nyuuzyou/wikis huggingface dataset:
Characterization of T0 Space by Distinct Closures of Singletons

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then
:$T$ is a $T_0$ space 
::$\forall x, y \in S: x \ne y \implies \set x^- \ne \set y^-$
where $\set y^-$ denotes the closure of $\set y$.
== Proof ==
=== Sufficient Condition ===
Let $T$ be a $T_0$ space.
Let $x, y \in S$ such that
:$x \ne y$
By definition of $T_0$ space:
:$\paren {\exists U \in \tau: x \in U \land y \notin U} \lor \paren {\exists U \in \tau: x \notin U \land y \in U}$
, suppose
:$\exists U \in \tau: x \in U \land y \notin U$
By definition of singleton:
:$x \in \set x$
By Set is Subset of its Topological Closure:
:$\set x \subseteq \set x^-$
Then by definition of subset:
:$x \in \set x^-$
Hence by definition of intersection:
:$\set x^- \cap U \ne \O$
By definition of singleton:
:$\forall z: z \in \set y \implies z = y$
Then by definition of intersection:
:$\set y \cap U = \O$
Hence by Open Set Disjoint from Set is Disjoint from Closure
:$\set y^- \cap U = \O$
Thus:
:$\set x^- \ne \set y^-$
=== Necessary Condition ===
Assume that:
:$(1): \quad \forall x, y \in S: x \ne y \implies \set x^- \ne \set y^-$
By Characterization of $T_0$ Space by Closed Sets it suffices to prove that
:$\leftparen {\exists F \subseteq S: F}$ is closed $\rightparen {\land x \in F \land y \notin F}$
or
:$\leftparen {\exists F \subseteq S: F}$ is closed $\rightparen {\land x \notin F \land y \in F}$
Then assume:
:$(2): \quad \forall F \subseteq S: F$ is closed $\land x \in F \implies y \in F$
Define $F := \set y^-$
By Topological Closure is Closed:
:$F$ is closed.
We will prove now that
:$x \notin F$
:$x \in F$
Then:
:$\set x \subseteq F$
By Topological Closure of Subset is Subset of Topological Closure:
:$\set x^- \subseteq F^-$
By Set is Closed iff Equals Topological Closure:
:$(3): \quad \set x^- \subseteq F$
By definition of singleton:
:$x \in \set x$
By Set is Subset of its Topological Closure:
:$\set x \subseteq \set x^-$
Then by definition of subset:
:$x \in \set x^-$
Because $\set x^-$ is closed therefore by $(2)$:
:$y \in \set x^-$
Then:
:$\set y \subseteq \set x^-$
By Topological Closure of Subset is Subset of Topological Closure:
:$F \subseteq \paren {set x^-}^-$
Then by Closure of Topological Closure equals Closure:
:$F \subseteq \set x^-$
Hence by $(3)$ and definition of set equality:
:$F = \set x^-$
This contradicts the assumption $(1)$.
Thus $x \notin F$.
By definition of singleton:
:$y \in \set y$
By Set is Subset of its Topological Closure:
:$\set y \subseteq \set y^- = F$
Thus by definition of subset:
:$y \in F$
Hence
:$\exists F \subseteq S: F$ is closed $\land x \notin F \land y \in F$
== Sources ==
* 
* 
Category:T0 Spaces
Characterization of T1 Space using Basis

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be a basis for $T$.
Then:
:$T$ is a $T_1$ Space
:$\forall x, y \in S : x \ne y$, both:
::$\exists B_x \in \BB : x \in B_x, y \notin B_x$
:and:
::$\exists B_y \in \BB : y \in B_y, x \notin B_y$
== Proof ==
=== Necessary Condition ===
From Basis induces Local Basis:
:$\forall x \in S : \BB_x = \set{B \in \BB : x \in B}$ is a local basis of $x$
By definition of local basis:
:$\forall x \in S : \BB_x$ is a neighborhood basis of open sets
From Characterization of T1 Space using Neighborhood Basis:
:$\forall x, y \in S : x \ne y$, both:
::$\exists B_x \in \BB_x : y \notin B_x$
:and:
::$\exists B_y \in \BB_y : x \notin B_y$
Hence:
:$\forall x, y \in S : x \ne y$, both:
::$\exists B_x \in \BB : x \in B_x, y \notin B_x$
:and:
::$\exists B_y \in \BB : y \in B_y, x \notin B_y$
=== Sufficient Condition ===
Let $T$ satisfy:
:$\forall x, y \in S : x \ne y$, both:
::$\exists B_x \in \BB : x \in B_x, y \notin B_x$
:and:
::$\exists B_y \in \BB : y \in B_y, x \notin B_y$
By definition of basis:
:$\BB \subseteq \tau$.
It follows that:
:$\forall x, y \in S : x \ne y$, both:
::$\exists B_x \in \tau : x \in B_x, y \notin B_x$
:and:
::$\exists B_y \in \tau : y \in B_y, x \notin B_y$
Hence $T$ is a $T_1$ Space by definition.
Category:T1 Spaces
Category:Topological Bases
Characterization of T1 Space using Neighborhood Basis

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
For each $x \in S$, let $\NN_x$ be a neighborhood basis at $x$.
Then:
:$T$ is a $T_1$ Space
:$\forall x, y \in S : x \ne y$, both: 
::$\exists N \in \NN_x : y \notin N$
:and:
::$\exists M \in \NN_y : x \notin M$
== Proof ==
=== Necessary Condition ===
Let $T$ be a $T_1$ Space.
Let $x, y \in S$ such that $x \ne y$.
By definition of $T_1$ Space:
:$\exists U \in \tau : x \in U, y \notin U$
From Set is Open iff Neighborhood of all its Points:
:$U$ is a neighborhood of $x$
By definition of neighborhood basis:
:$\exists N \in \NN_x : N \subseteq U$
It follows that:
:$y \notin N$
Similarly:
$\exists M \in \NN_y : x \notin M$
The result follows.
=== Sufficient Condition ===
Let $T$ satisfy:
:$\forall x, y \in S : x \ne y$, both: 
::$\exists N \in \NN_x : y \notin N$
:and:
::$\exists M \in \NN_y : x \notin M$
Let $x, y \in S$ such that $x \ne y$.
:
:$\exists N \in \NN_x : y \notin N$
By definition of neighborhood:
:$\exists U \in \tau : x \in U : U \subseteq N$
It follows that:
:$x \in U, y \notin U$
Similarly:
:$\exists V \in \tau : y \in V, x \notin V$
Since $x, y$ were arbitrary:
:$\forall x, y \in S$ such that $x \ne y$, both:
:: $\exists U \in \tau: x \in U, y \notin U$
:and:
:: $\exists V \in \tau: y \in V, x \notin V$
It follows that $T$ is a $T_1$ Space by definition.
Category:T1 Spaces
Category:Local Bases
Characterization of T3 Space

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space. 
:$(1) \quad T$ is a $T_3$ space
:$(2) \quad \forall F \subseteq S : S \setminus F \in \tau, y \in S \setminus F : \exists V \in \tau : y \in V, V^- \cap F = \O$
:$(3) \quad \forall U \in \tau, y \in U : \exists V \in \tau : y \in V, V^- \subseteq U$
where $V^-$ denotes the closure of $V$ in $T$
== Proof ==
=== Statement $(1)$ implies Statement $(2)$ ===
Let $T$ be a $T_3$ space.
Let $F \subseteq S : S \setminus F \in \tau$.
Let $y \in S \setminus F$.
By definition of $T_3$ space:
:$\exists V, W \in \tau : y \in V, F \subseteq W : V \cap W = \O$
From Subset of Set Difference iff Disjoint Set:
:$V \subseteq S \setminus W$
By definition of closed set:
:$S \setminus W$ is closed in $T$
From Closure of Subset of Closed Set of Topological Space is Subset:
:$V^- \subseteq S \setminus W$
From Set Difference with Subset is Superset of Set Difference:
:$S \setminus W \subseteq S \setminus F$
From Subset Relation is Transitive:
:$V^- \subseteq S \setminus F$
From Subset of Set Difference iff Disjoint Set:
:$V^- \cap F = \O$
The result follows.
=== Statement $(2)$ implies Statement $(3)$ ===
Let $T$ satisfy:
:$\forall F \subseteq S : S \setminus F \in \tau, y \in S \setminus F : \exists V \in \tau : y \in V, V^- \cap F = \O$
Let $U \in \tau$.
Let $y \in U$.
Let $F = S \setminus U$.
From Set Difference with Set Difference:
:$S \setminus F = U$
Hence:
:$S \setminus F \in \tau$
and
:$y \in S \setminus F$
From $(2)$:
:$\exists V \in \tau : y \in V, V^- \cap F = \O$
Hence:
:$V^- \cap S \setminus U = \O$
From Subset of Set Difference iff Disjoint Set:
:$V^- \subseteq U$
Since $U$ and $y$ were arbitrary, we have shown:
:$\forall U \in \tau, y \in U : \exists V \in \tau : y \in V, V^- \subseteq U$
=== Statement $(3)$ implies Statement $(1)$ ===
Let $T$ satisfy:
:$\forall U \in S, y \in U : \exists V \in \tau : y \in V, V^- \subseteq U$
Let $U \in \tau$.
Let $y \in U$.
From $(3)$:
:$\exists V \in \tau : y \in V, V^- \subseteq U$
Let $N = V^-$.
By definition of closure:
:$V \subseteq N$
From Topological Closure is Closed:
:$S \ N \in \tau$
Hence:
:$\exists N \subseteq S : S \setminus N \in \tau : \exists V \in \tau : y \in V \subseteq N \subseteq U$
Since $U$ and $y$ were arbitrary, we have shown:
:$\forall U \in \tau, y \in U : \exists N \subseteq S : S \setminus N \in \tau : \exists V \in \tau : y \in V \subseteq N \subseteq U$
By definition, $T$ is a $T_3$ space
Category:T3 Spaces
Characterization of Unit Tangent Bundle

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold with or without boundary.
Let $TM$ be the tangent bundle of $M$.
Let $UTM$ be the unit tangent bundle of $M$.
Let $\pi : UTM \to M$ be the canonical projection.
Then $UTM$ is a smooth, properly embedded submanifold of the codimension-$1$ with boundary $\map \partial {UTM} = \map {\pi^{-1}} {\partial M}$ in $TM$.
== Proof ==
== Sources ==
* : $\S 2$. Riemannian Metrics. Definitions
Category:Tangent Bundles
Category:Riemannian Geometry
Characterization of Unitary Operators

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $A$ be a bounded linear operator on $\HH$.
Then the following are equivalent:
:$(1): \quad A$ is a unitary operator
:$(2): \quad A^* A = A A^* = I$, where $A^*$ denotes the adjoint of $A$, and $I$ denotes the identity operator
:$(3): \quad A$ is a normal isometry
== Proof ==
=== $(3)$ implies $(2)$ ===
Suppose that $A$ is a normal isometry.
Then:
:$\innerprod {A x} {A x} = \innerprod x x$ for each $x \in \HH$.
From Adjoint is Involutive, we have:
:$\innerprod {A x} {\paren {A^\ast}^\ast x} = \innerprod x x$ 
So, from the definition of the adjoint, we have:
:$\innerprod {A^\ast A x} x = \innerprod {I x} x$ for all $x \in \HH$.
Hence from Operator with Zero Numerical Range is Zero Operator: Corollary, we have:
:$A^\ast A = I$ 
Since $A$ is normal, we also have:
:$A A^\ast = I$
=== $(1)$ implies $(2)$ ===
Let $A$ be a unitary operator.
From the proof that $(3)$ implies $(2)$, we have: 
:$A^\ast A = I$
It remains to show that:
:$A A^\ast = I$ 
Let $y \in \HH$, we aim to show:
:$A A^\ast y = y$
Since $A$ is a surjection, there exists $x \in \HH$ such that:
:$A x = y$
Then we have:
:$x = A^\ast A x = A^\ast y$ 
Then we have:
:$A A^\ast y = A x = y$ 
=== $(2)$ implies $(1)$ ===
Suppose that:
:$A^\ast A = A A^\ast = I$. 
By hypothesis, $A$ is linear. 
Since $A A^\ast = I$, we have $\map A {A^\ast x} = x$ for each $x \in \HH$. 
Hence $A$ is also surjective. 
Finally, since $A^\ast A = I$, we have:
:$\innerprod {A^\ast A x} y = \innerprod x y$ for each $x, y \in \HH$. 
From the definition of the adjoint, we have:
:$\innerprod {A x} {A^{\ast \ast} y} = \innerprod x y$ for each $x, y \in \HH$. 
From Adjoint is Involutive, we have:
:$\innerprod {A x} {A y} = \innerprod x y$ for each $x, y \in \HH$.
Hence $A$ is an isomorphism.
=== $(2)$ implies $(3)$ ===
Suppose that:
:$A^\ast A = A A^\ast = I$
In particular $A^\ast A = A A^\ast$, so $A$ is normal. 
From the fact that $(2)$ implies $(1)$, $A$ is a unitary operator.
Hence $A$ is a surjective isometry.
Hence $A$ is a normal isometry.
== Sources ==
* $\text {II}.2.18$
Category:Linear Transformations on Hilbert Spaces
Characterization of Zero Entropy of Open Cover

== Theorem ==
Let $X$ be a compact topological space.
Let $\alpha$ be an open cover of $X$.
Let $\map H \alpha$ be the entropy of $\alpha$.
Then $\map H \alpha = 0$ $X \in \alpha$.
== Proof ==
== Sources ==
* Chapter $7$: Topological Entropy
Category:Ergodic Theory
Category:Topology
Characterization of von Neumann-Boundedness in Hausdorff Locally Convex Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$.
Let $U \subseteq X$.
Then $U$ is von Neumann-bounded :
:for each $p \in \PP$ there exists $C_p > 0$ such that: 
::$\map p x < C_p$
:for each $x \in U$. 
== Proof ==
For each $p \in \PP$, let: 
:$B_p = \set {y \in X : \map p y < 1}$
Note that by the definition of the standard topology, $B_p$ is an open neighborhood of $\mathbf 0_X$. 
Let $r > 0$. 
From , for $y \in X$ we have: 
:$\map p y < 1$ 
:
:$\map p {r y} < r$ 
So, we have: 
:$r B_p = \set {y \in X : \map p y < r}$
=== Necessary Condition ===
Suppose that $U$ is von Neumann-bounded.
Then there exists $s > 0$ such that: 
:$U \subseteq t B_p = \set {y \in X : \map p y < t}$ 
for $t > s$
In particular: 
:$U \subseteq \set {y \in X : \map p y < s + 1}$ 
That is: 
:$\map p x < s + 1$
for all $x \in U$.
=== Sufficient Condition ===
Suppose that for each $p \in \PP$ there exists $C_p > 0$ such that: 
:$\map p x < C_p$
for each $x \in U$.
Let $V$ be an open neighborhood of $\mathbf 0_X$.
From Open Sets in Standard Topology of Locally Convex Space, there exists $p_1, \ldots, p_n \in \PP$ and $\epsilon > 0$ such that: 
:$\ds \bigcap_{k \mathop = 1}^n \epsilon B_{p_k} = \set {y \in X : \map {p_k} y < \epsilon \text { for each } 1 \le k \le n} \subseteq V$
Let: 
:$\ds r = \frac 1 \epsilon \max_{1 \le k \le n} C_{p_k}$
Then: 
:$\ds r \bigcap_{k \mathop = 1}^n \epsilon B_{p_k} = \bigcap_{k \mathop = 1}^n \paren {\max_{1 \le k \le n} C_{p_k} } B_{p_k} \subseteq r V$
and generally if $s > r$: 
:$\ds \bigcap_{k \mathop = 1}^n \paren {\max_{1 \le k \le n} C_{p_k} } B_{p_k} \subseteq s \bigcap_{k \mathop = 1}^n \epsilon B_{p_k} \subseteq s V$ 
Note that for $x \in U$, $1 \le k \le n$ and $s > r$ we then have: 
:$\ds \map {p_k} x < C_{p_k} \le \max_{1 \le k \le n} C_{p_k} < s \epsilon$
for $x \in U$, and hence: 
:$U \subseteq s V$
for $s > r$.
So $U$ is von Neumann-bounded.
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Category:Locally Convex Spaces
Characterization of von Neumann-Boundedness in Normed Vector Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $U \subseteq X$.
Then $U$ is von Neumann-bounded there exists $M > 0$ such that:
:$\norm x < M$
for all $x \in U$. 
== Proof ==
From Normed Vector Space is Hausdorff Locally Convex Space, $\struct {X, \norm {\, \cdot \,} }$ can be viewed as the Hausdorff locally convex space $\struct {X, \norm {\, \cdot \,} }$. 
The result is then immediate from Characterization of von Neumann-Boundedness in Hausdorff Locally Convex Space.
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Category:Normed Vector Spaces
Characterization of von Neumann-Boundedness in terms of Local Basis

== Theorem ==
Let $\struct {X, \tau}$ be a topological vector space.
Let $\mathcal B$ be a local basis for $\mathbf 0_X$ in $\struct {X, \tau}$.
Let $E \subseteq X$. 
Then $E$ is von Neumann-bounded :
:for each $V \in \mathcal B$ there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$. 
== Proof ==
=== Necessary Condition ===
Suppose that $E$ is von Neumann-bounded.
Then: 
:for each open neighbourhood $V$ of ${\mathbf 0}_X$, there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$
Since $\mathcal B$ consists of open neighbourhoods of ${\mathbf 0}_X$, we in particular have: 
:for each $V \in \mathcal B$ there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$. 
=== Sufficient Condition ===
Suppose that:
:for each $V \in \mathcal B$ there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$. 
Let $U$ be an open neighbourhood of ${\mathbf 0}_X$. 
Then there exists some $V \in \mathcal V$ such that $V \subseteq U$ and:
:there exists $s > 0$ such that $E \subseteq t V \subseteq t U$ for each $t > s$. 
So $E$ is von Neumann-bounded.
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Category:Local Bases
Characterizations of Metric Connections

== Theorem ==
Let $\struct {M, g}$ be a Riemannian or pseudo-Riemannian manifold (with or without boundary).
Let $TM$ be the tangent bundle of $TM$.
Let $\nabla$ be a connection on $TM$.
Let $\map {\mathfrak X} M$ be the space of smooth vector fields on $M$.
Let $\gamma$ be a smooth curve in $M$.
For all $i \in \N_{>0} : i \le \dim M$ let $\tuple {E_i}$ be a smooth local frame for $TM$.
Let $\set {\Gamma^k_{ij} }$ be the connection coefficients $\tuple {E_i}$.
Let $D_t$ be the covariant derivative along $\gamma$.
Then the following properties are equivalent:
=== Property $1$ ===
:$\forall X, Y, Z \in \map {\mathfrak X} M : \nabla_X \innerprod Y Z = \innerprod {\nabla_X Y} Z + \innerprod Y {\nabla_X Z}$
=== Property $2$ ===
:$\nabla g = 0$
=== Property $3$ ===
:$\Gamma^l_{ki} g_{lj} + \Gamma^l_{kj} g_{il} = \map {E_k} {g_{ij} }$
=== Property $4$ ===
:$\dfrac \d {\d t} \innerprod Y Z = \innerprod {D_t Y} Z + \innerprod Y {D_t Z}$
=== Property $5$ ===
:If $V, W \in \map {\mathfrak X} \gamma$, then $\innerprod V W$ is constant along $\gamma$
=== Property $6$ ===
:Parallel transport map along $\gamma$ is a linear isometry
=== Property $7$ ===
:Orthonormal basis at a point of $\gamma$ can be extended to a parallel orthonormal frame along $\gamma$
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Connections on Abstract Riemannian Manifolds
Category:Connections
Characterizing Property of Infimum of Subset of Real Numbers

== Theorem ==
Let $S \subset \R$ be a non-empty subset of the real numbers.
Let $S$ be bounded below.
Let $\alpha \in \R$.
:$(1): \quad \alpha$ is the infimum of $S$
:$(2): \quad \alpha$ is a lower bound for $S$
::and:
:::$\forall \epsilon \in \R_{> 0}$ there exists $x \in S$ with $x < \alpha + \epsilon$
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\alpha$ be the infimum of $S$.
Then by definition, $\alpha$ is a lower bound for $S$.
Let $\epsilon>0$.
Because $\alpha+\epsilon>\alpha$, it is not a lower bound for $S$.
Thus there exists $x\in S$ with $x < \alpha + \epsilon$.
=== $(2)$ implies $(1)$ ===
Let $\alpha$ be a lower bound of $S$ such that $\forall \epsilon > 0$ there exists $x \in S$ with $x < \alpha + \epsilon$.
Let $d \in \R$ be a lower bound of $S$.
We have to prove that $d \le \alpha$.
 $d > \alpha$.
Let $\epsilon = d - \alpha > 0$.
Then there exists $x \in S$ such that $x < \alpha + \left({d - \alpha}\right) = d$.
But this contradicts our assumption that $d$ is a lower bound of $S$.
== Also see ==
* Characterizing Property of Supremum of Subset of Real Numbers
Category:Infima
Category:Real Numbers
Characterizing Property of Supremum of Subset of Real Numbers

== Theorem ==
Let $S \subset \R$ be a non-empty subset of the real numbers.
Let $S$ be bounded above.
Let $\omega \in \R$.
:$(1): \quad \omega$ is the supremum of $S$
:$(2): \quad \omega$ is an upper bound for $S$
::and:
:::$\forall \epsilon \in \R_{> 0}$ there exists $x \in S$ with $x > \omega - \epsilon$
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\omega$ be the supremum of $S$.
Then by definition, $\omega$ is an upper bound for $S$.
Let $\epsilon > 0$.
Because $\omega - \epsilon < \omega$, it is not an upper bound for $S$.
Thus there exists $x\in S$ with $x > \omega - \epsilon$.
=== $(2)$ implies $(1)$ ===
Let $\omega$ be an upper bound of $S$ such that $\forall \epsilon > 0$ there exists $x \in S$ with $x > \omega - \epsilon$.
Let $d \in \R$ be an upper bound of $S$.
We have to prove that $d \ge \omega$.
 $d < \omega$.
Let $\epsilon = \omega - d > 0$.
Then there exists $x \in S$ such that $x > \omega - \left({\omega - d}\right) = d$.
But this contradicts our assumption that $d$ is an upper bound of $S$.
== Also see ==
* Characterizing Property of Infimum of Subset of Real Numbers
Category:Suprema
Category:Real Numbers
Charles Babbage's Conjecture/Historical Note

== Historical Note on Charles Babbage's Conjecture ==
The name Charles Babbage's Conjecture has been coined by as a convenient label to apply to a statement whose full description is unwieldy.
Note that the name Babbage's Conjecture is already used in the literature to refer to a conjecture suggested by .
According to 's $1997$ work , the refutation is the work of , who also determined that no other such counterexample exists for any $n$ less than $150 \, 000$.
However, this still needs to be corroborated by evidence on the internet.
 also suggests that any higher power of $16 \, 843$ is also a counterexample, which would be interesting, as it would suggest that, for example:
:$16 \, 843^4 \divides \dbinom {2 \times 16 \, 843^2 - 1} {16 \, 843^2 - 1} - 1$
It appears prudent, therefore, to ascertain whether is actually correct in his citation of $16 \, 843$ being the counterexample that actually discovered.
Hence research is invited into the literature, to find the original work and to confirm what is claimed here.
== Sources ==
* : $16,843$
* : $16,843$
Category:Historical Notes
Chebyshev's Sum Inequality/Also known as

== Chebyshev's Sum Inequality: Also known as==
Chebyshev's Sum Inequality is also known as Chebyshev's Inequality.
However, some sources use this name to mean the Bienaymé-Chebyshev Inequality, which is a completely different result.
Hence, to avoid confusion, the name Chebyshev's Inequality is not used on .
== Sources ==
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Chebyshev's Inequality: $3.2.7$
* : $\S 36$: Inequalities: Chebyshev's Inequality
* 
Category:Chebyshev's Sum Inequality
Chebyshev's Sum Inequality/Continuous

== Theorem ==
Let $u, v: \closedint 0 1 \to \R$ be integrable functions.
Let $u$ and $v$ both be either increasing or decreasing.
Then:
:$\ds \paren {\int_0^1 u \rd x} \cdot \paren {\int_0^1 v \rd x} \le \int_0^1 u v\rd x$
== Proof ==
== Sources ==
* 
Category:Chebyshev's Sum Inequality
Chebyshev's Sum Inequality/Discrete

== Theorem ==
Let $a_1, a_2, \ldots, a_n$ be real numbers such that:
:$a_1 \ge a_2 \ge \cdots \ge a_n$
Let $b_1, b_2, \ldots, b_n$ be real numbers such that:
:$b_1 \ge b_2 \ge \cdots \ge b_n$
Then:
:$\ds \dfrac 1 n \sum_{k \mathop = 1}^n a_k b_k \ge \paren {\dfrac 1 n \sum_{k \mathop = 1}^n a_k} \paren {\dfrac 1 n \sum_{k \mathop = 1}^n b_k}$
=== Condition for Equality ===
== Proof ==
== Also presented as ==
== Also known as ==
== Sources ==
* : $\S 36$: Inequalities: Chebyshev's Inequality: $36.10$
* 
Category:Chebyshev's Sum Inequality
Chebyshev's Sum Inequality/Discrete/Also presented as

== Chebyshev's Sum Inequality (Discrete): Also presented as ==
Some sources present Chebyshev's sum inequality as:
:$\ds n \sum_{k \mathop = 1}^n a_k b_k \ge \paren {\sum_{k \mathop = 1}^n a_k} \paren {\sum_{k \mathop = 1}^n b_k}$
where:
:$a_1, a_2, \ldots, a_n$ are real numbers such that:
::$a_1 \ge a_2 \ge \cdots \ge a_n$
:$b_1, b_2, \ldots, b_n$ are real numbers such that:
::$b_1 \ge b_2 \ge \cdots \ge b_n$
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Chebyshev's Inequality: $3.2.7$
* : $\S 36$: Inequalities: Chebyshev's Inequality: $36.11$
Category:Chebyshev's Sum Inequality
Chebyshev's Sum Inequality/Discrete/Equality

== Theorem ==
Let $a_1, a_2, \ldots, a_n$ be real numbers such that:
:$a_1 \ge a_2 \ge \cdots \ge a_n$
Let $b_1, b_2, \ldots, b_n$ be real numbers such that:
:$b_1 \ge b_2 \ge \cdots \ge b_n$
Then equality in Chebyshev's Sum Inequality (Discrete), that is:
:$\ds \dfrac 1 n \sum_{k \mathop = 1}^n a_k b_k = \paren {\dfrac 1 n \sum_{k \mathop = 1}^n a_k} \paren {\dfrac 1 n \sum_{k \mathop = 1}^n b_k}$
holds :
:$\forall j, k \in \set {1, 2, \ldots, n}: a_j = a_k, b_j = b_k$
== Proof ==
== Also known as ==
== Sources ==
* 
Category:Chebyshev's Sum Inequality
Chebyshev's Sum Inequality/Discrete/Proof

== Theorem ==
== Proof ==
We have that the sequences $\sequence {a_k}$ and $\sequence {b_k}$ are both decreasing.
For $j, k \in \set {1, 2, \ldots, n}$, consider:
:$\paren {a_j - a_k} \paren {b_j - b_k}$
Therefore $a_j - a_k$ and $b_j - b_k$ have the same sign for all $j, k \in \set {1, 2, \ldots, n}$.
So:
:$\forall j, k \in \set {1, 2, \ldots, n}: \paren {a_j - a_k} \paren {b_j - b_k} \ge 0$
Hence:
The result follows.
== Also presented as ==
== Also known as ==
Category:Chebyshev's Sum Inequality