nyuuzyou/wikis huggingface dataset:
Partition of Integer into Powers of 2 for Consecutive Integers

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map b n$ denote the number of ways $n$ can be partitioned into (integer) powers of $2$.
Then:
:$\map b {2 n} = \map b {2 n + 1}$
== Proof ==
We prove the theorem by establishing a bijection between the set of partitions of $2 n$ with that of $2 n + 1$, under the constraint where each partition is an integer power of $2$.
We have that $2^k$ is even for all $k > 0$.
Also, we have that $2 n + 1$ is odd for all $n$.
So, for each partition of $2 n + 1$ into integer powers of $2$, there is at least one part of size $1$.
Removing this part gives a partition of $2 n$ into integer powers of $2$.
The inverse of this is adding a part of size $1$ to a partition of $2 n$.
This will give a partition of $2 n + 1$ into integer powers of $2$.
Hence both removing and adding are well-defined mappings.
Thus they are bijections.
Therefore:
:$\map b {2 n} = \map b {2 n + 1}$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $12$
Category:Partition Theory
Partition of Integer where Part Appears no more than Twice/Examples/4

== Examples of Partition of Integer where Part Appears no more than Twice ==
The integer $4$ can be partitioned such that no part appears more than twice as follows:
:$4$
:$3 + 1$
:$2 + 2$
:$2 + 1 + 1$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $14$
Category:Examples of Integer Partitions
Partition of Integer where Part Appears no more than Twice/Table

== Table of Partition of Integer where Part Appears no more than Twice ==
The following table presents a list of the number of ways a positive integer $n$ can be partitioned such that no part appears more than twice for all $n$ from $1$ to $15$.
In the following, $\map v n$ denotes the number of such partitions for $n$.
:$\begin{array} {|r|r|}
\hline
n & \map v n \\
\hline
1 & 1 \\
2 & 2 \\
3 & 2 \\
4 & 4 \\
5 & 5 \\
6 & 6 \\
7 & 9 \\
8 & 13 \\
9 & 16 \\
10 & 22 \\
11 & 27 \\
12 & 36 \\
13 & 44 \\
14 & 57 \\
15 & 70 \\
\hline
\end{array}$
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $14$
Category:Examples of Integer Partitions
Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
:$A^* := \ds \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$
:$B^* := \ds \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Let $A^*$, $B^*$ and $\relcomp S {A^* \cup B^*}$ be expressed as the union of convex components of $S$:
:$\ds A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$
where $\relcomp S X$ denotes the complement of $X$ with respect to $S$.
Then the set $M = \set {A_\alpha, B_\beta, C_\gamma}$ inherits a linear ordering from $S$, and so is a linearly ordered set.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $5$
Category:Linearly Ordered Spaces
Partition of Non-Regular Prime Stellated Cyclic Polygons into Rotation Classes

== Theorem ==
Let $p$ be an odd prime.
Let $C$ be a circle whose center is $O$.
Consider the set $P$ of $p$ points on the circumference of $C$ dividing it into $p$ equal arcs.
Let $S$ be the set of all non-regular stellated $p$-gons whose vertices are the elements of $P$.
Let $\sim$ denote the equivalence relation on $S$ defined as:
:$\forall \tuple {a, b} \in S \times S: a \sim b \iff$ there exists a plane rotation about $O$ transforming $a$ to $b$.
Then the $\sim$-equivalence classes of $S$ into which $S$ can thereby be partitioned all have cardinality $p$.
== Proof ==
== Examples ==
== Sources ==
* : $\text {3-3}$ Wilson's Theorem: Theorem $\text {3-5}$
Category:Stellated Polygons
Category:Set Partitions
Category:Examples of Equivalence Classes
Category:Partition of Non-Regular Prime Stellated Cyclic Polygons into Rotation Classes
Partition of Non-Regular Prime Stellated Cyclic Polygons into Rotation Classes/Examples/Pentagons

== Examples of Use of Partition of Non-Regular Prime Stellated Cyclic Polygons into Rotation Classes ==
The equivalence classes by rotation of the non-regular stellated pentagons whose vertices are equally spaced on the circumference of a circle are depicted thus.
:
Thus there are $2$ equivalence classes, each with $5$ elements.
 suggests that these equivalence classes could be nicknamed fish and bat.
== Sources ==
* : $\text {3-3}$ Wilson's Theorem: Theorem $\text {3-5}$
Category:Partition of Non-Regular Prime Stellated Cyclic Polygons into Rotation Classes
Partition of Singletons yields Discrete Topology

== Theorem ==
Let $S$ be a set which is non-empty.
Let $\PP$ be the (trivial) partition of singletons on $S$:
:$\PP = \set {\set x: x \in S}$
Then the partition topology on $\PP$ is the discrete topology.
== Proof ==
From Basis for Discrete Topology it is shown that $\PP$ as defined here forms the basis of the discrete topology.
== Sources ==
* : Part $\text {II}$: Counterexamples: $5$. Partition Topology: $2$
Category:Discrete Topology
Category:Partition Topology
Partition of Spectrum of Densely-Defined Linear Operator

== Theorem ==
Let $\HH$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a densely-defined linear operator.
Let $\map \sigma T$ be the spectrum of $T$. 
Then: 
:$\map \sigma T = \map {\sigma_p} T \cup \map {\sigma_s} T \cup \map {\sigma_r} T$
where:
:$\map {\sigma_p} T$ is the point spectrum of $T$
:$\map {\sigma_s} T$ is the continuous spectrum of $T$.
:$\map {\sigma_r} T$ is the residual spectrum of $T$
== Proof ==
Let $\lambda \in \map \sigma T$.
Then, from the definition of the resolvent set of $T$, at least one of the following is false: 
:$(1) \quad$ $T - \lambda I$ is injective
:$(2) \quad$ $\map {\paren {T - \lambda I} } {\map D T}$ is everywhere dense in $\HH$
:$(3) \quad$ $\paren {T - \lambda I}^{-1}$ is bounded.
Note that either $(1)$ is true and one of $(2)$ or $(3)$ is false, or $(1)$ is false.
So we have three cases:
:$(\text A) \quad$ $(1)$ is false
:$(\text B) \quad$ $(1)$ is true, $(2)$ is true, $(3)$ is false
:$(\text C) \quad$ $(1)$ is true, $(2)$ is false
Note that the last case includes both the case that $(3)$ is true and $(3)$ is false.
Clearly only one of $(\text A)$, $(\text B)$ or $(\text C)$ can be true.
$(\text A)$ is precisely the definition of the point spectrum, $(\text B)$ is precisely the definition of the continuous spectrum and $(\text C)$ is precisely the definition of the residual spectrum.
So we have the result.
== Sources ==
* $25.3$: The Spectrum of Closed Unbounded Self-Adjoint Operators
Category:Spectra (Densely-Defined Linear Operators)
Pascal's Mystic Hexagram

== Theorem ==
Let $ABCDEF$ be a hexagram whose $6$ vertices lie on an ellipse such that the order of vertices along the ellipse is $AECFBD$.
:
Then the points of intersection of the sides of $ABCDEF$ lie on a straight line.
== Proof ==
== Also see ==
* Pappus's Hexagon Theorem
* Desargues' Theorem
* Pascal's Theorem
== Sources ==
* : $9$
* : $9$
Category:Pascal's Theorem
Category:Conic Sections
Category:Projective Geometry
Pascal's Rule

== Theorem ==
Let $\dbinom n k$ be a binomial coefficient.
For positive integers $n, k$ with $1 \le k \le n$:
:$\dbinom n {k - 1} + \dbinom n k = \dbinom {n + 1} k$
This is also valid for the real number definition:
:$\forall r \in \R, k \in \Z: \dbinom r {k - 1} + \dbinom r k = \dbinom {r + 1} k$
Thus the binomial coefficients can be defined using the following recurrence relation:
:$\dbinom n k = \begin{cases}
1 & : k = 0 \\
0 & : k > n \\
\dbinom {n - 1} {k - 1} + \dbinom {n - 1} k & : \text{otherwise}
\end{cases}$
=== Complex Numbers ===
== Direct Proof ==
== Combinatorial Proof ==
== Proof for Real Numbers ==
== Also presented as ==
== Examples ==
== Also known as ==
== Also see ==
* Definition:Pascal's Triangle
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Binomial Coefficients: $3.1.4$: Binomial Coefficients
* : Chapter $\text {III}$: The Natural Numbers: $\S 19$: Combinatorial Analysis: Theorem $19.10$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.6$
* : $\S 1.18$: Sequences Defined Inductively: Exercise $3 \ \text{(c)}$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $12$
* : $35$
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.30$
* : $35$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: $3.6.$
* 
* 
Category:Pascal's Rule
Category:Binomial Coefficients
Pascal's Rule/Also presented as

== Pascal's Rule: Also presented as == 
Some sources present Pascal's rule as:
:$\dbinom n k + \dbinom n {k + 1} = \dbinom {n + 1} {k + 1}$
Others present it in the form:
:$\dbinom {n - 1} {k - 1} + \dbinom {n - 1} k = \dbinom n k$
== Sources ==
* : $\S 1.2$: Problem $31.1$
* : $\text I$. Algebra: Permutations and Combinations: Two important relations
* : Chapter $\text {III}$: The Natural Numbers: $\S 19$: Combinatorial Analysis: Theorem $19.10$
* : $\S 0.6$: Theorem $8: \ 3$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.6$
* : $\text {3-1}$ Permutations and Combinations: Exercise $3$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $12$
* : $35$
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.30$
* : $\S 1.2.6$: Binomial Coefficients: $\text{D} \ (9)$
* : $35$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: $3.6.$
Category:Pascal's Rule
Pascal's Rule/Combinatorial Proof

== Theorem ==
== Proof ==
Suppose you were a member of a club with $n + 1$ members (including you).
Suppose it were time to elect a committee of $k$ members from that club.
From Cardinality of Set of Subsets, there are $\dbinom {n + 1} k$ ways to select the members to form this committee.
Now, you yourself may or may not be elected a member of this committee.
Suppose that, after the election, you are not a member of this committee.
Then, from Cardinality of Set of Subsets, there are $\dbinom n k$ ways to select the members to form such a committee.
Now suppose you are a member of the committee. Apart from you, there are $k - 1$ such members.
Again, from Cardinality of Set of Subsets, there are $\dbinom n {k - 1}$ ways of selecting the other $k - 1$ members so as to form such a committee.
In total, then, there are $\dbinom n k + \dbinom n {k - 1}$ possible committees.
Hence the result.
== Sources ==
* : $\S 1.2$: Problem $31.1$
* : $\text I$. Algebra: Permutations and Combinations: Two important relations
* : $\S 0.6$: Theorem $8: \ 3$
* : $\text {3-1}$ Permutations and Combinations: Exercise $3$
Category:Pascal's Rule
Pascal's Rule/Complex Numbers

== Theorem ==
For all $z, w \in \C$ such that it is not the case that $z$ is a negative integer and $w$ an integer:
:$\dbinom z {w - 1} + \dbinom z w = \dbinom {z + 1} w$
where $\dbinom z w$ is a binomial coefficient.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $42$ (Solution)
Category:Pascal's Rule
Pascal's Rule/Direct Proof

== Theorem ==
== Proof ==
Let $n, k \in \N$ with $1 \le k \le n$.
== Sources ==
* : $\text I$. Algebra: Permutations and Combinations: Two important relations
* : $\S 3$: Natural Numbers: Exercise $\S 3.11 \ (4)$
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem
Category:Pascal's Rule
Category:Direct Proofs
Pascal's Rule/Mistake/First Edition

== Source Work ==
:
:The Dictionary
::$35$
== Mistake ==
:From the rule for constructing the triangle it follows that
::$\paren {\dfrac n m} + \paren {\dfrac n {m + 1} } = \paren {\dfrac {n + 1} {m + 1} }$
== Correction ==
The notation is wrong.
It ought to be:
:$\dbinom n m + \dbinom n {m + 1} = \dbinom {n + 1} {m + 1}$
This has been partially corrected in , but the $+$ and $=$ signs have been omitted.
== Sources ==
* : $35$
Category:Mistakes/Curious and Interesting Numbers
Pascal's Rule/Mistake/Second Edition

== Source Work ==
:
:The Dictionary
::$35$
== Mistake ==
:From the rule for constructing the triangle it follows that
::$\dbinom n m \ \ \dbinom n {m + 1} \ \ \dbinom {n + 1} {m + 1}$
== Correction ==
The operators have been omitted.
It ought to be:
:$\dbinom n m + \dbinom n {m + 1} = \dbinom {n + 1} {m + 1}$
== Sources ==
* : $35$
Category:Mistakes/Curious and Interesting Numbers
Pascal's Theorem

== Theorem ==
Let $ABCDEF$ be a hexagon whose $6$ vertices lie on a conic section and whose opposite sides are not parallel.
Then the points of intersection of the opposite sides, when produced as necessary, all lie on a straight line.
== Proof ==
:
:
== Also known as ==
== Also see ==
* Pascal's Mystic Hexagram, a variant of this
* Brianchon's Theorem
* Pappus's Hexagon Theorem
* Desargues' Theorem
== Historical Note ==
== Sources ==
* : Chapter $\text{V}$: "Greatness and Misery of Man"
* : $6$
* 
* : Chapter $\text {A}.16$: Pascal ()
* : $6$
* 
* 
* 
Category:Pascal's Theorem
Category:Conic Sections
Category:Projective Geometry
Pascal's Theorem/Historical Note

== Historical Note on Pascal's Theorem ==
Pascal's Theorem was discovered by when he was in his mid-teens, in the wake of his encounter with .
He published it in his , which contains $400$ or so corollaries deduced from it, formed by allowing pairs of the six points involved to merge into coincidence.
 called this theorem:
:a sort of cat's cradle.
== Sources ==
* : Chapter $\text{V}$: "Greatness and Misery of Man"
Category:Historical Notes
Pascal's Triangle/Graphical Presentation/Modulo 2

== Pascal's Triangle Modulo $2$ ==
Entries $n$ are colour coded as follows:
:$n \equiv 0 \pmod 2$: $\color {black} {\text {White} }$
:$n \equiv 1 \pmod 2$: $\color {black} {\text {Black} }$
:
:
Category:Pascal's Triangle
Pascal's Triangle/Graphical Presentation/Modulo 3

== Pascal's Triangle Modulo $3$ ==
Entries $n$ are colour coded as follows:
:$n \equiv 0 \pmod 3$: $\color {black} {\text {White} }$
:$n \equiv 1 \pmod 3$: 
:$n \equiv 2 \pmod 3$: 
:
Category:Pascal's Triangle
Pascal's Triangle/Graphical Presentation/Modulo 4

== Pascal's Triangle Modulo $4$ ==
Entries $n$ are colour coded as follows:
:$n \equiv 0 \pmod 4$: $\color {black} {\text {White} }$
:$n \equiv 1 \pmod 4$: 
:$n \equiv 2 \pmod 4$: 
:$n \equiv 3 \pmod 4$: 
:
Category:Pascal's Triangle
Pascal's Triangle/Graphical Presentation/Modulo 5

== Pascal's Triangle Modulo $5$ ==
Entries $n$ are colour coded as follows:
:$n \equiv 0 \pmod 5$: $\color {black} {\text {White} }$
:$n \equiv 1 \pmod 5$: 
:$n \equiv 2 \pmod 5$: 
:$n \equiv 3 \pmod 5$: 
:$n \equiv 4 \pmod 5$: 
:
Category:Pascal's Triangle
Pascal's Triangle/Graphical Presentation/Modulo 6

== Pascal's Triangle Modulo $6$ ==
Entries $n$ are colour coded as follows:
:$n \equiv 0 \pmod 6$: $\color {black} {\text {White} }$
:$n \equiv 1 \pmod 6$: 
:$n \equiv 2 \pmod 6$: 
:$n \equiv 3 \pmod 6$: 
:$n \equiv 4 \pmod 6$: 
:$n \equiv 5 \pmod 6$: 
:
Category:Pascal's Triangle
Pascal's Wager

== Philosophical Argument ==
God either exists, or He does not.
If God exists, then when I die, I will go either to Hell or to Heaven.
If God does not exist, then when I die, my consciousness will simply vanish, and I will know nothing more.
If God exists, and I do not believe in Him, then I will go to Hell.
If God exists, and I do believe in Him, then I will go to Heaven.
If God does not exist, then whether or not I believe in Him, I will vanish into oblivion.
Therefore, if I do not believe in God, I will either vanish or go to Hell.
If I do believe in God, I will either vanish or go to Heaven.
Whichever way round it is, I stand to lose more by not believing in God than I do by believing in Him.
Therefore, I will believe in God.
== Refutations ==
:$(1): \quad$ There are many gods. How do I know I picked the right one to worship? If I picked the wrong one, then the god who is really in charge is going to be angry with me, and I will be punished by him as a result.
:$(2): \quad$ By all accounts, if God exists, then he is omniscient, so he knows my mind. He is not going to be impressed with a conscious rational decision based upon hope for a favourable personal outcome.
:$(3): \quad$ I cannot make myself believe in a philosophical system which my rational faculties discount as being at best extremely improbable, and at worst ludicrously childish. Belief is an internal process based upon the consideration of evidence, and not an emotional reaction to a story which may or may not be aesthetically appealing.
:$(4): \quad$ I intend to lead a good and moral life, based upon whatever humanistic codes of conduct are appropriate for the time and the place in which I live. If God exists, then he is either just (in which case he will judge me by my conduct and either bless me or curse me accordingly), or he is unjust (in which case he will condemn me solely for my intellectual reaction to the evidence presented for the truth that I have rejected). In the latter case, I would not want to be a party to such unjustness and would rather suffer for eternity than to compromise my intellectual integrity.
== Historical Note ==
== Sources ==
* : Chapter $\text{V}$: "Greatness and Misery of Man"
* 
Category:Philosophical Questions
Pasch's Theorem

== Theorem ==
Let $a, b, c, d$ be points on a line.
Let $\tuple {a, b, c}$ denote that $b$ lies between $a$ and $c$.
Then $\tuple {a, b, c}$ and $\tuple {b, c, d}$ together imply that $\tuple {a, b, d}$.
That is, if:
:$b$ is between $a$ and $c$
and:
:$c$ is between $b$ and $d$
then:
:$b$ is between $a$ and $d$.
== Proof ==
== Also see ==
* Outer Transitivity of Betweenness
== Historical Note ==
Category:Geometry
Category:Ordered Geometry
Pasting Lemma

== Theorem ==
=== Pair of Continuous Mappings on Open Sets ===
=== Union of Open Sets ===
=== Pair of Continuous Mappings on Closed Sets ===
=== Finite Union of Closed Sets ===
=== Counterexample of Infinite Union of Closed Sets ===
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 33$
Category:Pasting Lemma
Category:Continuous Mappings (Topology)
Category:Named Theorems
Pasting Lemma/Continuous Mappings on Closed Sets

== Theorem ==
Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be a finite indexing set.
Let $\family {C_i}_{i \mathop \in I}$ be a finite family of closed sets of $T$.
Let $\family {f_i : C_i \to Y}_{i \mathop \in I}$ be a finite family of continuous mappings.
Let $C = \ds \bigcup_{i \mathop \in I} C_i$.
Let $f = \ds \bigcup_{i \mathop \in I} f_i : C \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of relations.
Let for all $i, j \in I$, $f_i$ and $f_j$ agree on $C_i \cap C_j$.
Then $f$ is a continuous mapping on $C = \ds \bigcup_{i \mathop \in I} C_i$.
== Proof ==
From Union of Family of Mappings which Agree is Mapping:
:$f$ is a mapping from $C$ to $Y$.
From Restriction of Union of Mappings which Agree Equals Mapping:
:$\forall i \in I : f \restriction_{C_i} = f_i$
Hence:
:$\forall i \in I : f \restriction_{C_i}$ is continuous
From Pasting Lemma for Finite Union of Closed Sets
:$f$ is continuous
== Also see ==
* Pasting Lemma for Continuous Mappings on Open Sets for an analogous statement for open sets.
Category:Pasting Lemma
Category:Closed Sets
Pasting Lemma/Continuous Mappings on Open Sets

== Theorem ==
Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be an indexing set.
Let $\family {C_i}_{i \mathop \in I}$ be a family of open sets of $T$.
Let $\family {f_i : C_i \to Y}_{i \mathop \in I}$ be a family of continuous mappings.
Let $C = \ds \bigcup_{i \mathop \in I} C_i$.
Let $f = \ds \bigcup_{i \mathop \in I} f_i : C \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of relations.
Let for all $i, j \in I$, $f_i$ and $f_j$ agree on $C_i \cap C_j$.
Then $f$ is a continuous mapping on $C = \ds \bigcup_{i \mathop \in I} C_i$.
== Proof ==
From Union of Family of Mappings which Agree is Mapping:
:$f$ is a mapping from $C$ to $Y$.
From Restriction of Union of Mappings which Agree Equals Mapping:
:$\forall i \in I : f \restriction_{C_i} = f_i$
Hence:
:$\forall i \in I : f \restriction_{C_i}$ is continuous
From Pasting Lemma for Union of Open Sets
:$f$ is continuous.
== Also see ==
* Pasting Lemma for Continuous Mappings on Closed Sets for an analogous statement for closed sets.
Category:Pasting Lemma
Category:Open Sets
Pasting Lemma/Counterexample of Infinite Union of Closed Sets

== Theorem ==
Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be an infinite indexing set.
For all $i \in I$, let $C_i$ be closed in $T$.
Let $f: X \to Y$ be a mapping such that the restriction $f \restriction_{C_i}$ is continuous for all $i$.
Then it does not necessarily hold that $f$ is continuous on $C = \ds \bigcup_{i \mathop \in I}C_i$.
That is, it is not necessarily the case that $f \restriction_C$ is continuous.
== Proof ==
Let $f : \closedint {-1} 1 \to \R$ be the function on the closed interval $\closedint {-1} 1$ defined by:
:$\map f x = \begin{cases}
1 & : x \ge 0 \\
-1 & : x < 0 
\end{cases}$
The function $f$ is discontinuous at $0$.
The function $f$ restricted to the closed intervals:
:$\closedint {-1} {-\dfrac 1 2}, \closedint {-1} {-\dfrac 1 3}, \ldots, \closedint {-1} {-\dfrac 1 n}, \ldots, \closedint 0 1$
are constant functions.
From Constant Function is Continuous, the function $f$ restricted to the closed intervals:
:$\closedint {-1} {-\dfrac 1 2}, \closedint {-1} {-\dfrac 1 3}, \ldots, \closedint {-1} {-\dfrac 1 n}, \ldots, \closedint 0 1$
are continuous.
We have:
:$\closedint {-1} 1 = \paren {\ds \bigcup_{n \mathop \in \N_{>1} } \closedint {-1} {- \dfrac 1 n} } \cup \closedint 0 1$
The result follows.
Category:Pasting Lemma
Category:Closed Sets
Pasting Lemma/Finite Union of Closed Sets

== Theorem ==
Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be a finite indexing set.
Let $\family {C_i}_{i \mathop \in I}$ be a finite family of closed sets of $T$.
Let $f: X \to Y$ be a mapping such that the restriction $f \restriction_{C_i}$ is continuous for all $i \in I$.
Then $f$ is continuous on $C = \ds \bigcup_{i \mathop \in I} C_i$, that is, $f \restriction_C$ is continuous.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $V \subset S$ be a closed set.
By Continuity Defined from Closed Sets, we have that $U_i = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is also closed.
From the definition of a restriction, we have that $U_i = C_i \cap f^{-1} \sqbrk V$.
Therefore, we can compute:
That is, $U = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is the union of finitely many closed sets.
Therefore, $U$ is itself closed by definition of a topology.
It follows by Continuity Defined from Closed Sets that $f \restriction_C$ is also continuous.
== Also known as ==
This theorem is sometimes referred to as the pasting lemma.
== Also see ==
* Pasting Lemma for Union of Open Sets for an analogous statement for open sets.
== Sources ==
* : Problems: Section $1: \ 5$
* : $2$: Topological Spaces and Continuous Functions: $\S 18$: Continuous Functions: Theorem $18.3$: The pasting lemma
Category:Pasting Lemma
Category:Closed Sets
Pasting Lemma/Pair of Continuous Mappings on Closed Sets

== Theorem ==
Let $X$ and $Y$ be topological spaces.
Let $A$ and $B$ be closed in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be continuous mappings that agree on $A \cap B$.
Let $f \cup g$ be the union of the mappings $f$ and $g$:
:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$
Then the mapping $f \cup g : A \cup B \to Y$ is continuous.
== Proof ==
Follows directly from Pasting Lemma for Continuous Mappings on Closed Sets
== Also see ==
* Pasting Lemma for Pair of Continuous Mappings on Open Sets for an analogous statement for open sets.
Category:Pasting Lemma