nyuuzyou/wikis huggingface dataset:
Countably Compact Metric Space is Sequentially Compact

== Theorem ==
Let $M$ be a countably compact metric space.
Then $M$ is sequentially compact.
== Proof ==
This follows directly from the results:
:Metric Space is First-Countable
:Countably Compact First-Countable Space is Sequentially Compact
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Countably Compact Spaces
Category:Sequentially Compact Spaces
Countably Compact Space is Countably Paracompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countably compact space.
Then $T$ is countably paracompact.
== Proof ==
From the definition, $T$ is countably compact every countable open cover of $S$ has a finite subcover.
From Subcover is Refinement of Cover, it follows that every countable open cover of $S$ has an open refinement which is locally finite.
This is precisely the definition of countably paracompact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Countably Compact Spaces
Category:Countably Paracompact Spaces
Countably Compact Space is Pseudocompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countably compact space.
Then $T$ is a pseudocompact space.
== Proof ==
Let $T = \struct {S, \tau}$ be a countably compact space.
Then every countable open cover of $S$ has a finite subcover.
By definition:
:$T$ is pseudocompact every continuous real-valued function on $S$ is bounded.
Let $f: S \to \R$ be such that $f$ is continuous.
Consider the sets:
:$S_n = \set {x \in S: \size {\map f x} < n}$
$S_n$ forms a countable cover of $S$ whose finite subcover yields a bound for the absolute value of $f$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Countably Compact Spaces
Category:Pseudocompact Spaces
Countably Compact Space is Weakly Countably Compact

== Theorem ==
Every countably compact space is a weakly countably compact space.
== Proof ==
Let $T = \struct {S, \tau}$ be a countably compact space.
By definition:
:$T$ is weakly countably compact every infinite subset of $S$ has a limit point in $S$.
By definition of countably compact space:
:every infinite subset of $S$ has an $\omega$-accumulation point in $S$.
By definition, an $\omega$-accumulation point is a limit point.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Countably Compact Spaces
Category:Weakly Countably Compact Spaces
Countably Compact Space satisfies Countable Finite Intersection Axiom

== Theorem ==
=== Definition by Finite Subcover ===
=== Definition by Countable Finite Intersection Axiom ===
== Proof ==
Let every countable open cover of $S$ have a finite subcover.
Let $\AA$ be any set of closed sets of $T$ satisfying $\bigcap \AA = \O$.
We define the set:
:$\VV := \set {S \setminus A : A \in \AA}$
which is clearly an open cover of $S$.
From De Morgan's Laws: Difference with Union:
:$\ds S \setminus \bigcup \VV = \bigcap \set {S \setminus V : V \in \VV} = \bigcap \set{A : A \in \AA} = \O$
and therefore $S = \bigcup \VV$.
By definition, there exists a finite subcover $\tilde \VV \subseteq \VV$.
We define:
:$\tilde \AA := \set {S \setminus V : V \in \tilde \VV}$
Then $\tilde \AA \subseteq \AA$ by definition of $\VV$.
Because $\tilde \VV$ covers $S$, it follows directly that:
:$\ds \bigcap \tilde \AA = \bigcap \set {S \setminus V : V \in \tilde \VV} = S \setminus \bigcup \tilde \VV = \O$
Thus, in every countable set $\AA$ of closed sets of $T$ satisfying $\ds \bigcap \AA = \O$ exists a finite subset $\tilde \AA$ such that $\ds \bigcap \tilde \AA = \O$.
That is, $S$ satisfies the Countable Finite Intersection Axiom.
The converse works exactly as the previous, but with the roles of the open cover and $\AA$ reversed.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Countably Compact Spaces
Countably Infinite Set has Enumeration

== Theorem ==
Let $S$ be a countably inifnite set.
Then there exists a countably infinite enumeration $\set{s_1, s_2, s_3, \ldots}$ of $S$.
== Proof ==
By definition of countably inifnite set:
:there exists a bijection $f:S \to \N$
From Inverse of Bijection is Bijection:
:$f^{-1} : \N \to S$ is a bijection
Let $s = f^{-1}$.
It follows that $s : \N \to S$ is a countably infinite enumeration $\set{s_1, s_2, s_3, \ldots}$ by definition.
Countably Infinite Set in Countably Compact Space has Omega-Accumulation Point

== Theorem ==
Let $\struct {X, \tau}$ be a countably compact topological space.
Let $A \subseteq X$ be countably infinite.
Then $A$ has an $\omega$-accumulation point in $X$.
=== Corollary ===
== Proof ==
Proof by Contradiction.
 that $A$ does not have an $\omega$-accumulation point in $X$.
Let $\SS \subseteq \powerset A$ be the set of all finite subsets of $A$.
By Set of Finite Subsets of Countable Set is Countable, we have that $\SS$ is countable.
For all (finite) $F \in \SS$, define:
:$U_F = \paren {F \cup \paren {X \setminus A} }^\circ$
where ${}^\circ$ denotes the interior.
By Image of Countable Set under Mapping is Countable, we have that $\CC = \set {U_F: F \in \SS}$ is countable.
By the definition of an $\omega$-accumulation point, it follows that:
:$\forall x \in X: \exists U \in \tau: x \in U: \exists F \in \SS: U \cap A = F$
By Set Difference Union Intersection, we have:
:$U = \paren {U \cap A} \cup \paren {U \setminus A} \subseteq F \cup \paren {X \setminus A}$
By Set Interior is Largest Open Set, it follows that $\CC$ is a countable open cover for $X$.
Hence, by the definition of a countably compact space, there exists a finite subcover $\CC'$ of $\CC$ for $X$.
By the Principle of Finite Choice, there exists an indexed family $\family {G_V}_{V \mathop \in \CC'}$ of elements of $\SS$ such that:
:$\forall V \in \CC': V = U_{G_V} = \paren {G_V \cup \paren {X \setminus A} }^\circ$
Define:
:$\ds B = \bigcup_{V \mathop \in \CC'} G_V \subseteq A$
By Union of Finite Sets is Finite, it follows that $B$ is finite.
We can use the definition of a cover to conclude that:
:$X \subseteq B \cup \paren {X \setminus A}$
By Relative Complement inverts Subsets, it follows that:
:$\paren {X \setminus B} \cap A = \O$
Hence, $A \subseteq B$.
But then by Subset of Finite Set is Finite, it follows that $A$ is finite, a contradiction.
== Also see ==
* Infinite Set in Compact Space has $\omega$-Accumulation Point
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Countably Compact Spaces
Category:Infinite Sets
Category:Countable Sets
Countably Metacompact Lindelöf Space is Metacompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a Lindelöf space which is also countably metacompact.
Then $T$ is metacompact.
== Proof ==
By the definitions:
:If $T = \struct {S, \tau}$ is a Lindelöf space then every open cover of $S$ has a countable subcover.
:If $T = \struct {S, \tau}$ is a countably metacompact space then every countable open cover of $S$ has an open refinement which is point finite.
It follows trivially that every open cover of $S$ has an open refinement which is point finite.
Hence the result by definition of metacompact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Countably Metacompact Spaces
Category:Metacompact Spaces
Category:Lindelöf Spaces
Countably Paracompact Lindelöf Space is Paracompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a Lindelöf space which is also countably paracompact.
Then $T$ is paracompact.
== Proof ==
By the definitions:
:If $T = \struct {S, \tau}$ is a Lindelöf space then every open cover of $S$ has a countable subcover.
:If $T = \struct {S, \tau}$ is a countably paracompact space then every countable open cover of $S$ has an open refinement which is locally finite.
It follows trivially that every open cover of $S$ has an open refinement which is locally finite.
Hence the result by definition of paracompact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Countably Paracompact Spaces
Category:Paracompact Spaces
Category:Lindelöf Spaces
Countably Paracompact Space is Countably Metacompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countably paracompact space.
Then $T$ is countably metacompact.
== Proof ==
From the definition, $T$ is countably paracompact every countable open cover of $X$ has an open refinement which is locally finite.
Consider some countable open cover $\UU$ of $X$.
Let $x \in X$.
Then there exists some neighborhood $\N_x$ of $x$ which intersects only finitely many elements of $\UU$.
Thus $x$ itself can be in only finitely many elements of $\UU$.
Hence $T$ must be, by definition, countably metacompact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Countably Paracompact Spaces
Category:Countably Metacompact Spaces
Counterexample/Examples/Sum of Cubes of Digits

== Example of Counterexample ==
Let $P$ be the statement:
:There exists no integer which is the sum of the cubes of its digits.
A counterexample to $P$ is the number $153$, as can be seen in Pluperfect Digital Invariants: $3$ Digits:
== Sources ==
* 
* 
Category:Counterexamples
Counterexample/Examples/Sum of Sine and Cosine equals 1

== Example of Counterexample ==
Let $P$ be the statement:
:$\forall x \in \R: \cos x + \sin x = 1$
A counterexample to $P$ is the real number $\dfrac \pi 4 \in \R$:
== Sources ==
* 
Category:Counterexamples
Counting Measure is Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Then the counting measure $\size {\,\cdot\,}$ on $\struct {X, \Sigma}$ is a measure.
== Proof ==
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\size {\,\cdot\,}$.
=== Proof of $(1)$ ===
The values that $\size {\,\cdot\,}$ can take are the natural numbers $\N$ and $+\infty$.
All of these are positive, whence:
:$\forall S \in \Sigma: \size S \ge 0$
=== Proof of $(2)$ ===
It is to be shown that (for a sequence $\sequence {S_n}_{n \mathop \in \N}$ of pairwise disjoint sets):
:$\ds \sum_{n \mathop = 1}^\infty \size {S_n} = \size {\bigcup_{n \mathop = 1}^\infty S_n}$
Suppose that the cardinality of at least one $S_i$ is infinite.
Then the cardinality of:
:$\ds \bigcup_{n \mathop = 1}^\infty S_n$
is infinite by Subset of Finite Set is Finite.
Hence:
:$\ds \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = +\infty$
Now as:
:$\map \mu {S_i} = +\infty$
it follows by the definition of extended real addition that:
:$\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = +\infty$
Suppose now that all $S_i$ are finite, and $\ds \sum_{n \mathop = 1}^\infty \size {S_n}$ converges.
Then by Convergent Series of Natural Numbers, it follows that for some $N \in \N$:
:$\forall m \ge N: \size {S_m} = 0$
which by Cardinality of Empty Set means $S_m = \O$.
Therefore we conclude that:
:$\ds \sum_{n \mathop = 1}^\infty \size {S_n} = \sum_{n \mathop = 1}^N \size {S_n}$
and by Union with Empty Set:
:$\ds \bigcup_{n \mathop = 1}^\infty S_n = \bigcup_{n \mathop = 1}^N S_n$
By Cardinality of Set Union: Corollary:
:$\ds \size {\bigcup_{n \mathop = 1}^N S_n} = \sum_{n \mathop = 1}^N \size {S_n}$
and combining this with the above yields the desired identity.
Suppose finally that $\ds \sum_{n \mathop = 1}^\infty \size {S_n}$ diverges.
To establish the desired identity, it is to be shown that $\ds \bigcup_{n \mathop = 1}^\infty S_n$ is infinite.
Suppose to the contrary that it has finite cardinality, say $k$.
By Cardinality of Set Union: Corollary, for each $N \in \N$:
:$\ds \sum_{n \mathop = 1}^N \size {S_n} = \size {\bigcup_{n \mathop = 1}^N S_n}$
Now since:
:$\ds \bigcup_{n \mathop = 1}^N S_n \subseteq \bigcup_{n \mathop = 1}^\infty S_n$
it follows by Cardinality of Subset of Finite Set that, for each $N \in \N$:
:$\ds \size {\bigcup_{n \mathop = 1}^N S_n} \le k$
which contradicts the assumption that $\ds \sum_{n \mathop = 1}^\infty \size {S_n}$ diverges.
Therefore $\ds \bigcup_{n \mathop = 1}^\infty S_n$ is infinite and the identity follows.
=== Proof of $(3')$ ===
By Cardinality of Empty Set:
:$\size \O = 0$
Having verified the axioms, it follows that $\size {\,\cdot\,}$ is a measure.
Category:Measures
Category:Counting Measure
Counting Measure on Natural Numbers is Sigma-Finite

== Theorem ==
Let $\mu$ be the counting measure on $\struct {\N, \map \PP \N}$. 
Then $\mu$ is $\sigma$-finite.
== Proof ==
For each $n \in \N$, define: 
:$X_n = \N \cap \closedint 1 n = \set {1, 2, \ldots, n}$
Since $X_n \subseteq \N$ for each $n$ from Intersection is Subset, we have that:
:$X_n$ is $\map \PP \N$-measurable for each $n$. 
We show that $\sequence {X_n}_{n \mathop \in \N}$ is an exhausting sequence in $\map \PP \N$ and that: 
:$\map \mu {X_n} < \infty$ for each $n \in \N$. 
From the definition of counting measure, we also have: 
:$\map \mu {X_n} = n < \infty$ 
for each $n$. 
We have that: 
:$\N \cap \closedint 1 n \subseteq \N \cap \closedint 1 {n + 1}$ 
from Set Intersection Preserves Subsets, so:
:$X_n \subseteq X_{n + 1}$ 
So $\sequence {X_n}_{n \mathop \in \N}$ is an increasing sequence of $\map \PP \N$-measurable sets.
It now only remains to verify: 
:$\ds \N = \bigcup_{n \mathop = 1}^\infty X_n$
We clearly have: 
:$\ds \bigcup_{n \mathop = 1}^\infty X_n \subseteq \N$
For each $k \in \N$, we have $k \in X_k$, so we also have: 
:$\ds \N \subseteq \bigcup_{n \mathop = 1}^\infty X_n$ 
so that: 
:$\ds \N = \bigcup_{n \mathop = 1}^\infty X_n$
So $\sequence {X_n}_{n \mathop \in \N}$ is exhausting sequence in $\map \PP \N$ and that: 
:$\map \mu {X_n} < \infty$ for each $n \in \N$.
So:
:$\mu$ is $\sigma$-finite.
Category:Counting Measure
Counting Theorem

== Theorem ==
Every well-ordered set is order isomorphic to a unique ordinal.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Motivation ==
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Counting Theorem
Category:Ordinals
Category:Well-Orderings
Category:Named Theorems
Counting Theorem/Corollary

== Corollary to Counting Theorem ==
Every properly well-ordered proper class is order isomorphic to the class of all ordinals.
== Proof ==
Let $A$ be a properly well-ordered class.
Let $\On$ denote the class of all ordinals.
By the Axiom of Replacement, neither $A$ nor $\On$ can be order isomorphic to a proper lower section of the other.
Hence it must be that $A$ is order isomorphic to $\On$.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 4$ The counting theorem: Theorem $4.1$ (The counting theorem)
Category:Counting Theorem
Counting Theorem/Motivation

== Motivation for the Counting Theorem ==
What we have achieved with the Counting Theorem is that for any properly well-ordered collection $A$, we can assign ordinal numbers as indices of the elements of $A$, treating the latter as a sequence:
:the $1$st, the $2$nd, $\ldots$, the $\alpha$th, $\ldots$ elements of $A$
and moreover, we can assign these indices in an order-preserving way.
That is, for all ordinal numbers $\alpha$ and $\beta$, $\alpha < \beta$ the $\alpha$th element comes before the $\beta$th element.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 4$ The counting theorem
Category:Counting Theorem
Counting Theorem/Proof 1

== Theorem ==
== Proof ==
=== Existence ===
Let $\struct {S, \preceq}$ be a woset.
From Condition for Woset to be Isomorphic to Ordinal‎, it is enough to show that for every $a \in S$, the initial segment $S_a$ of $S$ determined by $a$ is order isomorphic to some ordinal.
Let:
:$E = \set {a \in S: S_a \text{ is not isomorphic to an ordinal} }$
We will show that $E = \O$.
 that $E \ne \O$.
Let $a$ be the minimal element of $E$.
This is bound to exist by definition of woset.
So, if $x \prec a$, it follows that $S_x$ is isomorphic to an ordinal.
But for $x \prec a$, we have $S_x = \paren {S_a}_x$ from definition of an ordinal.
So every segment of $S_a$ is isomorphic to an ordinal.
Hence from Condition for Woset to be Isomorphic to Ordinal‎, $S_a$ itself is isomorphic to an ordinal.
This contradicts the supposition that $a \in E$.
Hence $E = \O$ and existence has been proved.
=== Uniqueness ===
Uniqueness follows from Isomorphic Ordinals are Equal.
Hence the result.
== Also presented as ==
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.12$
Category:Counting Theorem
Counting Theorem/Proof 2

== Theorem ==
== Proof ==
Let $A$ be a properly well-ordered class.
Let $\On$ denote the class of all ordinals.
By the Comparability Theorem, either:
:$A$ is order isomorphic to a lower section of $\On$
or:
:$\On$ is order isomorphic to a lower section of $A$.
Let $A$ be a set.
From Well-Ordering on Set is Proper Well-Ordering, $A$ is a properly well-ordered class.
As $A$ is a set, every lower section of $A$ is a set.
 $\On$ is order isomorphic to a lower section $L$ of $A$.
Then $\On$ would then be in one-to-one correspondence with that lower section.
Thus there would be a mapping $\phi: L \to \On$.
By the Axiom of Replacement, $\phi \sqbrk L = \On$ is then a set.
But from Class of All Ordinals is Proper Class, $\On$ is a proper class.
From this contradiction it follows that $\On$ cannot be order isomorphic to a lower section $L$ of $A$.
Hence $A$ is order isomorphic to a lower section of $\On$.
From Lower Section of Class of All Ordinals is Ordinal, that means $A$ is order isomorphic to an ordinal.
From Distinct Ordinals are not Order Isomorphic it follows that $A$ is order isomorphic to exactly one such ordinal.
== Also presented as ==
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 4$ The counting theorem: Theorem $4.1$ (The counting theorem)
Category:Counting Theorem
Couriers' Meeting

== Classic Problem ==
The Holy Father sent a courier from Rome to Venice, commanding him that he reach Venice in $7$ days.
The Most Illustrious Signorina of Venice sent a courier from Venice to Rome, directing him to reach Rome in $9$ days.
It is $250$ miles between Rome and Venice.
It so happened that both couriers started at exactly the same time.
In how many days do they meet?
== Solution ==
:$3 \frac {15} {16}$ days.
== Proof ==
Let $A$ and $B$ denote the couriers starting from Rome and Venice respectively.
Let $t$ be the time in days after they set out when they meet.
Let $x$ be the number of miles from Rome where that happens.
We have that:
:$A$ travels at $\dfrac {250} 7$ miles a day
:$B$ travels at $\dfrac {250} 9$ miles a day.
Thus we have:
Hence the result.
== Sources ==
* 
* : The Couriers Meeting: $94$
Category:Classic Problems
Covariance as Expectation of Product minus Product of Expectations

== Theorem ==
Let $X$ and $Y$ be random variables. 
Let the expectations of $X$ and $Y$ exist and be finite. 
Then the covariance of $X$ and $Y$ is given by: 
:$\cov {X, Y} = \expect {X Y} - \expect X \expect Y$
== Proof ==
== Sources ==
* 
* 
Category:Covariance
Covariance of Sums of Random Variables

== Theorem ==
Let $n$ be a strictly positive integer.
Let $\sequence {X_i}_{1 \mathop \le i \mathop \le n}$, $\sequence {Y_j}_{1 \mathop \le j \mathop \le n}$ be sequences of random variables.
Then: 
:$\ds \cov {\sum_{i \mathop = 1}^n X_i, \sum_{j \mathop = 1}^n Y_j} = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \cov {X_i, Y_j}$
 
== Proof ==
Category:Covariance
Category:Covariance of Sums of Random Variables
Covariance of Sums of Random Variables/Lemma

== Theorem ==
Let $n$ be a strictly positive integer.
Let $\sequence {X_i}_{1 \mathop \le i \mathop \le n}$ be a sequence of random variables.
Let $Y$ be a random variable.
Then: 
:$\ds \cov {\sum_{i \mathop = 1}^n X_i, Y} = \sum_{i \mathop = 1}^n \cov {X_i, Y}$
 
== Proof ==
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition: 
:$\ds \cov {\sum_{i \mathop = 1}^n X_i, Y} = \sum_{i \mathop = 1}^n \cov {X_i, Y}$
=== Basis for the Induction ===
We have that: 
:$\ds \cov {\sum_{i \mathop = 1}^1 X_i, Y} = \cov {X_1, Y} = \sum_{i \mathop = 1}^1 \cov {X_i, Y}$
We therefore have that $\map P 1$ is true. 
This is our basis for the induction.
=== Induction Hypothesis ===
Suppose that $\map P n$ is true for some fixed $n \in \N$. 
That is: 
:$\ds \cov {\sum_{i \mathop = 1}^n X_i, Y} = \sum_{i \mathop = 1}^n \cov {X_i, Y}$
We aim to show that it logically follows that $\map P {n + 1}$ is true. 
That is: 
:$\ds \cov {\sum_{i \mathop = 1}^{n + 1} X_i, Y} = \sum_{i \mathop = 1}^{n + 1} \cov {X_i, Y}$
=== Induction Step ===
This is our induction step:
We have: 
Hence the result by induction.
Category:Covariance
Category:Covariance of Sums of Random Variables
Covariant Derivative of Smooth Vector Field along Smooth Vector Field in Smooth Local Frame

== Theorem ==
Let $M$ be a smooth manifold with or without boundary.
Let $TM$ be the tangent bundle of $M$.
Let $\nabla$ be a connection in $TM$.
Let $U \subseteq M$ be an open subset.
Suppose $\tuple {E_i}$ is a smooth local frame over $U$.
Let $\set {\Gamma^k_{ij}}$ be the connection coefficients of $\nabla$ $\tuple {E_i}$.
Let $\map {\mathfrak X} U$ be the space of smooth vector fields on $U$.
Let $X, Y \in \map {\mathfrak X} U$ be smooth vector fields.
Suppose $X$ and $Y$ read as $X = X^i E_i$ and $Y = Y^i E_i$ $\tuple {E_i}$.
Then:
:$\nabla_X Y = \paren {\map X {Y^k} + X^i Y^j \Gamma^k_{ij}} E_k$
== Proof ==
== Sources ==
* : $\S 4$: Connections. Connections in the Tangent Bundle
Category:Smooth Manifolds
Category:Connections
Category:Covariant Derivatives
Covariant Hom Functor is Continuous

== Theorem ==
Let $\mathbf{Set}$ be the category of sets.
Let $\mathbf C$ be a locally small category.
Let $C$ be an object of $\mathbf C$, and let $\hom \paren {C, \cdot}: \mathbf C \to \mathbf{Set}$ be the covariant hom functor based at $C$.
Then $\hom \paren {C, \cdot}$ is a continuous functor.
== Proof ==
== Sources ==
* : $\S 5.5$: Proposition $5.25$
Category:Category Theory
Category:Functors
Covariant Hom Functor is Functor

== Theorem ==
Let $\mathbf{Set}$ be the category of sets.
Let $\mathbf C$ be a locally small category.
Let $C \in \mathbf C_0$ be an object of $\mathbf C$.
Let $\operatorname{Hom}_\mathbf C \paren {C, \cdot}: \mathbf C \to \mathbf{Set}$ be the covariant hom functor based at $C$.
Then $\operatorname{Hom}_\mathbf C \paren {C, \cdot}$ is a functor.
== Proof ==
For brevity and readability, let us write $\operatorname{Hom}$ for $\operatorname{Hom}_\mathbf C$.
For any object $D$ of $\mathbf C$ and morphism $f: C \to D \in \operatorname{Hom} \paren {C, D}$, we have:
:$\operatorname{Hom} \paren {C, \operatorname{id}_D} \paren f = \operatorname{id}_D \circ f = f$
and so:
:$\operatorname{Hom} \paren {C, \operatorname{id}_D} = \operatorname{id}_{\operatorname{Hom} \paren {C, D} }$
Thus:
:$\operatorname{Hom} \paren {C, \cdot}$ preserves identity morphisms
For composable morphisms $g$ and $f$ of $\mathbf C$ we furthermore have:
for any $x \in \operatorname{Hom} \paren {C, \operatorname{dom} f}$ composable with $f$.
Hence the result, by Equality of Mappings.
== Sources ==
* : $\S 2.7$
Category:Category Theory
Category:Category of Sets
Category:Functors
Covariant Tensor/Examples/Rank 1

== Example of Covariant Tensor ==
A covariant tensor of rank $1$ is a set satisfying, for all $i$:
:$\overline F_i = \ds \sum_{r \mathop = 1}^k \dfrac {\partial x^r} {\partial \overline x^i} F_r$
== Sources ==
* 
* 
Category:Examples of Covariant Tensors
Covariant Tensor/Examples/Rank 2

== Example of Covariant Tensor ==
A covariant tensor of rank $2$ is a set satisfying, for all $i$ and $j$:
:$\overline F_{ij} = \ds \sum_r \sum s \dfrac {\partial x^r} {\partial \overline x^i} \dfrac {\partial x^s} {\partial \overline x^j} F_{rs}$
== Sources ==
* 
* 
Category:Examples of Covariant Tensors
Cover is Cover of Subset

== Theorem ==
Let $S$ be a set.
Let $\CC$ be a cover of $S$.
Let $T \subseteq S$ be a subset of $S$.
Then, $\CC$ is a cover of $T$.
== Proof ==
By definition of a cover:
:$\ds S \subseteq \bigcup C$
But then, by Subset Relation is Transitive:
:$\ds T \subseteq \bigcup C$
Therefore, $C$ is a cover of $T$ by definition.
Category:Covers