nyuuzyou/wikis huggingface dataset:
Characteristics of Eulerian Graph

== Theorem ==
A finite (undirected) graph is Eulerian it is connected and each vertex is even.
Note that the definition of graph here includes:
* Simple graph
* Loop-graph
* Multigraph
* Loop-multigraph
but does not include directed graph.
== Proof of Necessary Condition ==
== Proof of Sufficient Condition ==
== Also see ==
* Characteristics of Traversable Graph
== Sources ==
* : $\S 3.1$: The Königsberg Bridge Problem: An Introduction to Eulerian Graphs: Theorem $3.2$
* : Chapter $\text {A}.21$: Euler ()
* 
* 
* 
Category:Characteristics of Eulerian Graph
Category:Eulerian Graphs
Characteristics of Eulerian Graph/Necessary Condition

== Theorem ==
Let $G$ be a finite (undirected) graph.
Let $G$ be Eulerian.
Then $G$ is connected and each vertex of $G$ is even.
Note that the definition of graph here includes:
* Simple graph
* Loop-graph
* Multigraph
* Loop-multigraph
but does not include directed graph.
== Proof ==
Suppose that $G$ is Eulerian.
Then $G$ contains an Eulerian circuit, that is, a circuit that uses each vertex and passes through each edge exactly once.
Since a circuit must be connected, $G$ is connected.
Beginning at a vertex $v$, follow the Eulerian circuit through $G$.
As the circuit passes through each vertex, it uses two edges: one going to the vertex and another leaving.
Each edge is used exactly once, so each of the vertices must be even.
Since the circuit must also end at $v$, it follows that $v$ is also even.
== Sources ==
* : $\S 3.1$: The Königsberg Bridge Problem: An Introduction to Eulerian Graphs: Theorem $3.2$
Category:Characteristics of Eulerian Graph
Characteristics of Eulerian Graph/Sufficient Condition

== Theorem ==
Let $G$ be a finite (undirected) graph which is connected
Let each vertex of $G$ be even.
Then $G$ is an Eulerian graph.
Note that the definition of graph here includes:
* Simple graph
* Loop-graph
* Multigraph
* Loop-multigraph
but does not include directed graph.
== Proof ==
=== Proof 1 ===
=== Proof 2 ===
Category:Eulerian Graphs
Characteristics of Eulerian Graph/Sufficient Condition/Proof 1

== Theorem ==
== Proof ==
Suppose that an (undirected) graph $G$ is connected and its vertices all have even degree.
If there is more than one vertex in $G$, then each vertex must have degree greater than $0$.
Begin at a vertex $v$.
From Graph with Even Vertices Partitions into Cycles, we know that $v$ will be on at least one cycle.
Since $G$ is connected, there must be an edge $\set {v, v_1}$ for some vertex $v_1 \ne v$.
Since $v_1$ has even degree greater than $0$, there is an edge $\set {v_1, v_2}$ where $v_2 \ne v_1$.
These two edges make a trail from $v$ to $v_2$.
Continue this trail, leaving each vertex on an edge that was not previously used, until returning to $v$.
This is always possible, because $v$ is on a cycle.
Call the circuit formed by this process $C_1$.
If $C_1$ covers all the edges of $G$, then the proof is complete.
Otherwise, remove all the edges that contribute to $C_1$ from $G$, leaving the graph $G_0$.
The remaining vertices are still even, and since $G$ is connected there is some vertex $u$ in both $G_0$ and $C_1$.
Repeat the same process as before, beginning at $u$.
The new circuit, $C_2$, can be added to $C_1$ by starting at $v$, moving along $C_1$ to $u$, travelling around $C_2$ back to $u$ and then along the remainder of $C_1$ back to $v$.
Repeat this process, adding each new circuit found to create a larger circuit.
Since $G$ is finite, this process must end at some point, and the resulting circuit will be an Eulerian circuit.
== Sources ==
* : $\S 3.1$: The Königsberg Bridge Problem: An Introduction to Eulerian Graphs: Theorem $3.2$
Category:Characteristics of Eulerian Graph
Characteristics of Eulerian Graph/Sufficient Condition/Proof 2

== Theorem ==
== Proof ==
Suppose that an (undirected) graph $G$ is connected and its vertices all have even degree.
From Graph with Even Vertices Partitions into Cycles, we can split $G$ into a number of cycles $\mathbb S = C_1, C_2, \ldots, C_k$.
Start at any vertex $v$ on cycle $C_1$ and traverse its edges until we encounter a vertex of another cycle of $\mathbb S$, $C_2$ say.
The edges of $C_2$ are then traversed, and then the traversal of $C_1$ is resumed when it is returned to.
As $C_1$ is traversed, the journey is interrupted so as to traverse any other cycles of $\mathbb S$ in the same way that $C_2$ is traversed.
Eventually the beginning of $C_1$ is reached, which is vertex $v$.
Thus there exists a circuit which includes $C_1$ and at least one other cycle (unless $C_1$ is the only cycle), as $G$ is connected.
If this circuit contains all the cycles $C_1, C_2, \ldots, C_k$, the required Eulerian circuit has been found.
If not, then the circuit just generated is traversed.
Because $G$ is connected, other cycles in $\mathbb S$ will be encountered.
These, again, are traversed as they are encountered.
This process is continued till all the cycles have been included in the circuit.
At this stage, the required Eulerian circuit has been found.
Hence $G$ is Eulerian.
== Sources ==
* : $\S 3.1$: Theorem $3.2$
Category:Characteristics of Eulerian Graph
Characteristics of Finite Tree/Condition 1

== Theorem ==
Let $T$ be a finite simple graph of order $n$.
Then:
:$T$ is a finite tree of order $n$ $T$ has $n - 1$ edges and has no circuits.
== Proof ==
First we recall that by definition of a tree, $T$ is:
:a simple connected graph
:with no circuits.
=== Sufficient Condition ===
Let $T$ be a finite tree.
From Finite Connected Simple Graph is Tree iff Size is One Less than Order, $T$ has $n - 1$ edges.
=== Necessary Condition ===
Let $T$ have $n - 1$ edges and no circuits.
From Simple Graph with no Circuits and Size One Less than Order is Connected $T$ is connected.
Hence $T$ is a tree by definition.
As $T$ is finite , $T$ is a finite tree.
Category:Characteristics of Finite Tree
Characteristics of Finite Tree/Condition 2

== Theorem ==
Let $T$ be a finite simple graph of order $n$.
Then:
:$T$ is a finite tree of order $n$ $T$ has $n - 1$ edges and is connected.
== Proof ==
This is an instance of:
:Finite Connected Simple Graph is Tree iff Size is One Less than Order.
Category:Characteristics of Finite Tree
Characteristics of Finite Tree/Condition 4

== Theorem ==
Let $T$ be a finite simple graph of order $n$.
Then:
:$T$ is a finite tree $T$ has no circuits, but adding one edge creates a cycle.
== Proof ==
=== Sufficient Condition ===
Let $T = \struct {V, E}$ be a finite tree.
Then by definition $T$ is connected and has no circuits.
Let $u, v \in V$ be any two vertices of $T$.
Let $P = \tuple {u, u_1, u_2, \ldots, u_{n - 1}, v}$ be a path from $u$ to $v$.
Let a new edges $\set {u, v}$ be added.
Then $\tuple {u, u_1, u_2, \ldots, u_{n - 1}, v, u}$ is now a cycle, which is by definition also a circuit, in $T$.
Note that this applies even when $P = \tuple {u, v}$: $\tuple {u, v, u}$ is still a cycle in $T$, but now $T$ is a multigraph.
=== Necessary Condition ===
Suppose $T$ has no circuits, but adding one edge creates a cycle, which is by definition also a circuit.
If $T$ were disconnected, then it would be possible to add an edge $e$ to connect two components of $T$.
By definition, $e$ would be a bridge.
From Condition for Edge to be Bridge, it follows that $e$ does not lie on a circuit.
So, if the only way to add an edge to $T$ forms a cycle, it follows that $T$ must be connected.
So $T$ is connected and has no circuits.
Hence by definition $T$ is a tree.
Then as $T$ is a finite , $T$ is a finite tree.
Category:Characteristics of Finite Tree
Characteristics of Floor and Ceiling Function

== Theorem ==
Let $f: \R \to \Z$ be an integer-valued function which satisfies both of the following:
:$(1): \quad \map f {x + 1} = \map f x + 1$
:$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$
Then either:
:$\forall x \in \Q: \map f x = \floor x$
or:
:$\forall x \in \Q: \map f x = \ceiling x$
=== Real Domain ===
== Proof ==
From $(1)$, by induction we have:
:$\forall n \in \N: \map f {x + n} = \map f x + n$
and
:$\forall n \in\N: \map f {x - n} = \map f x - n$
and therefore, in particular:
:$(3): \quad \forall n \in \Z: \map f n = \map f 0 + n$
From $(2)$, we get
Hence
:$\map f 0 = 0$
Thus from $(3)$ it follows that:
:$\forall n \in \Z: \map f n = n$
Suppose that $\map f {\dfrac 1 2} = k \le 0$.
Then:
We show that in this case, by induction:
:$\map f {\dfrac 1 n} = 0$ for all $n \in \N$
Induction hypothesis:
:$\map f {\dfrac 1 {n - 1} } = 0$
Then from $(1)$:
:$\map f {\dfrac n {n - 1} } = \map f {\dfrac 1 {n - 1} + 1} = 0 + 1 = 1$
so:
Finally, we show by induction on $m$ that even:
:$\map f {\dfrac m n} = 0$
for $m \in \set {1, \ldots, n - 1}$.
Above we have shown this for $m = 1$.
Let $1 \le m < n$. 
If $m \divides n$, then
:$\map f {\dfrac m n} = \map f {\dfrac 1 {n / m} } = 0$
Otherwise, write:
:$n = \paren {k - 1} m + r$
Then:
:$k > 1$
and:
:$1 \le r \le m - 1$
We therefore have:
:$k m = n + m - r$
so by the induction hypothesis:
:$\map f {\dfrac {k m} n} = \map f {1 + \dfrac {m - r} n} = 1 + \map f {\dfrac {m - r} n} = 1$
Then:
By $(1)$, this shows that
:$\map f {\dfrac 1 2} \le 0 \implies \map f x = \floor x$
for all rational $x$.
Suppose that $\map f {\dfrac 1 2} = k > 0$.
Then the integer-valued function $g: \R \to \Z$ satisfies:
:$\map g x = -\map f {-x}$
satisfies $(1)$ and $(2)$, and also:
Thus:
:$\map f {\dfrac 1 2} > 0 \implies \map f x = \ceiling x$
for all rational $x$.
== Historical Note ==
== Sources ==
* 
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $49$
Category:Floor Function
Category:Ceiling Function
Characteristics of Floor and Ceiling Function/Historical Note

== Historical Note on Characteristics of Floor and Ceiling Function ==
The result Characteristics of Floor and Ceiling Function was presented by and in $1990$.
The previous work done in $1905$ by to establish that the result does not necessarily hold in the real domain is to be noted.
== Sources ==
* 
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $49$: Answers to Exercises
Category:Historical Notes
Characteristics of Floor and Ceiling Function/Real Domain

== Theorem ==
Let $f: \R \to \Z$ be an integer-valued function which satisfies both of the following:
:$(1): \quad \map f {x + 1} = \map f x + 1$
:$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$
Then it is not necessarily the case that either:
:$\forall x \in \R: \map f x = \floor x$
or:
:$\forall x \in \R: \map f x = \ceiling x$
== Proof ==
Let $h: \R \to \R$ be a real function such that for all $x, y \in \R$:
Consider the integer-valued function $f: \R \to \Z$ defined as:
:$\map f x = \floor {\map h x}$
We claim that $f$ satisfies $(1)$ and $(2)$.
Proof for $(1)$:
We have that:
:$\map h {x + 1} = \map h x + \map h 1 = \map h x + 1$ 
by $(4)$ and $(3)$.
It follows that:
Thus $h$ satisfies $(1)$.
Proof for $(2)$:
Since $h$ satisfies $(4)$, it is an additive function.
By $(3)$ and since Additive Function is Linear for Rational Factors, this implies
:$(5): \quad \map h x = x$
for all $x \in \Q$.
Let $x \in \R$. 
Define $\alpha$ and $\beta$ by
:$(6): \quad \alpha := \floor {\map h x} = \map f x$
:$(7): \quad \beta := \map h x - \alpha$
Then:
and:
Thus $h$ satisfies $(2)$.
We have that:
:Rational Numbers form Subfield of Real Numbers
:Vector Space on Field Extension is Vector Space
Thus we can consider $\R$ as a vector space over $\Q$.
We also have that Square Root of 2 is Irrational
Hence the set $\set {1, \sqrt 2}$ is a linearly independent set in the vector space $\R$.
From Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set:
:there exists a basis $B$ of $\R$ which includes $1$ and $\sqrt 2$.
Then each $x \in \R$ can be written as a finite sum:
:$x := \ds \sum_{i \mathop = 1}^n b_i x_i$
where $b_i \in B$, $x_i \in \Q$ and $n$ depends on $x$.
Let $f$ be defined as:
:$\map f x = \ds \sum_{i \mathop = 1}^n \map f {b_i} x_i$
From Expression of Vector as Linear Combination from Basis is Unique, we have:
:$\map f x + \map f y = \map f {x + y}$
no matter how $\map f b$ is defined for $b \in B$.
Let $f$ be further defined as:
:$\map f 1 = 1$
and, for example:
:$\map f {\sqrt 2} = 4$
Then $f$ satisfies $(1)$ and $(2)$.
But:
:$\map f {\sqrt 2} \notin \set {1, 2}$
== Historical Note ==
== Sources ==
* 
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $49$
Category:Floor Function
Category:Ceiling Function
Characteristics of Minimally Inductive Class under Progressing Mapping

== Theorem ==
Let $M$ be a class which is minimally inductive under a progressing mapping $g$.
Then for all $x, y \in M$:
=== Sandwich Principle ===
=== Image of Proper Subset is Subset ===
=== Mapping Preserves Subsets ===
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.10$
Category:Minimally Inductive Classes
Category:Progressing Mappings
Category:Characteristics of Minimally Inductive Class under Progressing Mapping
Characteristics of Minimally Inductive Class under Progressing Mapping/Image of Proper Subset is Subset

== Theorem ==
:$x \subset y \implies \map g x \subseteq y$
== Proof ==
From Minimally Inductive Class under Progressing Mapping induces Nest, we have that $M$ is a nest in which:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
Thus corollary $1$ of the Sandwich Principle applies directly.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.10 \ (2)$
Category:Characteristics of Minimally Inductive Class under Progressing Mapping
Characteristics of Minimally Inductive Class under Progressing Mapping/Mapping Preserves Subsets

== Theorem ==
:$x \subseteq y \implies \map g x \subseteq \map g y$
== Proof ==
From Minimally Inductive Class under Progressing Mapping induces Nest, we have that $M$ is a nest in which:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
Thus corollary $2$ of the Sandwich Principle applies directly.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.10 \ (3)$
Category:Characteristics of Minimally Inductive Class under Progressing Mapping
Characteristics of Minimally Inductive Class under Progressing Mapping/Sandwich Principle

== Theorem ==
:$x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$
== Proof 1 ==
== Proof 2 ==
Category:Sandwich Principle
Category:Characteristics of Minimally Inductive Class under Progressing Mapping
Category:Characteristics of Minimally Inductive Class under Progressing Mapping/Sandwich Principle
Characteristics of Minimally Inductive Class under Progressing Mapping/Sandwich Principle/Proof 1

== Theorem ==
== Proof ==
From Minimally Inductive Class under Progressing Mapping induces Nest, we have that $M$ is a nest in which:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
Thus the Sandwich Principle applies directly.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.10 \ (1)$
Category:Characteristics of Minimally Inductive Class under Progressing Mapping/Sandwich Principle
Characteristics of Minimally Inductive Class under Progressing Mapping/Sandwich Principle/Proof 2

== Theorem ==
== Proof ==
By definition of minimally inductive class, $M$ is minimally closed under $g$ with respect to $\O$.
The result is then seen to be a direct application of Sandwich Principle for Minimally Closed Class.
Category:Characteristics of Minimally Inductive Class under Progressing Mapping/Sandwich Principle
Characteristics of Regular 4-Dimensional Polytopes

== Theorem ==
The $4$-dimensional regular polytopes have the following characteristics:
Name 	No. of cells	No. of faces	No. of edges	No. of vertices	Dual
Pentatope 	$5$ 	$10$ 	$10$ 	$5$ 	Self-dual
Tesseract 	$8$ 	$24$ 	$32$ 	$16$ 	$16$-cell
$16$-cell 	$16$ 	$32$ 	$24$ 	$8$ 	Tesseract
$24$-cell 	$24$ 	$96$ 	$96$ 	$24$ 	Self-dual
$120$-cell	$120$ 	$720$ 	$1200$ 	$600$ 	$600$-cell
$600$-cell	$600$ 	$1200$ 	$720$ 	$120$ 	$120$-cell
== Proof ==
== Sources ==
* : $6$
* : $6$
Category:Polytopes
Characteristics of Traversable Graph

== Theorem ==
A finite loop-multigraph is traversable it is connected and no more than two vertices are odd.
An Eulerian trail which is not an Eulerian circuit must start and end at an odd vertex.
== Proof ==
Let $G$ be a graph.
Suppose all the vertices are even, that is, there are no odd vertices.
Then $G$ is Eulerian, and the result holds.
Similarly, by the same result, if $G$ is Eulerian, it is by definition traversable.
So the question of graphs all of whose vertices are even is settled.
=== Sufficient Condition ===
Suppose $G$ is a connected graph for which exactly two vertices $u, v$ are odd.
Let $G'$ be the graph formed by adding the edge $u v$.
Note that if there is already an edge $u v$ in $G$, that will mean there is now more than one edge $u v$ in $G'$.
Thus $G'$ will then be a multigraph or loop-multigraph, and $uv$ a multiple edge.
Then $G'$ will have all its vertices even, and thus by the above result be Eulerian and by definition traversable.
Such an Eulerian circuit that traverses $G'$ can be written, for example, $P' = \tuple {v, w, x, \ldots, t, u, v}$.
Let us then remove the final edge $u v$ from $P'$ to get the path $P = \tuple {v, w, x, \ldots, t, u}$.
It follows that the path $P = \tuple {v, w, x, \ldots, t, u}$ is a path in $G$ which includes all edges in $G$.
Hence $P$ traverses $G$ and so $G$ is traversable.
We note that $u$ and $v$ are the odd vertices of $G$.
=== Necessary Condition ===
Suppose a graph $G$ is traversable.
Then it has a Eulerian trail $P$.
If $P$ is a circuit, then $G$ is Eulerian and therefore has all even vertices.
Now, suppose $P = \tuple {v, w, x, \ldots, t, u}$ is not a circuit.
Let $G'$ be the graph formed by adding the edge $uv$.
Then the path $P' = \tuple {v, w, x, \ldots, t, u, v}$ is an Eulerian circuit and so $G$ is Eulerian.
Hence all the vertices of $G'$ are even.
So the degrees of vertices $u$ and $v$ in $G$ (and no other) are odd.
Again, we note that $u$ and $v$ are the odd vertices of $G$.
Hence the result.
== Historical Note ==
== Sources ==
* : $\S 3.1$: The Königsberg Bridge Problem: An Introduction to Eulerian Graphs: Theorem $3.3$
* : Chapter $\text {A}.21$: Euler ()
* 
* 
* 
Category:Characteristics of Traversable Graph
Category:Semi-Eulerian Graphs
Characteristics of Vector in Plane

== Theorem ==
Let a Cartesian plane $\CC$ be established with origin $O$.
Let $\tuple {A_x, A_y}$ be an ordered pair of real numbers that can be used to represent a point in $\CC$.
Then:
:$\tuple {A_x, A_y}$ are the Cartesian coordinates of the terminal point of a position vector $\mathbf A$
:
:$\tuple {A_x, A_y}$ can be transformed into $\tuple { {A'}_x, {A'}_y}$ by rotating $\CC$ about $O$ through an angle $\varphi$ using:
== Proof ==
=== Sufficient Condition ===
Let $\tuple {A_x, A_y}$ represent the terminal point of a position vector $\mathbf A$.
Then from Rotation of Cartesian Axes around Vector:
=== Necessary Condition ===
Let $\tuple {A_x, A_y}$ fulfil the condition that it can be transformed into $\tuple { {A'}_x, {A'}_y}$ by rotating $\CC$ about $O$ through an angle $\varphi$ to $\CC'$ using the given equations.
Let the components $\tuple {A_x, A_y}$ of $\mathbf A$ be functions of the coordinates and perhaps of some other constant vector $\mathbf c$:
In the rotated plane $\CC'$, $\mathbf A$ has components $\tuple { {A'}_x, {A'}_y}$ which are also functions of the same things:
From Rotation of Cartesian Axes around Vector, the values $\tuple {x', y', {c'}_x, {c'}_y}$ can be replaced by $\tuple {x, y, c_x, c_y}$ and the angle of rotation $\varphi$.
In the special case where $\varphi = 0$, we have:
and so on.
It follows that:
Hence ${A'}_x$ is the same function of $\tuple {x', y', {c'}_x, {c'}_y}$ as $A_x$ is of $\tuple {x, y, c_x, c_y}$.
Similarly for ${A'}_y$ and $A_y$.
== Examples ==
== Sources ==
* : Chapter $1$ Vector Analysis $1.2$ Rotation of Coordinates
Category:Characteristics of Vector in Plane
Category:Cartesian Coordinate Systems
Category:Vectors
Characteristics of Vector in Plane/Examples/-y, x

== Examples of Use of Characteristics of Vector in Plane ==
Let a Cartesian plane $\CC$ be established with origin $O$.
The ordered pair $\tuple {-y, x}$ can be interpreted as the components of a position vector.
== Proof ==
We use Characteristics of Vector in Plane.
Let:
by setting $V_x = -y$ and $V_y = x$.
From Characteristics of Vector in Plane:
Using Rotation of Cartesian Axes around Vector it is seen that the equalities hold.
Hence the result.
== Sources ==
* : Chapter $1$ Vector Analysis $1.2$ Rotation of Coordinates: Example $1.2.1$
Category:Characteristics of Vector in Plane
Characteristics of Vector in Plane/Examples/x, -y

== Examples of Use of Characteristics of Vector in Plane ==
Let a Cartesian plane $\CC$ be established with origin $O$.
The ordered pair $\tuple {x, -y}$ cannot be interpreted as the components of a position vector.
== Proof ==
We use Characteristics of Vector in Plane.
Let:
These equations are inconsistent with Characteristics of Vector in Plane.
== Sources ==
* : Chapter $1$ Vector Analysis $1.2$ Rotation of Coordinates: Example $1.2.2$
Category:Characteristics of Vector in Plane
Characterization for Topological Evaluation Mapping to be Embedding

== Theorem ==
Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Then:
:$f$ is an embedding
:$(1)\quad$ the topology on $X$ is the initial topology with respect to $\family {f_i}_{i \mathop \in I}$
:$(2)\quad$ the family $\family {f_i}$ separates points
== Proof ==
=== Necessary Condition ===
Let $f$ be an embedding.
=== Sufficient Condition ===
Let:
:$(1)\quad$ the topology on $X$ be the initial topology with respect to $\family {f_i}_{i \mathop \in I}$
:$(2)\quad$ the family $\family {f_i}$ separate points
== Also see ==
* Characterization for Continuous Mappings Separate Points from Closed Sets
* Evaluation Mapping on T1 Space is Embedding iff Separates Points from Closed Sets
== Sources ==
* : Chapter $3$: New Space from Old: $\S8$: Product Spaces, Weak Topologies: Theorem $8.12$
Category:Evaluation Mappings (Topological Spaces)
Category:Embeddings (Topology)
Category:Characterization for Topological Evaluation Mapping to be Embedding