2): \quad$ From the definition of monomorphism as a homomorphism which is an injection.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Theorem $15.4$
Category:Total Orderings
Category:Monomorphisms (Abstract Algebra)
Monomorphism that is Split Epimorphism is Split Monomorphism

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be a morphism in $\mathbf C$ such that $f$ is a monomorphism and a split epimorphism.
Then $f: C \to D$ is a split monomorphism.
== Proof ==
Let $g: D \to C$ be the right inverse of $f$:
:$f \circ g = \operatorname{id}_D$
which is guaranteed to exist by definition of split epimorphism.
Therefore:
:$f \circ g \circ f = \operatorname{id}_D \circ f = f \circ \operatorname{id}_C$
by the property of the identity morphism.
Since $f$ is left cancellable, by the definition of monomorphism, we have:
:$g \circ f = \operatorname{id}_C$
Hence $f$ is a split monomorphism with left inverse $g$.
Category:Category Theory
Monotone Additive Function is Linear

== Theorem ==
Let $f: \R \to \R$ be a monotone real function which is additive, that is:
:$\forall x, y \in \R: \map f {x + y} = \map f x + \map f y$
Then:
:$\exists a \in \R: \forall x \in \R: \map f x = a x$
That is, $f$ is a linear function.
== Proof 1 ==
== Proof 2 ==
Category:Monotone Additive Function is Linear
Category:Monotone Real Functions
Category:Additive Functions
Category:Linear Real Functions
Monotone Additive Function is Linear/Proof 1

== Theorem ==
== Proof ==
Denote $a = \map f 1$.
As Additive Function is Linear for Rational Factors:
:$ \forall r \in \Q: \map f r = r \, \map f 1 = a r$
Let $x \in \R \setminus \Q$.
Let $\sequence {r_n}$ be an increasing sequence, with $r_n \in \Q$ for each $n \in \N$, such that $\ds \lim_{n \mathop \to \infty} r_n = x$.
Likewise, let $\sequence {s_n}$ be decreasing, with $s_n \in \Q$ for each $n \in \N$, such that $\ds \lim_{n \mathop \to \infty} s_n = x$.
From the Peak Point Lemma, it is always possible to construct sequences like these, for $\Q$ is dense in $\R$.
Now, by passing to $g = -f$ if necessary, we can assume that $f$ is increasing.
Then we have:
:$\map f {r_n} \le \map f x \le \map f {s_n}$
for all $n \in \N$.
As we have $\map f {r_n} = a r_n$ and $\map f {s_n} = a s_n$, it follows that:
:$a r_n \le \map f x \le a s_n$
Since both $a r_n$ and $a s_n$ converge to $a x$, we have $\map f x = a x$ by the Squeeze Theorem for Real Sequences.
Category:Monotone Additive Function is Linear
Monotone Additive Function is Linear/Proof 2

== Theorem ==
== Proof ==
We use a Proof by Contraposition.
To that end, suppose $f$ is not linear.
We know that Graph of Nonlinear Additive Function is Dense in the Plane.
Therefore $f$ is not bounded on any nonempty open interval.
But then $f$ is certainly not monotone.
Hence, by Rule of Transposition, if $f$ is monotone, then it is linear.
Category:Monotone Additive Function is Linear
Category:Proofs by Contraposition
Monotone Convergence Theorem (Real Analysis)

== Theorem ==
Let $\sequence {x_n}$ be a bounded monotone sequence sequence in $\R$.
Then $\sequence {x_n}$ is convergent.
=== Increasing Sequence ===
=== Decreasing Sequence ===
=== Graphical Illustration ===
== Examples ==
== Sources ==
* : $3.14$
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Theorem $1.2.6$
Category:Limits of Sequences
Category:Named Theorems
Category:Real Analysis
Category:Monotone Convergence Theorem
Category:Monotone Convergence Theorem (Real Analysis)
Monotone Convergence Theorem (Real Analysis)/Decreasing Sequence

== Theorem ==
Let $\sequence {x_n}$ be a decreasing real sequence which is bounded below.
Then $\sequence {x_n}$ converges to its infimum.
== Proof ==
Let $\sequence {x_n}$ be decreasing and bounded below.
Then $\sequence {-x_n}$ is increasing and bounded above.
Thus the Monotone Convergence Theorem for Increasing Sequence applies and the proof follows.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $4$
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Theorem $1.2.6$
* : $\S 4$: Convergent Sequences: $\S 4.17 \ \text{(ii)}$
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Monotone Convergence Theorem (Real Analysis)
Monotone Convergence Theorem (Real Analysis)/Examples/Power of Real Number between Zero and One

== Example of Use of Monotone Convergence Theorem (Real Analysis) ==
Let $x \in \R$ such that $0 < x < 1$.
The sequence $\sequence {a_n}_{n \mathop \ge 1}$ defined as:
:$a_n = x^n$
is convergent to the limit $0$.
== Proof ==
From Power of Real Number between Zero and One is Bounded, $\sequence {a_n}$ is bounded below with supremum $0$.
As $x < 1$, it follows from Real Number Ordering is Compatible with Multiplication that:
:$x^{k + 1} < x^k$
The result follows from the Monotone Convergence Theorem (Real Analysis).
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.18$: Examples: $\text{(ii)}$
Category:Examples of Monotone Convergence Theorem (Real Analysis)
Monotone Convergence Theorem (Real Analysis)/Examples/n - 1 over n

== Example of Use of Monotone Convergence Theorem (Real Analysis) ==
The sequence $\sequence {a_n}_{n \mathop \ge 1}$ defined as:
:$a_n = \dfrac {n - 1} n$
is convergent to the limit $1$.
== Proof ==
From Set of Numbers of form n - 1 over n is Bounded Above, $\sequence {a_n}$ is bounded above with supremum $1$.
Then we have that:
When $n = 1$ we have:
So:
:$\forall n \in \N_{>0}: a_{n + 1} - a_n > 0$
Thus $\sequence {a_n}$ is strictly increasing.
The result follows from the Monotone Convergence Theorem (Real Analysis).
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.18$: Examples: $\text{(i)}$
Category:Examples of Monotone Convergence Theorem (Real Analysis)
Monotone Convergence Theorem (Real Analysis)/Graphical Illustration

== Graphical Illustration of Monotone Convergence Theorem (Real Analysis) ==
The following diagram illustrates the Monotone Convergence Theorem:
:
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.17$
* : Chapter $1$: Normed and Banach spaces
Category:Monotone Convergence Theorem (Real Analysis)
Monotone Convergence Theorem (Real Analysis)/Increasing Sequence

== Theorem ==
Let $\sequence {x_n}$ be an increasing real sequence which is bounded above.
Then $\sequence {x_n}$ converges to its supremum.
== Proof ==
Suppose $\sequence {x_n}$ is increasing and bounded above.
By the Continuum Property, it has a supremum, $B$.
We need to show that $x_n \to B$ as $n \to \infty$.
Let $\epsilon \in \R_{>0}$.
By the definition of supremum, $B - \epsilon$ is not an upper bound.
Thus:
:$\exists N \in \N: x_N > B - \epsilon$
But $\sequence {x_n}$ is increasing.
Hence:
:$\forall n > N: x_n \ge x_N > B - \epsilon$
But $B$ is still an upper bound for $\sequence {x_n}$.
Then:
Hence the result.
== Sources ==
* : $3.14$
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $4$
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Theorem $1.2.6$
* : $\S 4$: Convergent Sequences: $\S 4.17 \ \text{(i)}$
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Monotone Convergence Theorem (Real Analysis)
Monotone Function is of Bounded Variation

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a monotone function.
Then $f$ is of bounded variation.
== Proof ==
We use the notation from the definition of bounded variation.
Let $P = \set {x_0, x_1, \ldots, x_n}$ be a finite subdivision of $\closedint a b$.
As $f$ is monotone, it is either increasing or decreasing.
First consider the case of $f$ increasing, then: 
:$\map f {x_i} \ge \map f {x_{i - 1} }$
for all $i \in \N$ with $i \le n$.
By the definition of the absolute value, we have: 
:$\size {\map f {x_i} - \map f {x_{i - 1} } } = \map f {x_i} - \map f {x_{i - 1} }$
Therefore:
for all finite subdivisions $P$.
Note that $\map f b - \map f a$ is independent of the subdivision $P$.
Therefore if $f$ is increasing, it is of bounded variation.
Now consider the case of $f$ decreasing.
We instead have:
:$\map f {x_i} \le \map f {x_{i - 1} }$
for all $i \in \N$ with $i \le n$.
Hence, by the definition of the absolute value, we instead have:
:$\size {\map f {x_i} - \map f {x_{i - 1} } } = -\paren {\map f {x_i} - \map f {x_{i - 1} } }$
Therefore:
for all finite subdivisions $P$.
Note again that $\map f a - \map f b$ is independent of the subdivision $P$. 
Therefore if $f$ is decreasing, it is of bounded variation.
Hence, if $f$ is either increasing or decreasing, it is of bounded variation.
Hence if $f$ is monotone, it is of bounded variation.
== Sources ==
* : $\S 6.3$: Functions of Bounded Variation: Theorem $6.5$
Category:Bounded Variation
Monotone Real Function is Darboux Integrable

== Theorem ==
Let $\closedint a b$ be a closed real interval, where $a < b$.
Let $f: \closedint a b \to \R$ be a monotone real function.
Then $f$ is Darboux integrable over $\closedint a b$.
== Proof ==
We consider the case where $f$ is increasing; the case where $f$ is decreasing is handled similarly.
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
By the Axiom of Archimedes, there exists a natural number $n$ such that:
:$n > \dfrac {\paren {b - a} \paren {\map f b - \map f a} } \epsilon$
For $k \in \set {0, 1, 2, \ldots, n}$, define:
:$x_k = a + k \dfrac {b - a} n$
Then:
:$a = x_0 < x_1 < \cdots < x_n = b$
and hence $P = \set {x_0, x_1, \ldots, x_n}$ is a subdivision of $\closedint a b$.
The lower Darboux sum and upper Darboux sum of $f$ belonging to the subdivision $P$ are, respectively:
Therefore:
Since $\epsilon \in \R_{>0}$ is arbitrary, the result follows.
== Also see ==
* Continuous Real Function is Darboux Integrable
== Sources ==
* : $6.9$
Category:Integral Calculus
Category:Monotone Real Functions
Monotone Real Function is Measurable

== Theorem ==
Let $I \subseteq \R$ be an open interval.
Let $\map \BB I$ and $\map \BB \R$ be the Borel $\sigma$-algebras of $I$ and $\R$ respectively.
Let $F : \R \to \R$ be a monotone function.
Then $F$ is $\map \BB I/\map \BB \R$-measurable.
== Proof ==
We first assume that $F$ is increasing.
From Borel Sigma-Algebra of Subset is Trace Sigma-Algebra, $\map \BB I$ is the trace $\sigma$-algebra of $I$ in $\map \BB \R$. 
From Characterization of Measurable Functions, it is enough to show that:
:$J_t = \set {x \in I : \map F x \le t} \in \map \BB I$ for each $t \in \R$. 
From Measure of Interval is Length and the definition of the trace $\sigma$-algebra, it is enough to show that $J_t$ is an interval for each $t \in \R$.
Fix $t \in \R$. 
Suppose that $x \in J_t$ and $y \in \R$ with $y \le z$. 
Since $F$ is increasing, we have:
:$\map F z \le \map F y \le t$. 
In particular, $\map F z \le t$ and $z \in J_t$.
Hence $J_t$ is a real interval that is unbounded below.
Hence $J_t \in \map \BB I$ from Measure of Interval is Length and the definition of the trace $\sigma$-algebra.
Hence by Characterization of Measurable Functions, $F$ is $\map \BB I/\map \BB \R$-measurable.
Now suppose that $F$ is decreasing.
Then $-F$ is increasing.
Hence from the previous case, $-F$ is $\map \BB I/\map \BB \R$-measurable.
From Pointwise Scalar Multiple of Measurable Function is Measurable, $F$ is $\map \BB I/\map \BB \R$-measurable.
Category:Measurable Functions
Category:Monotone Real Functions
Monotone Real Function with Everywhere Dense Image is Continuous

== Theorem ==
Let $I$ and $J$ be real intervals.
Let $f: I \to J$ be a monotone real function.
Let $f \sqbrk I$ be everywhere dense in $J$, where $f \sqbrk I$ denotes the image of $I$ under $f$.
Then $f$ is continuous on $I$.
== Proof ==
, let $f$ be increasing.
 $f$ is discontinuous at $c \in I$.
Let:
:$L^-$ denote $\ds \lim_{x \mathop \to c^-} \map f x$
:$L$ denote $\map f c$
:$L^+$ denote $\ds \lim_{x \mathop \to c^+} \map f x$
From Discontinuity of Monotonic Function is Jump Discontinuity, $L^-$ and $L^+$ are (finite) real numbers.
From Limit of Monotone Real Function: Corollary:
:$L^- \le L \le L^+$
Let $S = \openint {L^-} {L^+}$.
If $L^- = L = L^+$, then $c$ is not a discontinuity, a contradiction.
Thus $S$ is non-degenerate.
From Non-Degenerate Real Interval is Infinite, $S$ is infinite.
From Relative Difference between Infinite Set and Finite Set is Infinite, $S \setminus \set L$ is non-empty.
From Finite Subset of Metric Space is Closed, $\set L$ is closed.
From Open Set minus Closed Set is Open, $S \setminus \set L$ is open.
=== Lemma ===
Thus:
Thus there exists a non-empty open subset of $f \sqbrk I$ that is disjoint from $Y$.
From Open Set Characterization of Denseness, $f \sqbrk I$ is not everywhere dense in $J$.
This contradicts the hypothesis that $f \sqbrk I$ is everywhere dense in $J$.
Therefore $I$ contains no discontinuities, by Proof by Contradiction.
The same argument, mutatis mutandis, proves the case where $f$ is decreasing
Hence the result, by Proof by Cases.
Category:Continuous Real Functions
Category:Monotone Real Functions
Category:Real Analysis