nyuuzyou/wikis huggingface dataset:
Product with Inverse on Homomorphic Image is Group Homomorphism

== Theorem ==
Let $G$ be a group.
Let $H$ be an abelian group.
Let $\theta: G \to H$ be a (group) homomorphism.
Let $\phi: G \times G \to H$ be the mapping defined as:
:$\forall \tuple {g_1, g_2} \in G \times G: \map \phi {g_1, g_2} = \map \theta {g_1} \map \theta {g_2}^{-1}$
Then $\phi$ is a homomorphism.
== Proof ==
First note that from Group Homomorphism Preserves Inverses:
:$\map \theta {g_2}^{-1} = \paren {\map \theta {g_2} }^{-1} = \map \theta { {g_2}^{-1} }$
and so $\map \theta {g_1} \map \theta {g_2}^{-1}$ is not ambiguous:
:$\map \theta {g_1} \map \theta {g_2}^{-1} = \map \theta {g_1} \paren {\map \theta {g_2} }^{-1} = \map \theta {g_1} \map \theta { {g_2}^{-1} }$
From External Direct Product of Groups is Group, $G \times G$ is a group.
Let $a_1, a_2, b_1, b_2 \in G$.
We have:
Hence the result by definition of homomorphism.
== Examples ==
== Sources ==
* : Chapter $8$: The Homomorphism Theorem: Exercise $3$
Category:Group Homomorphisms
Category:Group Direct Products
Category:Product with Inverse on Homomorphic Image is Group Homomorphism
Product with Inverse on Homomorphic Image is Group Homomorphism/Examples/Mapping from D3 to Parity Group

== Examples of Use of Product with Inverse on Homomorphic Image is Group Homomorphism ==
Let $D_3$ denote the symmetry group of the equilateral triangle:
:
Let $G$ denote the parity group, defined as:
:$\struct {\set {1, -1}, \times}$
where $\times$ denotes conventional multiplication.
Let $\theta: D_3 \to G$ be the homomorphism defined as:
:$\forall x \in D_3: \map \theta x = \begin{cases} 1 & : \text{$x$ is a rotation} \\ -1 & : \text{$x$ is a reflection} \end{cases}$
Let $\phi: D_3 \times D_3 \to G$ be the mapping defined as:
:$\forall \tuple {g_1, g_2} \in D_3 \times D_3: \map \phi {g_1, g_2} = \map \theta {g_1} \map \theta {g_2}^{-1}$
Then the kernel $\map \ker \phi$ is the set of all pairs $\tuple {g_1, g_2}$ of elements of $D_3$ such that:
::$g_1$ and $g_2$ are both rotations
::$g_1$ and $g_2$ are both reflections.
== Proof ==
Thus $\map \ker \phi$ is the set of all pairs $\tuple {g_1, g_2}$ of elements of $D_3$ such that:
:$g_1$ and $g_2$ are both rotations
:$g_1$ and $g_2$ are both reflections.
== Sources ==
* : Chapter $8$: The Homomorphism Theorem: Exercise $3$
Category:Product with Inverse on Homomorphic Image is Group Homomorphism
Product with Inverse on Homomorphic Image is Group Homomorphism/Mistake

== Source Work ==
:
:Chapter $8$: The Homomorphism Theorem:
::Exercise $3$
== Mistake ==
:Suppose that $H$ is an abelian group and let $\vartheta: G \to H$ be a homomorphism. Define a map $\phi: G \times G \to H$ by
::$\map \phi {g_1, g_2} = \map \phi {g_1} \map \phi {g_2}^{-1}$.''
:Prove that $\phi$ is a homomorphism.
== Correction ==
The condition is wrong.
It should read:
::$\map \phi {g_1, g_2} = \map \vartheta {g_1} \map \vartheta {g_2}^{-1}$
== Sources ==
* : Chapter $8$: The Homomorphism Theorem: Exercise $3$
Category:Mistakes/A Course in Group Theory
Product with Matrix Exponential of Commutative Matrices

== Theorem ==
Let $\mathbf A$ and $\mathbf B$ be square matrices of order $m$ for some $m \in \Z_{\ge 1}$.
Let $t \in \R$ be a real number.
Let $e^{\mathbf A t}$ denote the matrix exponential of $\mathbf A$.
Let $\mathbf A \mathbf B = \mathbf B \mathbf A$.
Then:
:$e^{\mathbf A t} \mathbf B = \mathbf B e^{\mathbf A t}$
== Proof ==
Let:
:$\map \Phi t = e^{\mathbf A t} \mathbf B - \mathbf B e^{\mathbf A t}$
Category:Matrix Exponential
Category:Commutativity
Product with Repdigit can be Split into Parts which Add to Repdigit

== Theorem ==
Let $n$ be a positive integer with $d_1$ digits.
Let $m$ be a repdigit number with $d_2$ digits such that $d_2 > d_1$.
Let $r$ consist of the result when the rightmost $d_2$ digits of $m n$ is cut off and added to the remaining left hand portion.
Then $r$ is a repdigit number.
== Proof ==
Let $b > 1$ be the base we are working on.
Let $m = \sqbrk {aaa \dots a}_b$.
Let $R = \dfrac m a = \sqbrk {111 \dots 1}_b = \dfrac {b^{d_2} - 1} {b - 1}$.
Let the rightmost $d_2$ digits of $m n$ be $y$ and the remaining left hand portion be $x$.
Then we have:
so $r$ is divisible by $R$.
It remains to show that:
:$r \le b^{d_2} - 1 = \paren {b - 1} R$
because $\paren {b - 1} R$ is the largest $d_2$-digit integer.
This condition forces $r$ to have $d_2$ digits, so it remains a repdigit number.
Since $r$ is divisible by $R$, this is equivalent to:
:$r < b R$
We have:
Hence the result.
== Examples ==
== Sources ==
* : $6666$
* : $6666$
Category:Repdigit Numbers
Category:Product with Repdigit can be Split into Parts which Add to Repdigit
Product with Repdigit can be Split into Parts which Add to Repdigit/Mistake

== Source Work ==
:
:The Dictionary
::$6666$
:
:The Dictionary
::$6666$
== Mistake ==
:... if a number is multiplied by a number whose digits are all the same, for example, let $894$ be multiplied by $22,222$, then in this case the right-hand $5$ digits, added to the left-hand portion, form another number with equal digits: $894 \times 22,222 = 19866468$ and $198 + 66,468 = 66666$.
This is only guaranteed to work when the repdigit is strictly longer than the non-repdigit part.
For example:
It sometimes works, but you can't guarantee it:
== Sources ==
* : $6666$
* : $6666$
Category:Mistakes/Curious and Interesting Numbers
Product with Ring Negative

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Then:
:$\forall x, y \in \struct {R, +, \circ}: \paren {-x} \circ y = -\paren {x \circ y} = x \circ \paren {-y}$
where $\paren {-x}$ denotes the negative of $x$.
=== Corollary ===
== Proof ==
We have:
So from as applied to $\struct {R, +}$:
:$\paren {-x} \circ y = -\paren {x \circ y}$
The proof that $x \circ \paren {-y} = -\paren {x \circ y}$ follows identical lines.
== Also see ==
* Product of Ring Negatives
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.9$
* : Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $2 \ \text{(iv)}$
* : Chapter $1$: Rings - Definitions and Examples: $3$: Some special classes of rings: Lemma $1.2 \ \text{(ii)}$
* : $\S 54.2$ The definition of a ring and its elementary consequences: $\text{(i)}$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $4$
* : $\S 1$: Exercise $6$
Category:Ring Theory
Product with Ring Negative/Corollary

== Corollary to Product with Ring Negative ==
Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.
Then:
:$\forall x \in R: \paren {-1_R} \circ x = -x$
== Proof ==
== Sources ==
* : Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $2 \ \text{(vi)}$
Category:Ring Theory
Category:Rings with Unity
Product with Sum of Scalar

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group whose identity is $e$.
Let $\struct {R, +_R, \times_R}$ be a ring whose zero is $0_R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $x \in G, \lambda \in R$.
Let $\sequence {\lambda_m}$ be a sequence of elements of $R$, that is scalars.
Then:
:$\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$
== Proof ==
This follows by induction from , as follows.
For all $m \in \N_{>0}$, let $\map P m$ be the proposition:
:$\ds \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$
=== Basis for the Induction ===
Hence $\map P 1$ is true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then $\map P {n + 1}$ is true.
So this is our induction hypothesis:
:$\ds \paren {\sum_{k \mathop = 1}^n \lambda_k} \circ x = \sum_{k \mathop = 1}^n \paren {\lambda_k \circ x}$
Then we need to show:
:$\ds \paren {\sum_{k \mathop = 1}^{n + 1} \lambda_k} \circ x = \sum_{k \mathop = 1}^{n + 1} \paren {\lambda_k \circ x}$
=== Induction Step ===
This is our induction step:
So $\map P n \implies \map P {n + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m \in \N_{>0}: \paren {\sum_{k \mathop = 1}^m \lambda_k} \circ x = \sum_{k \mathop = 1}^m \paren {\lambda_k \circ x}$
== Also see ==
* Basic Results about Modules
* Basic Results about Unitary Modules
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Theorem $26.2 \ (4)$
Category:Module Theory
Products of 2-Digit Pairs which Reversed reveal Same Product

== Theorem ==
The following positive integers can be expressed as the product of $2$ two-digit numbers in $2$ ways such that the factors in one of those pairs is the reversal of each of the factors in the other:
:$504, 756, 806, 1008, 1148, 1209, 1472, 1512, 2016, 2208, 2418, 2924, 3024, 4416$
== Proof ==
Let $n \in \Z_{>0}$ such that:
:$n = \sqbrk {a b} \times \sqbrk {c d} = \sqbrk {b a} \times \sqbrk {d c}$
where $\sqbrk {a b}$ denotes the two-digit positive integer:
: $10 a + b$ for $0 \le a, b \le 9$
from the Basis Representation Theorem.
We have:
Thus the problem boils down to finding all the sets of one-digit integers $\set {a, b, c, d}$ such that $a c = b d$, and so that:
:$n = \sqbrk {a b} \times \sqbrk {c d} = \sqbrk {b a} \times \sqbrk {d c}$
and also:
:$n = \sqbrk {a d} \times \sqbrk {b c} = \sqbrk {d a} \times \sqbrk {c b}$
Thus we investigate all integers whose divisor count is $3$ or more, and find all those which have the product of single-digit integers in $2$ ways, as follows:
Further integers $n$ such that $\map {\sigma_0} n \le 4$ need not be investigated, as one of the pairs of factors will be greater than $9$.
== Sources ==
* : $504$
* : $504$
Category:Recreational Mathematics
Category:Reversals
Products of 2-Digit Pairs which Reversed reveal Same Product/Mistake

== Source Work ==
:
:The Dictionary
::$504$
:
:The Dictionary
::$504$
== Mistake ==
:$504$ is equal to both $12 \times 42$ and $21 \times 24$. There are thirteen such $2$-digit pairs, the largest being $36 \times 84 = 63 \times 48 = 3024$.
There are in fact $14$ such, the largest being $4416 = 46 \times 96 = 64 \times 69$.
== Sources ==
* : $504$
* : $504$
Category:Mistakes/Curious and Interesting Numbers
Products of Consecutive Integers in 2 Ways

== Theorem ==
The following integers are the product of consecutive integers in $2$ ways:
:$-720, 720, 5040$
== Proof ==
From 720 is Product of Consecutive Numbers in Two Ways:
:$720 = 6 \times 5 \times 4 \times 3 \times 2 = 10 \times 9 \times 8$
From 5040 is Product of Consecutive Numbers in Two Ways:
:$5040 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 10 \times 9 \times 8 \times 7$
Then:
:$-720 = \left({-6}\right) \left({-5}\right) \left({-4}\right) \left({-3}\right) \left({-2}\right) = \left({-10}\right) \left({-9}\right) \left({-8}\right)$
The same trick cannot be used for $-5040$ because there are four divisors in $10 \times 9 \times 8 \times 7$, and negating them makes the product positive.
Hence the result.
Category:Factorials
Products of Homeomorphic Spaces are Homeomorphic

== Theorem ==
Let:
:$\sequence {T_i}_{i \mathop \in I}$
:$\sequence {T'_i}_{i \mathop \in I}$
be indexed families of topological spaces, with indexing set $I$.
Let:
:$\sequence {\phi_i}_{i \mathop \in I}$
be an indexed family of homeomorphisms $\phi_i$ from $T_i$ to $T'_i$.
Define:
:$\ds T = \prod_{i \mathop \in I} T_i$
:$\ds T' = \prod_{i \mathop \in I} T'_i$
where $\ds \prod_{i \mathop \in I} T_i$ denotes the product space.
Then, the mapping $\phi : T \to T'$ defined as:
:$\map \phi x = \sequence {\map {\phi_i \circ \pr_i} x}_{i \mathop \in I}$
is a homeomorphism from $T$ to $T'$.
== Proof ==
We have that:
:$\map {\phi^{-1}} x = \sequence {\map {\phi_i^{-1} \circ \pr_i} x}_{i \mathop \in I}$
For:
It follows that $\phi$ is a bijection.
Additionally, by:
* Projection from Project Topology is Continuous
* Composite of Continuous Mappings is Continuous
* Continuous Mapping to Product Space
it follows that both $\phi$ and $\phi^{-1}$ are continuous.
Therefore, $\phi$ is a homeomorphism by definition.
Category:Homeomorphisms (Topological Spaces)
Category:Product Topology
Products of Open Sets form Local Basis in Product Space

== Theorem ==
Let $T_1 = \struct {A_1, \tau_1}$ and $T_2 = \struct {A_2, \tau_2}$ be topological spaces.
Let $\struct {T, \tau} = T_1 \times T_2$ be the product space of $T_1$ and $T_2$.
Let $\tuple {x, y} \in A_1 \times A_2$.
Let $W \in \tau$ be an open set of $T$ such that $\tuple {x, y} \in W$.
Then:
:$\exists U_1 \in \tau_1, U_2 \in \tau_2: \tuple {x, y} \in U_1 \times U_2 \subseteq W$
That is, products of open sets from $T_1$ and $T_2$ form a local basis at $\tuple {x, y}$.
== Proof ==
Let $W \in \tau$ such that $\tuple {x, y} \in W$.
From Natural Basis of Product Topology of Finite Product, $\tau$ is the topology with basis:
:$\BB = \set{U_1 \times U_2: U_1 \in \tau_1, U_2 \in \tau_2}$
Thus $W$ is the union of sets of the form $U_1 \times U_2$ where $U_1 \in \tau_1$ and $U_2 \in \tau_2$.
That is:
:$\exists U_1 \in \tau_1, U_2 \in \tau_2: \tuple{x, y} \in U_1 \times U_2 \subseteq W$
which is what was to be proved.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.5$: Products: Proposition $3.5.2$
Category:Product Topology
Products of Products are Homeomorphic to Collapsed Products

== Theorem ==
Let $I$ be an index set, and for each $i \in I$ let $J_i$ be an index set.
Let the sets $J_i$ be pairwise disjoint.
Let $\ds J = \bigcup_{i \mathop \in I} J_i$
For each $j \in J$, let $X_j$ be a topological space.
Then $\ds \prod_{j \mathop \in J} X_j$ is homeomorphic to $\ds \prod_{i \mathop \in I} \prod_{j \mathop \in J_i} X_j$.
== Proof ==
Category:Product Spaces
Products of Repdigit Numbers

== Fun Fact ==
 opens a mathematical miscellany with this delightful series:
and so on.
== Exploration ==
If we use the technique of long multiplication on that last line, we will be able to see the pattern:
 888888
x 8
-------
 64
 640
 6400
 64000
 640000
6400000
-------
7111104
-------
All we have to do is add a final $13$ to make the final number symmetrical.
A little experimentation shows that this sort of thing happens whatever digits you care to choose, for example:
and so on.
You can do this trick in any number base, for example:
and so on.
== Analysis ==
Let us pick two digits $r, s$ and a base $b$ such that $r, s < b$.
We can express a repdigit number $\sqbrk {rrr \ldots rrr}_b$ to any base $b$ as:
:$\ds \sqbrk {rrr \ldots rrr}_b = \sum_{k \mathop = 0}^n r b^k$
Thus:
Let $p = r s$.
There are two cases to consider here.
$(1) \quad$ Suppose $p < b$. Then:
:$\sqbrk {rrr \ldots rrr}_b \times s = \sqbrk {ppp \ldots ppp}_b$
and the pattern is obvious and boring.
An example here (using conventional decimal notation):
and so on.
$(2) \quad$ Now suppose $p \ge b$.
Note that $p < b^2$ as both $r < b, s < b$.
So we can express $p$ as $p = p_1 b + p_2$.
So:
Let $p_1 + p_2 = q$.
Again, there are two cases to consider:
$(2a) \quad$ Suppose $q < b$. Then:
:$\sqbrk {rrr \ldots rrr}_b \times s = \sqbrk {p_1 qqq \ldots qqq p_2}_b$
All we need to do is to add (or subtract) the difference between $p_1$ and $p_2$, and we get:
:$\sqbrk {rrr \ldots rrr}_b \times s + \paren {p_1 - p_2} = \sqbrk {p_1 qqq \ldots qqq p_1}_b$
An example here (using conventional decimal notation):
and so on.
$(2b) \quad$ Suppose $q \ge b$.
Then from the Division Theorem $q$ can be expressed as $q' b + t$ where $0 \le t < b$.
We have that $p_1, p_2 < b$ so $q = p_1 + p_2 < 2 b$, so it immediately follows that $q' = 1$.
It can also be seen that $t < b - 1$ and so (this is important in a bit) $t + 1 < b$.
Similarly we can see that $p_1 < b - 1$ and so (similarly important in a bit) $p_1 + 1 < b$.
So $q = b + t$, and so:
Putting $p_1 + 1 = p_3$ and $u = t + 1$, and noting that from above $u < b$, we have: 
:$\sqbrk {rrr \ldots rrr}_b \times s = \sqbrk {p_3 uuu \ldots uu t p_2}_b$
The final step is to calculate the difference between $\sqbrk {p_3 uuu \ldots uu t p_2}_b$ and $\sqbrk {p_3 uuu \ldots uuu p_3}_b$ in order to form a symmetrical pattern.
An example of this case is the $888 \ldots 88 \times 8 + 13 = 7111 \ldots 1117$ given at the start of this page.
 
== Sources ==
* : Calculator Curiosity 1
Category:Recreational Mathematics
Progressing Function Lemma

== Theorem ==
Let $A$ be a class.
Let $g$ be a progressing mapping on $A$.
Let $\RR$ be the relation defined as:
:$\map \RR {x, y} \iff \map g x \subseteq y \lor y \subseteq x$
where $\lor$ denotes disjunction (inclusive "or").
Then:
where $\land$ denotes conjunction ("and").
== Proof ==
In the following, let $x$ and $y$ be arbitrary elements of $A$ in the domain of $g$.
From Empty Set is Subset of All Sets:
:$\O \subseteq y$
Thus by the Rule of Addition:
:$\map g y \subseteq \O \lor \O \subseteq y$
and so it is seen that $\map \RR {y, \O}$.
That is, $(1)$ holds.
Let $\map \RR {x, y}$ and $\map \RR {y, x}$ both hold.
Thus by definition of $\RR$, both of the following hold:
:$(\text a): \quad \map g x \subseteq y$ or $y \subseteq x$
:$(\text b): \quad \map g y \subseteq x$ or $x \subseteq y$
We are to show that $\map \RR {x, \map g y}$ holds.
That is:
:$\map g x \subseteq \map g y$ or $\map g y \subseteq x$
Let $\map g x \subseteq y$.
Because $g$ is a progressing mapping:
:$y \subseteq \map g y$
and so:
:$\map g x \subseteq \map g y$
Thus $\map \RR {x, \map g y}$ holds.
Let $\map g y \subseteq x$.
Then by definition $\map \RR {x, \map g y}$ holds.
If neither $\map g x \subseteq y$ nor $\map g y \subseteq x$, then by $(\text a)$ and $(\text b)$:
:$y \subseteq x$ and $x \subseteq y$
and so by definition of set equality:
:$x = y$
Hence:
:$\map g x = \map g y$
Thus, again by definition of set equality:
:$\map g x \subseteq \map g y$
and so again, by definition, $\map \RR {x, \map g y}$ holds.
Thus it has been shown that in all cases, if $\map \RR {x, y}$ and $\map \RR {y, x}$ both hold, then $\map \RR {x, \map g y}$ holds.
Thus condition $(2)$ has been satisfied.
Hence the result.
== Linguistic Note ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Lemma $4.6$
Category:Progressing Mappings
Category:Named Theorems
Progressing Function Lemma/Linguistic Note

== Linguistic Note on Progressing Function Lemma ==
The name progressing function lemma comes from by and , where they refer to an inflationary mapping as a progressing function.
Hence it is reasonable to assume that this result could also be called:
:inflationary mapping lemma
:progressive mapping lemma
:progressing mapping lemma
and so on, with function replacing mapping as desired.
However, the author of this page has seen no evidence of this result anywhere but in that particular work, with any sort of name assigned to it.
Category:Linguistic Notes
Projection from Box Topology is Continuous

== Theorem ==
Let $\family {\struct{X_i, \tau_i}}_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$, that is:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\tau$ be the box topology on $X$.
For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$th projection on $X$:
:$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$
Then $\pr_i: \struct {X, \tau} \to \struct {X_i, \tau_i}$ is continuous for all $i \in I$.
== Proof ==
Let $\tau'$ be the product topology on $X$.
From Projection from Product Topology is Continuous:
:$\forall i \in I : \pr_i : \struct{X, \tau'} \to \struct{X_i, \tau_i}$ is continuous
From Box Topology contains Product Topology, $\tau$ is a finer topology than $\tau'$ on $X$.
From Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous, 
:$\forall i \in I : \pr_i : \struct {X, \tau} \to \struct{X_i, \tau_i}$ is continuous
Category:Box Topology
Projection from Cartesian Product under Chebyshev Distance is Continuous

== Theorem ==
Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces.
Let $\ds \AA = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
For all $i \in \set {1, 2, \ldots, n}$, let $\pr_i: \AA \to A_i$ be the $i$th projection on $\AA$:
:$\forall a \in \AA: \map {\pr_i} a = a_i$
where $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
Then for all $i \in \set {1, 2, \ldots, n}$, $p_i$ is continuous on $\AA$.
== Proof ==
Let $\epsilon \in \R_{>0}$.
Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$.
Let $\map {B_\epsilon} {a_i; d_i}$ be the open $\epsilon$-ball of $a_i$ in $M_i$.
From Open Ball in Cartesian Product under Chebyshev Distance:
:$\ds \map {B_\epsilon} {a; d_\infty} = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}$
By definition of $i$th projection:
:$\forall x \in \map {B_\epsilon} {a; d_\infty}: \map {\pr_i} x \in \map {B_\epsilon} {a_i; d_i}$
Thus by definition of image of subset of $\AA$ under $\pr_i$:
:$\pr_i \sqbrk {\map {B_\epsilon} {a; d_\infty} } \subseteq \map {B_\epsilon} {a_i; d_i}$
Thus by definition of continuity at a point, $\pr_i$ is continuous at $a$.
As $a$ is arbitrary, it follows that $\pr_i$ is continuous at all points of $\AA$.
Hence, by definition, $\pr_i$ is continuous from $\AA$ to $A_i$.
As $i$ is arbitrary, it follows that the result holds for all $i \in \set {1, 2, \ldots, n}$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $7$
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 15$
Category:Chebyshev Distance
Category:Projections
Category:Continuous Mappings on Metric Spaces
Projection from Metric Space Product with Euclidean Metric is Continuous

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\AA := A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.
Let $\MM = \struct {\AA, d}$ denote the metric space on $\AA$ where $d: \AA \to \R$ is the Euclidean metric on $\AA$:
:$\map d {x, y} := \sqrt {\paren {\map {d_1} {x_1, y_1} }^2 + \paren {\map {d_2} {x_2, y_2} }^2}$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.
Let $\pr_1: \MM \to M_1$ and $\pr_2: \MM \to M_2$ denote the first projection and second projection respectively on $\MM$.
Then $\pr_1$ and $\pr_2$ are both ‎continuous on $\MM$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 15$
Category:Projections
Category:Euclidean Metric
Category:Continuous Mappings on Metric Spaces
Category:Projection from Metric Space Product with Euclidean Metric is Continuous
Projection from Metric Space Product with Euclidean Metric is Continuous/Proof 1

== Theorem ==
== Proof ==
The Euclidean metric is an instance of the $p$-product metric:
:$\map {d_p} {x, y} := \paren {\paren {\map {d_1} {x_1, y_1} }^p + \paren {\map {d_2} {x_2, y_2} }^p}^{1/p}$
where $p = 2$.
The result is therefore seen to be an instance of Projection from Metric Space Product with P-Product Metric is Continuous.
Category:Projection from Metric Space Product with Euclidean Metric is Continuous
Projection from Metric Space Product with Euclidean Metric is Continuous/Proof 2

== Theorem ==
== Proof ==
We want to show that, for all $a \in \AA$:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in \AA: \map d {z, a} < \delta \implies \map {d_1} {\map {\pr_1} z, \map {\pr_1} a} < \epsilon$
and:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in \AA: \map d {z, a} < \delta \implies \map {d_2} {\map {\pr_2} z, \map {\pr_2} a} < \epsilon$
Let $\epsilon \in \R_{>0}$ be arbitrary.
Let $a = \tuple {x_0, y_0} \in \AA$ also be arbitrary.
Let $\delta = \epsilon$.
Let $z = \tuple {x_1, y_1} \in \AA$ such that $\map d {x, a} < \delta$.
We have:
Then:
and similarly:
Hence:
and:
We have that $a$ and $\epsilon$ are arbitrary.
Hence the result by definition of ‎continuity.
Category:Projection from Metric Space Product with Euclidean Metric is Continuous