nyuuzyou/wikis huggingface dataset:
Cartesian Product of Countable Sets is Countable/Formal Proof 2/Mistake

== Source Work ==
:
:Introduction to Set Theory:
::$2$. Set Theoretical Equivalence and Denumerability
This mistake can be seen in the 2009 Dover edition: ISBN 0-486-47222-1
== Mistake ==
:If $A$ and $B$ are denumerable, then so are $A \cup B$ and $A \times B$.
:...
:For instance, if $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ are enumerations of $A$ and $B$, then the map $f$ given by the rule
::$\map f {\tuple {a_k, a_l} } = \dfrac {\paren {k + l - 1} \paren {k + l - 2} } 2 + \dfrac {l + \paren {-1}^{k + 1} } 2 k + \dfrac {1 + \paren {-1}^{k + l - 1} } 2 l$
:gives an enumeration of $\tuple {A \times B}$.
== Correction ==
The expression starts incorrectly. It should be:
::$\map f {\tuple {a_k, b_l} } = \dfrac {\paren {k + l - 1} \paren {k + l - 2} } 2 + \dfrac {l + \paren {-1}^{k + 1} } 2 k + \dfrac {1 + \paren {-1}^{k + l - 1} } 2 l$
because the second element of the ordered pair comes from $B$, not $A$.
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
Category:Mistakes/Point Set Topology
Cartesian Product of Countable Sets is Countable/Informal Proof

== Theorem ==
== Proof ==
Let $S = \set {s_0, s_1, s_2, \dotsc}$ and $T = \set {t_0, t_1, t_2, \dotsc}$ be countable sets.
If both $S$ and $T$ are finite, the result follows immediately.
Suppose either of $S$ or $T$ (or both) is countably infinite.
We can write the elements of $S \times T$ in the form of an infinite table:
$\quad \begin{array} {*{4}c}
\tuple {s_0, t_0} & \tuple {s_0, t_1} & \tuple {s_0, t_2} & \cdots \\
\tuple {s_1, t_0} & \tuple {s_1, t_1} & \tuple {s_1, t_2} & \cdots \\
\tuple {s_2, t_0} & \tuple {s_2, t_1} & \tuple {s_2, t_2} & \cdots \\
\vdots & \vdots & \vdots & \ddots \\
\end{array}
$
This table clearly contains all the elements of $S \times T$.
Now we can count the elements of $S \times T$ by processing the table diagonally.
First we pick $\tuple {s_0, t_0}$.
Then we pick $\tuple {s_0, t_1}, \tuple {s_1, t_0}$. Then we pick $\tuple {s_0, t_2}, \tuple {s_1, t_1}, \tuple {s_2, t_0}$.
We can see that all the elements of $S \times T$ will (eventually) be listed, and there is a specific number (element of $\N$) to index each of its elements with.
Thus we have the required one-to-one correspondence between $S \times T$ and $\N$, and our assertion is proved.
Category:Cartesian Product of Countable Sets is Countable
Cartesian Product of Equivalence Relations

== Theorem ==
Let $A$ and $B$ be sets.
Let $\RR$ and $\SS$ be equivalence relations on $A$ and $B$ respectively.
Let $\TT$ be the relation on $A \times B$ defined as:
:$\forall \tuple {u, v}, \tuple {x, y} \in A \times B: \tuple {u, v} \mathrel \TT \tuple {x, y} \iff u \mathrel \RR x \land v \mathrel \SS y$
Then $\TT$ is an equivalence relation on $A \times B$.
The equivalence class $\eqclass {\tuple {a, b} } \TT$ of an element $\tuple {a, b}$ of $\TT$ is:
:$\eqclass {\tuple {a, b} } \TT = \set {\tuple {x, y}: x \in \eqclass a \RR, y \in \eqclass b \SS}$
== Proof ==
We have that $\RR$ and $\SS$ are equivalence relations on $A$ and $B$ respectively.
Thus they are both reflexive, symmetric and transitive.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Thus $\TT$ is seen to be reflexive.
=== Symmetry ===
Let $\tuple {x_1, y_1}, \tuple {x_2, y_2} \in A \times B$ such that:
:$\tuple {x_1, y_1} \mathrel \TT \tuple {x_2, y_2}$
Then:
Thus $\TT$ is seen to be symmetric.
=== Transitivity ===
Let $\tuple {x_1, y_1}, \tuple {x_2, y_2}, \tuple {x_3, y_3} \in A \times B$ such that:
Then:
Thus $\TT$ is seen to be transitive.
$\TT$ has been shown to be reflexive, symmetric and transitive.
Hence by definition $\TT$ is an equivalence relation.
The equivalence class $\eqclass {\tuple {a, b} } \TT$ follows directly:
:$\eqclass {\tuple {a, b} } \TT = \set {\tuple {x, y}: x \in \eqclass a \RR, y \in \eqclass b \SS}$
Category:Equivalence Relations
Category:Cartesian Product
Cartesian Product of Family/Examples/1 and 2

== Example of Cartesian Product of Family ==
Let $A_\O := \set \O$ and $A_{\set \O} := \set {\O, \set \O}$.
Thus $A_\O$ and $A_{\set \O}$ are the numbers $1$ and $2$ as defined by the Von Neumann construction.
Then:
:$A_\O \times A_{\set \O} = \set {\tuple {\O, \O}, \tuple {\O, \set \O} }$
while:
:$\ds \prod_{i \mathop \in A_{\set \O} } A_i = \set {\set {\tuple {\O, \O}, \tuple {\set \O, \O} }, \set {\tuple {\O, \O}, \tuple {\set \O, \set \O} } }$
== Proof ==
First we have:
Then:
The above is deconstructed as follows.
We have that:
Hence we have that:
Note that $(2)$ above is the set of all mappings from $\set {\O, \set \O}$ to $\set {\O, \set \O}$ as follows:
:Each such mapping is a set of $2$ ordered pairs of which the first coordinates are the elements of $\set {\O, \set \O}$
:From Cardinality of Set of All Mappings there are $2^2 = 4$ set of $2$ such ordered pairs.
Now we have to select the elements $f$ of $\ds \paren {\bigcup_{i \mathop \in A_{\set \O} } A_i}^{A_{\set \O} }$ such that:
:$\map f i \in A_i$
for all $i \in \set {\O, \set \O}$.
We have that:
Then:
Hence:
:$\ds \prod_{i \mathop \in A_{\set \O} } A_i = \set {\set {\tuple {\O, \O}, \tuple {\set \O, \O} }, \set {\tuple {\O, \O}, \tuple {\set \O, \set \O} } }$
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $8$
Category:Examples of Cartesian Products
Cartesian Product of Family/Examples/Household Pets

== Example of Cartesian Product of Family ==
Informally it is like this.
We have a set of sets which are indexed by the indexing set can be anything, as long as it is ordered.
We think of it as numbers, but it could be anything.
The indexing set indentifies both sets, and elements in the tuple.
Let $I$ be the set of household pets:
:$I := \set {\text {Bruiser}, \text {Claude} }$
Let $T$ be the set of toys they play with:
:$T:= \set {\text {rubber bone}, \text {chewed tennis ball}, \text {missing slipper}, \text {jingly roller}, \text {fluffy mouse on a string}, \text {table tennis ball} }$
The set of toys played with by $\text {Bruiser}$ consist of: 
::$\set {\text {rubber bone}, \text {chewed tennis ball}, \text {missing slipper} }$
The set of toys played with by $\text {Claude}$ consist of: 
::$\set {\text {table tennis ball}, \text {jingly roller}, \text {fluffy mouse on a string} }$
Thus we have a tuple of sets:
:$\tuple {\set {\text {rubber bone}, \text {chewed tennis ball}, \text {missing slipper} }, \set {\text {table tennis ball}, \text {jingly roller}, \text {fluffy mouse on a string} } }$
Let $I = \set {\text {Bruiser}, \text {Claude} }$ be considered as an indexing set.
The tuple indexed by $\text {Bruiser}$ is:
:$\set {\text {rubber bone}, \text {chewed tennis ball}, \text {missing slipper} }$
The tuple indexed by $\text {Claude}$ is:
:$\set {\text {table tennis ball}, \text {jingly roller}, \text {fluffy mouse on a string} }$
The Cartesian product of the toys of $\text {Bruiser}$ and $\text {Claude}$ is:
:$\left \{ {\tuple {\text {rubber bone}, \text {jingly roller} }, \tuple {\text {chewed tennis ball}, \text {jingly roller} }, \tuple {\text {missing slipper}, \text {jingly roller} }, }\right.$
::$\left. {\tuple {\text {rubber bone}, \text {table tennis ball} }, \tuple {\text {chewed tennis ball}, \text {table tennis ball} }, \tuple {\text {missing slipper}, \text {table tennis ball} }, }\right.$
::$\left. {\tuple {\text {rubber bone}, \text {fluffy mouse on a string} }, \tuple {\text {chewed tennis ball}, \text {fluffy mouse on a string} }, \tuple {\text {missing slipper}, \text {fluffy mouse on a string} } }\right\}$
Each of these tuples is indexed by $\set {\text {Bruiser}, \text {Claude} }$.
So:
:$\tuple {\text {chewed tennis ball}, \text {fluffy mouse on a string} }$ indexed by $\text {Bruiser}$ is $\text {chewed tennis ball}$
:$\tuple {\text {chewed tennis ball}, \text {fluffy mouse on a string} }$ indexed by $\text {Claude}$ is $\text {fluffy mouse on a string}$.
Category:Examples of Cartesian Products
Cartesian Product of Group Actions

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S$ and $T$ be sets.
Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be group actions.
Then the operation $*: G \times \paren {S \times T} \to S \times T$ defined as:
:$\forall \tuple {g, \tuple {s, t} } \in G \times \paren {S \times T}: g * \tuple {s, t} = \tuple {g *_S s, g *_T t}$
is a group action.
== Proof ==
The group action axioms are investigated in turn.
Let $g, h \in G$ and $s, t \in S$.
Thus:
demonstrating that holds.
Then:
demonstrating that holds.
The group action axioms are thus seen to be fulfilled, and so $*$ is a group action.
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions: Exercise $3$
Category:Group Actions
Category:Cartesian Product
Cartesian Product of Homeomorphisms is Homeomorphism

== Theorem ==
Let $S_1, S_2, T_1, T_2$ be topological spaces.
Let $f_1: S_1 \to T_1$ and $f_2: S_2 \to T_2$ be mappings.
Let:
:$f_1 \times f_2: S_1 \times S_2 \to T_1 \times T_2$
be defined as:
:$\forall \tuple {x, y} \in S_1 \times S_2: \map {\paren {f_1 \times f_2} } {x, y} = \tuple {\map {f_1} x, \map {f_2} y}$
where $S_1 \times S_2$ denotes the product space of $S_1$ and $S_2$, and similarly for $T_1 \times T_2$.
Let $f_1$ and $f_2$ be homeomorphisms.
Then $f_1 \times f_2$ is also a homeomorphism.
== Proof ==
From Cartesian Product of Mappings is Continuous iff Factor Mappings are Continuous:
:$f_1 \times f_2$ is continuous.
From Cartesian Product of Bijections is Bijection:
:$f_1 \times f_2$ is a bijection.
From Cartesian Product of Mappings is Continuous iff Factor Mappings are Continuous:
:$\paren {f_1 \times f_2}^{-1} = f_1^{-1} \times f_2^{-1}$ is continuous.
So, by definition, $f_1 \times f_2$ is a homeomorphism.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 20$
Category:Cartesian Product
Category:Homeomorphisms (Topological Spaces)
Cartesian Product of Infinite Set Equivalent to Itself implies Axiom of Choice

== Theorem ==
Suppose that, for every Dedekind-infinite set $A$, it holds that:
:$A \sim A \times A$
where $\sim$ denotes set equivalence, and $\times$ denotes the cartesian product.
Then, the holds.
== Proof ==
By Zermelo's Well-Ordering Theorem is Equivalent to Axiom of Choice, it suffices to show that the hypothesis implies Zermelo's Well-Ordering Theorem.
Since every finite set is well-orderable, it suffices to consider only infinite sets.
Let $B$ be an arbitrary infinite set.
, assume that $B \cap \On = \O$.
By Existence of Ordinal with no Surjection from Set, let $\beta \in \On \setminus \set \O$ such that there is no surjection from $B$ to $\beta$.
From Set is Infinite iff exist Subsets of all Finite Cardinalities, it follows that $\beta$ is infinite.
We have:
:$\omega \subseteq \beta \subseteq B \cup \beta$
so by Equivalent Conditions for Dedekind-Infinite Set:
:$B \cup \beta$ is Dedekind-infinite.
Therefore, , there exists a bijection:
:$\phi : B \cup \beta \to \paren {B \cup \beta} \times \paren {B \cup \beta}$
For each $x \in B$, define:
:$S_x = \set {\gamma \in \beta : \map \phi \gamma \in \beta \times \set x}$
 for some $x_0 \in B, S_{x_0} = \O$.
Then, since $\phi$ is a bijection, it follows that $\phi^{-1} \sqbrk {\beta \times \set {x_0}} \subseteq B$.
Hence, the mapping $\psi : B \to \beta$ defined as:
:$\map \psi y = \begin{cases} \gamma & : \map \phi y = \tuple {\gamma, x_0} \\ 0 & : \text{otherwise} \end{cases}$
is a surjection.
Since the contradicts the definition of $\beta$, we conclude that each $S_x \ne \O$.
Finally, we define $\chi : B \to \beta$ as:
:$\map \chi x = \min S_x$
Suppose that, for $x, y \in B$, we have:
:$\map \chi x = \map \chi y$
Then, since:
:$\map \phi {\map \chi x} \in \beta \times \set x$
:$\map \phi {\map \chi y} \in \beta \times \set y$
it follows from the equality that $x = y$.
Therefore, $\chi$ is an injection.
Thus, by Injection Induces Well-Ordering:
:$B$ is well-orderable
Since $B$ was arbitrary, it follows that Zermelo's Well-Ordering Theorem holds.
Category:Cartesian Product
Category:Equivalents of Axiom of Choice
Category:Set Equivalence
Cartesian Product of Intersections

== Theorem ==
:$\paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2} = \paren {S_1 \times T_1} \cap \paren {S_2 \times T_2}$
where $S_1, S_2, T_1, T_2$ are sets.
=== Corollary 1 ===
=== Corollary 2 ===
=== General Case ===
== Proof ==
== Sources ==
* : $\S 6$: Ordered Pairs: Exercise $\text{(ii)}$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 9 \alpha$
* : $3$: Continuity generalized: topological spaces: $3.5$: Products: Definition $3.5.1$
* : Chapter $1$: Sets and Logic: Exercise $14$
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts: Exercise $1.2 \ \text{(n)}$
Category:Cartesian Product
Category:Set Intersection
Category:Cartesian Product of Intersections
Cartesian Product of Intersections/Corollary 1

== Corollary to Cartesian Product of Intersections ==
:$A \times \paren {B \cap C} = \paren {A \times B} \cap \paren {A \times C}$
== Proof ==
Take the result Cartesian Product of Intersections:
:$\paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2} = \paren {S_1 \times T_1} \cap \paren {S_2 \times T_2}$
Put $S_1 = S_2 = A, T_1 = B, T_2 = C$:
== Sources ==
* : Chapter $1$. Sets: Exercise $8 \ \text{(i)}$
* : $\S 1.9$: Cartesian Product: Theorem $9.2$
* : $\S 3$. Ordered pairs; cartesian product sets: Exercise $3 \ (1)$
Category:Cartesian Product of Intersections
Cartesian Product of Intersections/Corollary 2

== Corollary to Cartesian Product of Intersections ==
:$\paren {A \times B} \cap \paren {B \times A} = \paren {A \cap B} \times \paren {A \cap B}$
== Proof ==
Take the result Cartesian Product of Intersections:
:$\paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2} = \paren {S_1 \times T_1} \cap \paren {S_2 \times T_2}$
Put $S_1 = A, S_2 = B, T_1 = B, T_2 = A$:
== Sources ==
* : Chapter $1$. Sets: Exercise $10$
* : $\S 3$. Ordered pairs; cartesian product sets: Exercise $3 \ (3)$
Category:Cartesian Product of Intersections
Cartesian Product of Intersections/General Case

== Theorem ==
Let $I$ be an index set.
Let $\family {S_i}_{i \mathop \in I}$ and $\family {T_i}_{i \mathop \in I}$ be indexed families of sets over the same index set.
Then:
:$\ds \paren {\prod_{i \mathop \in I} S_i} \cap \paren {\prod_{i \mathop \in I} T_i} = \prod_{i \mathop \in I} \paren {S_i \cap T_i}$
where $\ds \prod$ denotes Cartesian product.
== Proof ==
Let $\family {x_i}_{i \mathop \in I} \in \ds \paren {\prod_{i \mathop \in I} S_i} \cap \paren {\prod_{i \mathop \in I} T_i}$.
By definition of intersection, this is equivalent to the conjunction of:
:$\family {x_i}_{i \mathop \in I} \in \ds \prod_{i \mathop \in I} S_i$
:$\family {x_i}_{i \mathop \in I} \in \ds \prod_{i \mathop \in I} T_i$
By definition of Cartesian product, this means, for all $i \in I$:
:$x_i \in S_i$ and $x_i \in T_i$
Again by definition of intersection, this is equivalent to, for all $i \in I$:
:$x_i \in S_i \cap T_i$
Finally, by definition of Cartesian product, this is equivalent to:
:$\family {x_i}_{i \mathop \in I} \in \ds \prod_{i \mathop \in I} \paren {S_i \cap T_i}$
The result follows by definition of set equality.
Category:Cartesian Product of Intersections
Cartesian Product of Intervals is Convex Set

== Theorem ==
Let $n \in \N$.
For all $k \in \set {1, \ldots, n}$, let $\Bbb I_k$ be a real interval of any of the real interval types.
Then the cartesian product $\Bbb I_1 \times \ldots \times \Bbb I_n$ is a convex set.
== Proof ==
Let $\mathbf x, \mathbf y \in \Bbb I_1 \times \ldots \times \Bbb I_n$ with:
where $x_k , y_k \in \Bbb I_k$ for all $k \in \set {1, \ldots, n}$.
Let $t \in \closedint 0 1$.
Suppose that $x_k \le y_k$.
It follows that:
By definition of real interval, it follows that $t x_k + \paren {1-t} y_k \in \Bbb I_k$.
If instead $x_k \ge y_k$, it follows by similar calculations that:
:$x_k \ge t x_k + \paren {1-t} y_k \ge y_k$
By definition of real interval, it follows that $t x_k + \paren {1-t} y_k \in \Bbb I_k$.
It follows that $t \mathbf x + \paren {1-t} \mathbf y \in \Bbb I_1 \times \ldots \times \Bbb I_n$.
By definition of convex set, it follows that $\Bbb I_1 \times \ldots \times \Bbb I_n$ is a convex set.
Category:Convex Sets (Vector Spaces)
Cartesian Product of Intervals is Simply Connected

== Theorem ==
Let $n \in \N$.
For all $k \in \set {1, \ldots, n}$, let $\Bbb I_k$ be a real interval of any of the real interval types.
Let $\tau_0$ denote the subspace topology on the cartesian product $\Bbb I_1 \times \ldots \times \Bbb I_n$, induced by the Euclidean topology on $\R^n$.
Then $\struct {\Bbb I_1 \times \ldots \times \Bbb I_n, \tau_0}$ is simply connected.
== Proof ==
The result follows from Cartesian Product of Intervals is Convex Set and Convex Set is Simply Connected.
Category:Simply Connected Spaces
Cartesian Product of Mappings is Continuous iff Factor Mappings are Continuous

== Theorem ==
Let $S_1, S_2, T_1, T_2$ be topological spaces.
Let $f_1: S_1 \to T_1$ and $f_2: S_2 \to T_2$ be mappings.
Let:
:$f_1 \times f_2: S_1 \times S_2 \to T_1 \times T_2$
be defined as:
:$\forall \tuple {x, y} \in S_1 \times S_2: \map {\paren {f_1 \times f_2} } {x, y} = \tuple {\map {f_1} x, \map {f_2} y}$
where $S_1 \times S_2$ denotes the product space of $S_1$ and $S_2$, and similarly for $T_1 \times T_2$.
Then:
:$f_1 \times f_2$ is continuous
:
:both $f_1$ and $f_2$ are continuous.
== Proof ==
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 15$
Category:Cartesian Product
Category:Continuous Mappings
Cartesian Product of Natural Numbers with Itself is Countable

== Theorem ==
The Cartesian product $\N \times \N$ of the set of natural numbers $\N$ with itself is countable.
== Proof ==
This is simply a special case of Cartesian Product of Countable Sets is Countable.
== Sources ==
* : Chapter $1$: Mappings: $\S 15$
* : $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets: Example $2$
* : $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets: Corollary $7.4$
* : $2.5 \ \text{(iv)}$
* : Appendix $\text{A}$: Set Theory: Countable Sets
* : Chapter $1$: General Background: $\S 2$ Countable or uncountable?
Category:Set Theory
Category:Cartesian Product
Category:Countable Sets
Cartesian Product of Preimage with Image of Relation is Correspondence

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Then the restriction of $\RR$ to $\Preimg \RR \times \Img \RR$ is a correspondence.
== Proof ==
By the definition of a correspondence it will be shown that $\RR$ is both left-total and right-total.
$\RR$ is left-total :
:$\forall x \in S: \exists y \in T: x \mathrel \RR y$
By the definition of the pre-image of $\RR$:
:$\Preimg \RR = \set {x \in S: \exists y \in T: x \mathrel \RR y}$
Therefore $\RR$ is left-total.
$\RR$ is right-total :
:$\forall x \in T: \exists y \in S: x \mathrel \RR y$
By the definition of the image of $\RR$:
:$\Img \RR = \set {x \in T: \exists y \in S: x \mathrel \RR y}$
Therefore $\RR$ is right-total.
Hence $\RR$ is a correspondence.
Category:Relation Theory
Cartesian Product of Projections is Projection on Cartesian Product of Mappings

== Theorem ==
Let $I$ be an indexing set.
Let $\family {S_\alpha}_{\alpha \mathop \in I}$ and $\family {T_\alpha}_{\alpha \mathop \in I}$ be families of sets both indexed by $I$.
For each $\alpha \in I$, let $f_\alpha: S_\alpha \to T_\alpha$ be a mapping.
There exists a unique mapping:
:$\ds f: \prod_{\alpha \mathop \in I} S_\alpha \to \prod_{\alpha \mathop \in I} T_\alpha$
such that:
:$\forall \alpha \in I: \pr_\alpha \circ f = f_\alpha \circ \pr_\alpha$
where:
:$\circ$ denotes composition of mappings
:$\pr_\alpha$ denotes the $\alpha$th projection on either $\ds \prod_{\alpha \mathop \in I} S_\alpha$ or $\ds \prod_{\alpha \mathop \in I} T_\alpha$ as appropriate.
== Proof ==
=== Proof of Existence ===
Let $\mathbf x \in \ds \prod_{\alpha \mathop \in I} S_\alpha$ be arbitrary:
:$\mathbf x = \family {x_\alpha \in S_\alpha}_{\alpha \mathop \in I}$
Let $\ds f: \prod_{\alpha \mathop \in I} S_\alpha \to \prod_{\alpha \mathop \in I} T_\alpha$ be defined as:
:$\forall \mathbf x \in \ds \prod_{\alpha \mathop \in I} S_\alpha: \map f {\mathbf x} = \family {\map {f_\alpha} {x_\alpha} }_{\alpha \mathop \in I}$
We have:
Then:
and it is seen that $f$ is such that:
:$\forall \alpha \in I: \pr_\alpha \circ f = f_\alpha \circ \pr_\alpha$
as required.
Hence the existence of $f$ as specified.
=== Proof of Uniqueness ===
Let $f$ be as defined.
Let $\ds g: \prod_{\alpha \mathop \in I} S_\alpha \to \prod_{\alpha \mathop \in I} T_\alpha$ also be a mapping such that:
:$\forall \alpha \in I: \pr_\alpha \circ g = f_\alpha \circ \pr_\alpha$
Let $\mathbf x \in \ds \prod_{\alpha \mathop \in I} S_\alpha$ as before.
Let:
and it is seen that $g = f$.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 10$: Arbitrary Products: Exercise $2$
Category:Projections
Cartesian Product of Semirings of Sets

== Theorem ==
Let $\SS$ and $\TT$ be semirings of sets.
Then $\SS \times \TT$ is also a semiring of sets.
Here, $\times$ denotes Cartesian product.
== Proof ==
Recall the conditions for $\SS \times \TT$ to be a semiring of sets:
:$(1): \quad \O \in \SS \times \TT$
:$(2): \quad \SS \times \TT$ is $\cap$-stable
:$(3'):\quad$ If $A, B \in \SS \times \TT$, then there exists a finite sequence of pairwise disjoint sets $A_1, A_2, \ldots, A_n \in \SS \times \TT$ such that $\ds A \setminus B = \bigcup_{k \mathop = 1}^n A_k$.
=== Proof of $(1)$ ===
From Empty Set is Subset of All Sets:
:$\O \in \SS$
and:
:$\O \in \TT$
So:
:$\O \times \O \in \SS \times \TT$
From Cartesian Product is Empty iff Factor is Empty:
:$\O \times \O = \O$
=== Proof of $(2)$ ===
Let $S_1 \times T_1$ and $S_2 \times T_2$ be in $\SS \times \TT$.
Then from Cartesian Product of Intersections:
:$\paren {S_1 \times T_1} \cap \paren {S_2 \times T_2} = \paren {S_1 \cap S_2} \times \paren {T_1 \cap T_2}$
Since $\SS$ and $\TT$ are $\cap$-stable, $S_1 \cap S_2 \in \SS$ and $T_1 \cap T_2 \in \TT$.
Hence $\paren {S_1 \times T_1} \cap \paren {S_2 \times T_2} \in \SS \times \TT$.
=== Proof of $(3')$ ===
Let $S_1 \times T_1$ and $S_2 \times T_2$ be in $\SS \times \TT$.
Let $\sqcup$ signify union of disjoint sets.
Then:
Now recall that $\SS$ and $\TT$ are semirings of sets.
Thence the expressions $S_1 \setminus S_2$ and $T_1 \setminus T_2$ may be written as finite disjoint unions.
Applying Cartesian Product Distributes over Union again, each of the three $\sqcup$-operands in the expression above may thus be written as a finite disjoint union.
This yields the same fact for $\paren {S_1 \times T_1} \setminus \paren {S_2 \times T_2}$ as well, completing the proof.
== Sources ==
* : $13.1$
Category:Cartesian Product
Category:Semirings of Sets