nyuuzyou/wikis huggingface dataset:
Subsets Satisfy Formulation 5 Matroid Base Axiom Iff Complements Satisfy Formulation 1

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$.
Let $\mathscr B^* = \set{S \setminus B : B \in \mathscr B}$.
Then:
:$\mathscr B$ satisfies formulation $5$ base axiom:
:$\mathscr B^*$ satisfies formulation $1$ base axiom:
== Proof ==
Let $\mathscr B^{**} = \set{S \setminus B^* : B^* \in \mathscr B^*}$.
Then:
:$\mathscr B^{**} = \set{S \setminus \paren{S \setminus B} : B \in \mathscr B}$
From Relative Complement of Relative Complement:
:$\mathscr B^{**} = \mathscr B$
From Subsets Satisfy Formulation 1 Matroid Base Axiom Iff Complements Satisfy Formulation 5:
:$\mathscr B^*$ satisfies formulation $1$ base axiom
:$\mathscr B^{**}$ satisfies formulation $5$ base axiom
That is:
:$\mathscr B^*$ satisfies formulation $1$ base axiom
:$\mathscr B$ satisfies formulation $5$ base axiom
== Also see ==
* Subsets Satisfy Formulation 1 Matroid Base Axiom Iff Complements Satisfy Formulation 5
Category:Matroid Bases
Subsets in Increasing Union

== Theorem ==
Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be a nested sequence of sets, that is:
:$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$
Let $S$ be the increasing union of $S_0, S_1, S_2, \ldots, S_i, \ldots$:
:$\ds S = \bigcup_{i \mathop \in \N} S_i$
Then:
:$\forall s \in S: \exists k \in \N: \forall j \ge k: s \in S_j$
== Proof ==
Let $k \in \N$.
Let $j \ge k$.
Then by as many applications as necessary of Subset Relation is Transitive, we have:
:$S_k \subseteq S_j$
Now $s \in S$ means, by definition of set union, that:
:$\exists S_k \subseteq S: s \in S_k$
Then from above:
:$j \ge k \implies S_k \subseteq S_j$
it follows directly that:
:$\forall s \in S: \exists k \in \N: \forall j \ge k: s \in S_j$
from the definition of subset.
== Sources ==
* : Appendix $\text A$: Sets and Functions: $\text{A}.2$: Boolean Operations
Category:Subsets
Category:Set Union
Subsets of Disjoint Sets are Disjoint

== Theorem ==
Let $S$ and $T$ be disjoint sets.
Let $S' \subseteq S$ and $T' \subseteq T$.
Then $S'$ and $T'$ are disjoint.
== Proof ==
Let $S \cap T = \O$.
Let $S' \subseteq S$ and $T' \subseteq T$.
 $S' \cap T' \ne \O$.
Then:
From this contradiction:
:$S' \cap T' = \O$
Hence the result by definition of disjoint sets.
== Sources ==
* : Appendix $\text{A}.1$: Sets and Subsets: Problem Set $\text{A}.1$: $3$
Category:Subsets
Category:Disjoint Sets
Subsets of Equidecomposable Subsets are Equidecomposable

== Theorem ==
Let $A, B \subseteq \R^n$ be equidecomposable.
Let $S \subseteq A$.
Then there exists $T \subseteq B$ such that $S$ and $T$ are equidecomposable.
== Proof ==
Let $X_1, \dots, X_m$ be a decomposition of $A, B$ together with isometries $\mu_1, \ldots, \mu_m, \nu_1, \ldots, \nu_m: \R^n \to \R^n$ such that:
:$\ds A = \bigcup_{i \mathop = 1}^m \map {\mu_i} {X_i}$ 
and 
:$\ds B = \bigcup_{i \mathop = 1}^m \map {\nu_i} {X_i}$
Define:
:$Y_i = \mu_i^{-1} \paren {S \cap \map {\mu_i} {X_i} }$
Then:
and so $\sequence {Y_i}_{i \mathop = 1}^m$ forms a decomposition of $S$.
But for each $i$:
:$\paren {S \cap \map {\mu_i} {X_i} } \subseteq \map {\mu_i} {X_i}$
and so:
Hence:
:$\map {\nu_i} {Y_i} \subseteq \map {\nu_i} {X_i}$
and so:
Define:
:$\ds \bigcup_{i \mathop = 1}^m \map {\nu_i} {Y_i} = T$
Hence the result.
Category:Equidecomposable Sets
Category:Topology
Subspace Topology is Initial Topology with respect to Inclusion Mapping

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $Y$ be a non-empty subset of $X$.
Let $\iota: Y \to X$ be the inclusion mapping.
Let $\tau_Y$ be the initial topology on $Y$ with respect to $\iota$.
Then $\struct {Y, \tau_Y}$ is a topological subspace of $\struct {X, \tau}$.
That is:
:$\tau_Y = \set {U \cap Y: U \in \tau}$
== Proof ==
By Initial Topology with respect to Mapping equals Set of Preimages, it follows that:
:$\tau_Y = \set {\iota^{-1} \sqbrk U: U \in \tau}$
From Preimage of Subset under Inclusion Mapping, we have:
:$\forall S \subseteq X: \iota^{-1} \sqbrk S = S \cap Y$
Hence the result.
Category:Topological Subspaces
Category:Inclusion Mappings
Category:Initial Topology
Subspace Topology on Initial Topology is Initial Topology on Restrictions

== Theorem ==
Let $X$ be a set.
Let $I$ be an indexing set.
Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces indexed by $I$.
Let $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$.
Let $\tau$ be the initial topology on $X$ with respect to $\family {f_i}_{i \mathop \in I}$.
Let $\struct {Y, \tau_Y}$ be a topological subspace of $X$.
Then $\tau_Y$ is the initial topology on $Y$ with respect to $\family {f_i \restriction_Y}_{i \mathop \in I}$.
== Proof ==
From the definition of the initial topology, $\tau$ has sub-basis:
:$\set {f_i^{-1} \sqbrk U : i \in I, \, U \in \tau_i}$ 
From Sub-Basis for Topological Subspace, $\tau_Y$ has sub-basis:
:$\BB = \set {f_i^{-1} \sqbrk U \cap Y : i \in I, \, U \in \tau_i}$
That is:
:$\BB = \set {\paren {f_i \restriction_Y}^{-1} \sqbrk U : i \in I, \, U \in \tau_i}$
Then $\BB$ is a sub-basis for the initial topology on $Y$ with respect to $\family {f_i \restriction_Y}_{i \mathop \in I}$.
So $\tau_Y$ is the initial topology on $Y$ with respect to $\family {f_i \restriction_Y}_{i \mathop \in I}$.
Category:Initial Topology
Category:Topological Subspaces
Subspace of Complete Metric Space is Closed iff Complete

== Theorem ==
Let $\struct {M, d}$ be a complete metric space.
Let $\struct {S, d}$ be a subspace of $\struct {M, d}$.
Then $S$ is closed $S$ is complete.
== Proof ==
This will be proved by demonstrating the contrapositive:
:$S$ is not complete $S$ is not closed.
=== Necessary Condition ===
Suppose that $S$ is not complete.
Then there exists a Cauchy sequence $\sequence {x_n}$ in $S$ such that the limit $\ds x = \lim_{n \mathop \to \infty} x_n$, which exists in the complete metric space $M$, is not a member of $S$.
For all $\epsilon > 0$, there exists an $N \in \N$ such that for all $n \ge N$:
:$\map d {x, x_n} < \epsilon$
Hence $M \setminus S$ is not open.
Therefore, $S$ is not closed.
=== Sufficient Condition ===
Suppose that $S$ is not closed.
Then $M \setminus S$ is not open.
Therefore, there exists a $x \in M \setminus S$ such that for all $\epsilon > 0$, there exists a $y \in S$ such that $\map d {x, y} < \epsilon$.
So there exists a sequence $\sequence {y_n}$ in $S$ such that for all $n \in \N$:
:$\map d {x, y_n} < \dfrac 1 n$
Now, we show that $\sequence {y_n}$ is a Cauchy sequence.
Let $N \in \N$ be such that for all $n \ge N$:
:$\map d {x, y_n} < \dfrac \epsilon 2$
Let $m, n \ge N$.
Then, by :
:$\map d {y_m, y_n} \le \map d {x, y_m} + \map d {x, y_n} < \epsilon$
Hence $\sequence {y_n}$ is a Cauchy sequence.
Because $\struct {M, d}$ is a complete metric space by assumption, the limit $\ds \lim_{n \mathop \to \infty} y_n$ exists and is in $M$.
Denote this limit by $y$.
By the definition of $\sequence {y_n}$:
:$\ds \lim_{n \mathop \to \infty} \map d {x, y_n} = 0$
From Distance Function of Metric Space is Continuous and Composite of Continuous Mappings is Continuous:
:$\map d {x, y} = 0$
(we are composing the distance function with the map $y\mapsto(x,y)$.)
By definition of a metric, this implies that $x = y$.
Since $y \notin S$, $S$ is not complete.
Category:Metric Subspaces
Category:Complete Metric Spaces
Subspace of Either-Or Space less Zero is not Lindelöf

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Let $H = S \setminus \set 0$ be the set $S$ without zero.
Then the topological subspace $T_H = \struct {H, \tau_H}$ is not a Lindelöf space.
== Proof ==
By definition of topological subspace, $U \subseteq H$ is open in $T_H$ :
:$(1): \quad \set 0 \nsubseteq U$
or:
:$(2): \quad \openint {-1} 1 \subseteq U$
But for all $U \subseteq H$, condition $(1)$ holds as $0 \notin H$.
So $T_H$ is by definition a discrete space.
As $T_H$ is uncountable, we have that Uncountable Discrete Space is not Lindelöf holds.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $2$
Category:Lindelöf Spaces
Category:Either-Or Topology
Subspace of Euclidean Space is Closed

== Theorem ==
Let $\R^n$ be a Euclidean space.
Every subspace of $\R^n$ is closed in $\R^n$.
== Proof ==
First we note that from Euclidean Space is Normed Vector Space, $\R^n$ has a norm $\norm {\, \cdot \,}$.
Let $V$ be a subspace of $\R^n$.
Let $\set {\mathbf v_1, \ldots, \mathbf v_k}$ be a basis for $V$.
Extend this to basis $\set {\mathbf v_1, \ldots, \mathbf v_k, \mathbf v_{k + 1}, \ldots, \mathbf v_n}$ for $\R^n$.
By using Gram-Schmidt procedure, there exists a set of orthonormal vectors $\set {\mathbf u_1, \ldots, \mathbf u_n}$ such that:
:$\forall k \in \N_{> 0} : k \le n : \map \span {\mathbf v_1, \ldots, \mathbf v_k} = \map \span {\mathbf u_1, \ldots, \mathbf u_k}$
where $\span$ stands for the linear span.
Let $A \in \R^{\paren {n - k} \times n}$ be a matrix as follows:
:$A = \begin {bmatrix} {\mathbf u_{k + 1} }^\intercal \\ \vdots \\ {\mathbf u_n}^\intercal \end {bmatrix}$
where $\mathbf u^\intercal$ denotes the transpose of $\mathbf u$ when expressed as a column matrix.
=== $V \subseteq \ker A$ ===
By definition of orthonormality of $\set {\mathbf u_i}$ it follows that:
:$\forall i \in \N_{> 0} : i \le k : A \mathbf u_i = 0$.
Hence, any linear combination of $\mathbf u_1, \ldots, \mathbf u_k$ lies in the kernel of $A$.
That is, $V \subseteq \ker A$.
=== $\ker A \subseteq V$ ===
Suppose $\mathbf x \in \ker A$.
Let $\ds \mathbf x = \sum_{i \mathop = 1}^n \alpha_i \mathbf u_i$, where $\set {\alpha_i}$ are scalars.
By definition of kernel:
:$A \mathbf x = 0$.
Then:
Hence:
:$\forall i \in \N_{\ge k + 1} : i \le n : \alpha_i = 0$
So:
:$\ds \mathbf x = \sum_{i \mathop = 1}^k$
That is:
:$\mathbf x \in V$
Therefore by definition of subset:
:$\ker A \subseteq V$
By definition of set equality:
:$V = \ker A$
By Kernel of Linear Transformation between Finite-Dimensional Normed Vector Spaces is Closed, $V$ is closed.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Euclidean Space
Subspace of Finite Complement Topology is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then every topological subspace of $T$, including $T$ itself, is a compact space.
== Proof ==
Let $T_H = \struct {H, \tau_H}$ be a subspace of $T$.
Let $\CC$ be an open cover of $T_H$.
Let $U \in \CC$ be any set in $C$.
$U$ covers all but a finite number of points of $T_H$.
So for each of those points we pick an element of $\CC$ which covers each of those points.
Hence we have a finite subcover of $T_H$.
So by definition $T_H$ is a compact space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $2$
Category:Compact Spaces
Category:Finite Complement Topology
Subspace of Metric Space is Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$.
Let $d_H: H \times H \to \R$ be the restriction $d \restriction_{H \times H}$ of $d$ to $H$.
Let $\struct {H, d_H}$ be a metric subspace of $\struct {A, d}$.
Then $d_H$ is a metric on $H$.
== Proof ==
By definition of restriction:
:$\forall x, y \in H: \map {d_H} {x, y} = \map d {x, y}$
As $d$ is a metric, the metric space axioms are all fulfilled by all $x, y \in A$ under $d$.
As $H \subseteq A$, by definition of subset, all $x, y \in H$ are also elements of $A$.
Therefore the metric space axioms are all fulfilled by all $x, y \in H$ under $d_H$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces
Category:Metric Subspaces
Subspace of Metrizable Space is Metrizable Space

== Theorem ==
Let $\struct {X, \tau}$ be a metrizable topological space.
Let $\struct {Y, \tau_Y}$ be a subspace of $\struct{X, \tau}$.
Then:
:$\struct {Y, \tau_Y}$ is a metrizable topological space
== Proof ==
By definition of metrizable topological space:
:there exists a metric $d$ on $X$ such that the topology induced by $d$ is $\tau$
From Metric Subspace Induces Subspace Topology:
:$\tau_Y$ is the topology induced by the subspace metric $d_Y$ on $Y$
From Subspace of Metric Space is Metric Space:
:$\struct {Y, d_Y}$ is a metric space
By definition, $\struct {Y, \tau_Y}$ is a metrizable topological space.
Category:Metrizable Topologies
Category:Topological Subspaces
Subspace of Noetherian Topological Space is Noetherian

== Theorem ==
Let $X$ be a Noetherian topological space.
Let $Y \subseteq X$ be a subspace.
Then $Y$ is Noetherian.
== Proof ==
Let $Y_1 \subset Y_2 \subset \ldots \subset$ be an ascending chain of open sets in $Y$.
By definition of subspace topology, there exists open sets $X_1, X_2, \ldots$ such that
:$X_i \cap Y = Y_i$
for all $i$.
By taking the intersection $Z_i := \ds \bigcap_{j \mathop = i}^{\infty} X_j$, we have:
:$Z_i \cap Y = Y_i$
:$Z_1 \subset Z_2 \subset \dots$
Since $X$ is Noetherian, every ascending chain of open sets is eventually constant.
Hence $Z_i$ is eventually constant.
Then $Y_i = Z_i \cap Y$ is eventually constant.
Hence $Y$ is also Noetherian.
== Also see ==
* Space is Noetherian iff all Subsets are Compact
Category:Noetherian Topological Spaces
Subspace of Normed Vector Space with Induced Norm forms Normed Vector Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $Y \subseteq X$ be a vector subspace.
Let $\norm {\, \cdot \,}_Y$ be the induced norm on $Y$.
Then $\struct {Y, \norm {\, \cdot \,}_Y}$ is a normed vector space.
== Proof ==
=== Positive definiteness ===
By definition of induced norm:
:$\forall y \in Y : \norm {y}_Y = \norm {y}_X > 0$
Suppose $y \in Y: \norm y_Y = 0$.
Since $\norm {\, \cdot \,}_Y$ is an induced norm in $\struct {X, \norm {\, \cdot \,}_X}$:
:$\norm y_X = 0$
Therefore:
:$y = \mathbf 0 \in X$
By definition of a vector subspace, $Y$ satisfies :
:$\mathbf 0 \in Y$
Thus:
:$y = \mathbf 0 \in Y \subseteq X$
=== Positive homogeneity ===
Let $y \in Y$.
Let $\alpha \in \R$.
Then we have that:
=== Triangle inequality ===
Let $y_1, y_2 \in Y$.
By : 
:$y_1 + y_2 \in Y$
Furthermore:
== Sources ==
* : $\S 1.2$: Normed and Banach spaces. Normed spaces
Category:Normed Vector Spaces
Subspace of Product Space has Initial Topology with respect to Restricted Projections

== Theorem ==
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\XX$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds \XX := \prod_{i \mathop \in I} X_i$
For each $i \in I$, let $\pr_i: \XX \to X_i$ denote the projection of $\XX$ onto $X_i$.
Let $Y \subseteq \XX$ be a non-empty subset of $\XX$.
For each $i \in I$, let $\pr_i \restriction_Y$ denote the restriction of $\pr_i$ to $Y$.
Let $\tau_\XX$ be the product topology on $\XX$.
Let $\tau_Y$ be the subspace topology on $Y$.
Then:
:$\tau_Y$ is the initial topology on $Y$ with respect to the mappings $\family {\pr_i \restriction_Y : Y \to X_i}_{i \mathop \in I}$
== Proof ==
By definition of product topology:
:$\tau_\XX$ is the initial topology with respect to the mappings $\family {\pr_i : \XX \to X_i}_{i \mathop \in I}$
From Subspace Topology on Initial Topology is Initial Topology on Restrictions:
:$\tau_Y$ is the initial topology on $Y$ with respect to the mappings $\family {\pr_i \restriction_Y : Y \to X_i}_{i \mathop \in I}$
Category:Product Topology
Category:Topological Subspaces
Category:Initial Topology
Subspace of Product Space is Homeomorphic to Factor Space

== Theorem ==
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Suppose that $X$ is non-empty.
Then for each $i \in I$ there is a subspace $Y_i \subseteq X$ which is homeomorphic to $\struct {X_i, \tau_i}$.
Specifically, for any $z \in X$, let:
:$Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$
and let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
Then $\struct {Y_i, \upsilon_i}$ is homeomorphic to $\struct {X_i, \tau_i}$, where the homeomorphism is the restriction of the projection $\pr_i$ to $Y_i$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
Category:Subspace of Product Space is Homeomorphic to Factor Space
Category:Product Spaces
Category:Topological Subspaces
Category:Homeomorphisms (Topological Spaces)
Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton

== Theorem ==
Let $T_1$ and $T_2$ be non-empty topological spaces.
Let $b \in T_2$.
Let $T_1 \times T_2$ be the product space of $T_1$ and $T_2$.
Let $T_2 \times T_1$ be the product space of $T_2$ and $T_1$.
Then:
:$T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$
:$T_1$ is homeomorphic to the subspace $\set b \times T_1$ of $T_2 \times T_1$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 19$
Category:Subspace of Product Space is Homeomorphic to Factor Space
Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Lemma

== Theorem ==
Let $T_1$ and $T_2$ be non-empty sets.
Let $b \in T_2$.
Let $T_1 \times T_2$ be the Cartesian product of $T_1$ and $T_2$.
Let $f: T_1 \to T_1 \times \set b$ be the mapping defined by:
:$\map f x = \tuple {x, b}$
Then:
:$f$ is a bijection.
== Proof ==
=== $f$ is an Injection ===
Let $x, y \in T_1$.
Then $f$ is an injection by definition.
=== $f$ is a Surjection ===
Let $t \in T_1 \times \set b$
Then:
:$\exists x \in T_1: t = \tuple {x, b}$
So:
:$\map f x = \tuple {x, b} = t$
Then $f$ is a surjection by definition.
It follows that $f$ is a bijection by definition.
Category:Subspace of Product Space is Homeomorphic to Factor Space
Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Proof 1

== Theorem ==
== Proof ==
From Finite Cartesian Product of Non-Empty Sets is Non-Empty both $T_1 \times T_2$ and $T_2 \times T_1$ are both non-empty.
The conclusions follow immediately from Subspace of Product Space is Homeomorphic to Factor Space.
Category:Subspace of Product Space is Homeomorphic to Factor Space
Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Proof 2

== Theorem ==
== Proof ==
The conclusions are symmetrical.
, therefore, it will be shown that $T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$.
Let $f: T_1 \to T_1 \times \set b$ be defined as:
:$\map f x = \tuple {x, b}$
=== Lemma ===
$f^{-1}$ is a restriction to the subspace $T_1 \times \set b$ of the projection $\pr_1$ of $T_1 \times T_2$ onto $T_1$.
From Projection from Product Topology is Continuous , $\pr_1$ is a continuous.
From Restriction of Continuous Mapping is Continuous, $f^{-1}$ is a continuous.
It is to be shown that $f$ is continuous.
Let $x \in T_1$.
Let $U$ be an open set in $T_1 \times \set b$ such that:
:$\map f x \in U$
By definition of the subspace topology then for some $U'$ open in $T_1 \times T_2$:
:$U' \cap \paren {T_1 \times \set b} = U$
By the definition of the Natural Basis of Product Topology, there exist open sets $V_1$ and $V_2$ in $T_1$ and $T_2$, respectively, such that:
:$\map f x = \tuple {x, b} \in V_1 \times V_2 \subseteq U'$
Then for any $y \in V_1$:
:$\map f y = \tuple {y, b} \in V_1 \times V_2 \subseteq U'$
But:
:$\tuple {y, b} \in T_1 \times \set b$
so:
:$\tuple {y, b} \in U$
Thus if $y \in V_1$, it follows that $\map f y \in U$.
Then $f$ is continuous by definition.
Category:Subspace of Product Space is Homeomorphic to Factor Space
Subspace of Product Space is Homeomorphic to Factor Space/Proof 1

== Theorem ==
== Proof ==
Let $z \in X$.
Let $i \in i$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$. 
For all $j \in I$ let:
::$Z_j = \begin{cases} X_i & i = j \\
\set{z_j} & j \ne i \end{cases}$
=== Lemma 1 ===
From Product Space of Subspaces is Subspace of Product Space , $\struct {Y_i, \upsilon_i}$ is a product space.
Consider the projection:
:$p_i: \struct {Y_i, \upsilon_i} \to \struct {X_i, \tau_i}$
From Projection from Product of Family is Injection iff Other Factors are Singletons, $p_i$ is injective.
From Projection from Product of Family is Surjective, $p_i$ is surjective.
 
From Projection from Product Topology is Continuous:General Result, $p_i$ is continuous.
From Projection from Product Topology is Open:General Result, $p_i$ is open.
Thus, by definition, we have that $p_i$ is a homeomorphism.
Consider the projection:
:$\pr_i: \struct {X, \tau} \to \struct {X_i, \tau_i}$
and the restriction:
:$\pr_i {\restriction_{Y_i} }: \struct {Y_i, \upsilon_i} \to \struct {X_i, \tau_i}$
=== Lemma 2 ===
Thus $\pr_i {\restriction_{Y_i} }: Y_i \to X_i$ is a homeomorphism.
Category:Subspace of Product Space is Homeomorphic to Factor Space
Subspace of Product Space is Homeomorphic to Factor Space/Proof 1/Lemma 1

== Theorem ==
Let $\family {X_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the Cartesian product of $\family {X_i}_{i \mathop \in I}$.
Let $z \in X$.
Let $i \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$ 
For all for all $j \in I$ let:
::$Z_j = \begin{cases} X_i & i = j \\
\set{z_j} & j \ne i \end{cases}$
Then:
:$Y_i = \prod_{j \mathop \in I} Z_j$
== Proof ==
The result follows by definition of set equality.
Category:Subspace of Product Space is Homeomorphic to Factor Space
Subspace of Product Space is Homeomorphic to Factor Space/Proof 1/Lemma 2

== Theorem ==
Let $\family {X_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the Cartesian product of $\family {X_i}_{i \mathop \in I}$.
Let $z \in X$.
Let $i \in I$.
Let $\pr_i : X \to X_i$ be the $i$th-projection from $X$.
For all for all $j \in I$ let:
:$Z_j = \begin{cases} X_i & i = j \\ \set{z_j} & j \ne i \end{cases}$
Let $Y_i = \prod_{j \mathop \in I} Z_j$
Let $p_i : Y_i \to X_i$ be the $i$th-projection from $Y_i$.
Then:
:$\pr_i {\restriction_{Y_i} } = p_i$
== Proof ==
For all $y \in Y_i$:
By Equality of Mappings:
:$\pr_i {\restriction_{Y_i} } = p_i$
Category:Subspace of Product Space is Homeomorphic to Factor Space
Subspace of Product Space is Homeomorphic to Factor Space/Proof 2

== Theorem ==
== Proof ==
Let $z \in X$.
Let $i \in I$. 
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$. 
Let $p_i = \pr_i {\restriction_{Y_i}}$.
Note that by definitions of a restriction and a projection then:
:$\forall y \in Y_i: \map {p_i} y = y_i$ 
=== $p_i$ is an injection ===
=== $p_i$ is a surjection ===
=== $p_i$ is a continuous mapping ===
=== $p_i$ is an open mapping ===
Category:Subspace of Product Space is Homeomorphic to Factor Space