nyuuzyou/wikis huggingface dataset:
Pasting Lemma/Pair of Continuous Mappings on Open Sets

== Theorem ==
Let $X$ and $Y$ be topological spaces.
Let $A$ and $B$ be open in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be continuous mappings that agree on $A \cap B$.
Let $f \cup g$ be the union of the mappings $f$ and $g$:
:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$
Then the mapping $f \cup g : A \cup B \to Y$ is continuous.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Pasting Lemma for Pair of Continuous Mappings on Closed Sets for an analogous statement for closed sets.
Category:Pasting Lemma
Category:Pasting Lemma for Pair of Continuous Mappings on Open Sets
Pasting Lemma/Pair of Continuous Mappings on Open Sets/Proof 2

== Theorem ==
== Proof ==
From Union of Mappings which Agree is Mapping
:$f \cup g$ is well-defined.
By :
:$f \cup g$ is continuous $\paren {f \cup g}^{-1} \sqbrk U$ is open in $A \cup B$ for every open $U$ in $Y$.
Let $U$ be an arbitrary open subset in $Y$.
From Preimage of Union Mapping is Union of Preimages:
:$\paren {f \cup g}^{-1} \sqbrk U = f^{-1} \sqbrk U \cup g^{-1} \sqbrk U$
By :
:$f^{-1} \sqbrk U$ is open in $A$
:$g^{-1} \sqbrk U$ is open in $B$.
From Open Set in Open Subspace:
:$f^{-1} \sqbrk U$ and $g^{-1} \sqbrk U$ are open in $X$.
By :
:$A \cup B$ is open in $X$
:$f^{-1} \sqbrk U \cup g^{-1} \sqbrk U$ is open in $X$
From Open Set in Open Subspace:
:$f^{-1} \sqbrk U \cup g^{-1} \sqbrk U$ is open in $A \cup B$
Hence:
:$\paren {f \cup g}^{-1} \sqbrk U$ is open in $A \cup B$
Since $U$ was an arbitrary open subset of $Y$:
:for all open subsets $U$ of $Y$, $\paren {f \cup g}^{-1} \sqbrk U$ is open in $A \cup B$
By :
:$f \cup g$ is continuous.
Category:Pasting Lemma for Pair of Continuous Mappings on Open Sets
Pasting Lemma/Union of Open Sets

== Theorem ==
Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be an indexing set.
Let $\family {C_i}_{i \mathop \in I}$ be a family of open sets of $T$.
Let $f: X \to Y$ be a mapping such that the restriction $f \restriction_{C_i}$ is continuous for all $i \in I$.
Then $f$ is continuous on $C = \ds \bigcup_{i \mathop \in I} C_i$, that is, $f \restriction_C$ is continuous.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $V$ be an open set of $S$.
By assumption, we have that, for all $i \in I$, $U_i = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is also open in $T$.
By , we have that $U_i = C_i \cap f^{-1} \sqbrk V$.
Therefore, we can compute:
That is, $U = \paren {f \restriction_C}^{-1} \sqbrk V$ is a union of open sets.
Therefore, $U$ is itself open by definition of topology.
It follows that $f \restriction_C$ is also continuous by definition.
== Also see ==
* Pasting Lemma for Finite Union of Closed Sets for an analogous statement for closed sets.
Category:Pasting Lemma
Category:Open Sets
Path-Connected Set in Subspace

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq B \subseteq S$.
Let $T_B = \struct {B, \tau_B}$ be the topological space where $\tau_B$ is the subspace topology on $B$.
Then
:$A$ is path-connected in $T_B$ $A$ is path-connected in $T$.
== Proof ==
Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$.
Let $\tau'_A$ be the subspace topology on $A$ induced by $\tau_B$.
By the definition of a path-connected set, $A$ is path-connected in $T$ every two points in $A$ are path-connected in $\struct {A, \tau_A}$.
Similarly, $A$ is path-connected in $T_B$ every two points in $A$ are path-connected in $\struct {A, \tau'_A}$.
From Subspace of Subspace is Subspace, we have $\tau_A = \tau'_A$.
Hence every two points in $A$ are path-connected in $\struct {A, \tau_A}$ every two points in $A$ are path-connected in $\struct {A, \tau'_A}$.
Category:Path-Connected Spaces
Path-Connected Space is Connected

== Theorem ==
Let $T$ be a topological space which is path-connected.
Then $T$ is connected.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Closed Topologist's Sine Curve is not Path-Connected for a proof that the converse does not apply.
== Sources ==
* : $3$: Connectedness and Compactness: $\S 24$: Connected Subspaces of the Real Line
* : $6.4$: Comparison of Definitions: Proposition $6.4.1$
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Path-Connected Space is Connected
Path-Connected Space is Connected/Proof 1

== Theorem ==
== Proof ==
Suppose $X \mid Y$ is a separation of $T$.
As $X$ and $Y$ are non-empty, we can find $x \in X$ and $y \in Y$.
As $T$ is path-connected, there exists a path $f : \closedint 0 1 \to T$ with initial point $x$ and final point $y$.
Subset of Real Numbers is Interval iff Connected shows that $\closedint 0 1$ is connected.
Continuous Image of Connected Space is Connected shows that $\Img f$ is connected.
Connected Subspace Lie in One Component of Separation shows that either $\Img f \cap X = \O$, or $\Img f \cap Y = \O$.
As we have $x, y \in \Img f$, this is a contradiction.
It follows that there can be no separation of $T$, so $T$ is connected.
== Sources ==
* : $3$: Connectedness and Compactness: $\S 24$: Connected Subspaces of the Real Line
Category:Path-Connected Space is Connected
Path-Connected Space is Connected/Proof 2

== Theorem ==
== Proof ==
Let $D$ be the discrete space $\set {0, 1}$.
Let $T$ be path-connected.
Let $f: T \to D$ be a continuous surjection.
Let $x, y \in T: \map f x = 0, \map f y = 1$.
Let $I \subset \R$ be the closed real interval $\closedint 0 1$.
Let $g: I \to T$ be a path from $x$ to $y$.
Then by Composite of Continuous Mappings is Continuous it follows that $f \circ g: I \to D$ is a continuous surjection.
This contradicts the connectedness of $I$ as proved in Subset of Real Numbers is Interval iff Connected.
Hence the result.
Category:Path-Connected Space is Connected
Path-Connected Space is not necessarily Locally Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is path-connected.
Then it is not necessarily the case that $T$ is also locally path-connected.
== Proof ==
Let $T$ be the extended topologist's sine curve.
From Extended Topologist's Sine Curve is Path-Connected, $T$ is a path-connected space.
From Extended Topologist's Sine Curve is not Locally Path-Connected, $T$ is not a locally path-connected space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Locally Path-Connected Spaces
Category:Path-Connected Spaces
Path-Connected iff Path-Connected to Point

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $U \subseteq S$ be a non-empty subset of $T$.
Then $U$ is path-connected :
:$\exists p \in U: \forall q \in U: \exists f: \left[{0 \,.\,.\, 1}\right] \to U: f$ continuous, $f \left({0}\right) = p$ and $f \left({1}\right) = q$
== Proof ==
=== Necessary Condition ===
When $U$ is path-connected, any $p \in U$ satisfies the assertion.
Because $U$ is non-empty, such a $p$ exists.
=== Sufficient Condition ===
Suppose $p$ is such a point.
For $q \in U$, denote $p \sim q$ when $p$ and $q$ are path-connected.
To show that $U$ is path-connected, we have to show:
:$\forall q, r \in U: q \sim r$
We already know that $p \sim q$ and $p \sim r$.
From Path-Connectedness is Equivalence Relation, we find that hence $q \sim p$ by symmetry of $\sim$.
Subsequently, we conclude $q \sim r$ from transitivity of $\sim$.
Hence the result.
== Notes ==
One might think that this gives an equivalent definition for path-connectedness. 
However, the definition given incorporates the empty set by vacuous truth, whilst the condition given on this page does not. 
When desired, this may be overcome by handling the empty set separately.
Category:Path-Connected Spaces
Path-Connectedness is Equivalence Relation

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff a$ is path-connected to $b$
where $a, b \in S$.
Then $\sim$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
From Point is Path-Connected to Itself, we have that $a \sim a$.
So $\sim$ is reflexive.
=== Symmetry ===
If $a \sim b$ then $a$ is is path-connected to $b$ by definition.
We form the mapping $g: \closedint 0 1 \to \closedint 0 1$:
:$\map g x = 1 - x$
which is trivially continuous.
By Composite of Continuous Mappings is Continuous $f \circ g$ is continuous.
Putting it together we see that $f \circ g$ maps $0$ to $b$ and $1$ to $a$.
So $b \sim a$ and $\sim$ has been shown to be symmetric.
=== Transitivity ===
Follows directly from Joining Paths makes Another Path.
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Examples of Equivalence Relations
Category:Path-Connected Spaces
Path-Connectedness is Preserved under Homeomorphism

== Theorem ==
Let $\struct {T_1, \tau_1}, \struct {T_2, \tau_2}$ be topological spaces.
Let $\phi: T_1 \to T_2$ be a homeomorphism.
Let $S \subseteq T_1$ be a subset of $T_1$.
Let $S$ be path-connected in $\struct {T_1, \tau_1}$.
Then $\phi \sqbrk S$ is path-connected in $\struct {T_2, \tau_2}$.
That is, path-connectedness is a topological property.
== Proof ==
By definition of homeomorphism, $\phi$ is a continuous mapping.
The result now follows from Continuous Image of Path-Connected Set is Path-Connected.
Category:Topological Properties
Category:Path-Connected Spaces
Path-Homotopic Paths have Path Homotopy

== Theorem ==
Let $X$ be a topological space.
Let $f, g: \closedint 0 1 \to X$ be paths.
Then $f$ and $g$ are path-homotopic, there exists a path homotopy between $f$ and $g$.
== Proof ==
=== Necessary Condition ===
By , there exists a homotopy $H : \closedint 0 1 \times \closedint 0 1 \to X$ such that:
:$\forall s \in \closedint 0 1: \map H {s, 0} = \map f s$
:$\forall s \in \closedint 0 1: \map H {s, 1} = \map g s$
Homotopic Paths have Same Endpoints shows that:
:$\map f 0 = \map g 0$
and:
:$\map f 1 = \map g 1$
As $f$ and $g$ are homotopic relative to $\set{ 0 , 1 }$, we have:
:$\forall s' \in \set{ 0 , 1 } , t \in \closedint 0 1 : \map H { s' , t } = \map f { s' } = \map g { s' }$
By , it follows that $H$ is a .
=== Sufficient Condition ===
We are given that there exists a path homotopy between $f$ and $g$.
Let $H: \closedint 0 1 \times \closedint 0 1 \to X$ be a path homotopy between $f$ and $g$.
By , it follows that $f$ and $g$ are homotopic relative to $\set{ 0 , 1 }$.
== Sources ==
* : $\S 51$
Category:Homotopy Theory
Path Component is not necessarily Arc Component

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $P$ be a path component of $T$.
Then it is not necessarily the case that $P$ is also an arc component of $T$.
== Proof ==
Let $T = \struct {S, \tau_p}$ be a finite particular point space.
From Particular Point Space is Path-Connected, $T$ is path-connected.
Therefore $S$ is a path component in $T$.
But from Particular Point Space is not Arc-Connected, $T$ is not arc-connected.
Therefore $S$ is not an arc component in $T$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Arc Components
Category:Path Components
Path Component of Locally Path-Connected Space is Closed

== Theorem ==
Let $T = \struct {S, \tau}$ be a locally path-connected topological space.
Let $G$ be a path component of $T$.
Then $G$ is open in $T$.
== Proof ==
Let $x \in \partial G$, where $\partial G$ denotes the boundary of $G$.
By , it follows that $S \in \tau$.
As $x \in S$, it follows that $S$ is a neighborhood of $x$.
By definition of locally path-connected space, it follows that there exists a path-connected neighborhood $N$ of $x$ such that $N \subseteq S$.
By definition of boundary, there exists $y \in N$ such that $y \in G$.
As $N$ is path-connected, there exists a path between $x$ and $y$.
By definition of path component , it follows that $x \in G$.
Thus, $\partial G \subseteq G$.
From Set is Closed iff it Contains its Boundary, it follows that $G$ is closed.
== Sources ==
* : $3$: Connectedness and Compactness: $\S 25$: Components and Local Connectedness
Category:Path Components
Category:Locally Path-Connected Spaces
Path Component of Locally Path-Connected Space is Open

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a locally path-connected topological space.
Let $G$ be a path component of $T$.
Then $G$ is open in $T$.
== Proof ==
By definition of locally path-connected, $T$ has a basis of path-connected set.
Thus $S$ is a union of open path-connected sets of $T$.
By Path Components are Open iff Union of Open Path-Connected Sets, the path components of $T$ are open in $T$.
== Also see ==
* Component of Locally Connected Space is Open, an analogous result for connected components
* Equivalence of Definitions of Path Component where it is shown that a space is locally path-connected iff the path components of open subspaces are open.
Category:Path Components
Category:Locally Path-Connected Spaces
Path Components are Open iff Union of Open Path-Connected Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
:$(1): \quad$ The path components of $T$ are open.
:$(2): \quad S$ is a union of open path-connected sets of $T$.
== Proof ==
=== Condition (1) implies Condition (2) ===
=== Condition (2) implies Condition (1) ===
== Also see ==
* Components are Open iff Union of Open Connected Sets, an analogous result for components
Category:Path-Connected Spaces
Category:Path Components
Category:Path Components are Open iff Union of Open Path-Connected Sets
Path Components are Open iff Union of Open Path-Connected Sets/Lemma 1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $C$ be a path component of $T$.
Let $U$ be a path-connected set of $T$.
Then:
:$U \cap C \ne \O$ $U \ne \O$ and $U \subseteq C$
== Proof ==
=== Necessary Condition ===
Let $U \cap C \ne \O$.
From Union of Path-Connected Sets with Common Point is Path-Connected, $U \cup C$ is a path-connected set of $T$.
From Set is Subset of Union, $C \subseteq U \cup C$.
By definition of a path component, $C$ is a maximal path-connected set.
Hence $C = U \cup C$.
From Union with Superset is Superset, $U \subseteq C$.
Then:
The result follows.
=== Sufficient Condition ===
Let $U \ne \O$ and $U \subseteq C$.
Then:
The result follows.
Category:Path Components are Open iff Union of Open Path-Connected Sets
Path Components are Open iff Union of Open Path-Connected Sets/Path Components are Open implies Space is Union of Open Path-Connected Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let the path components of $T$ be open sets.
Then:
:$S$ is a union of open path-connected sets of $T$.
== Proof ==
By definition, the path components of $T$ are a partition of $S$.
Hence $S$ is the union of the open path components of $T$.
Since a path component is a maximal path-connected set by definition, then $S$ is a union of open path-connected sets of $T$
Category:Path Components are Open iff Union of Open Path-Connected Sets
Path Components are Open iff Union of Open Path-Connected Sets/Space is Union of Open Path-Connected Sets implies Path Components are Open

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $S$ be the union of open path-connected sets of $T$.
Then:
:The path components of $T$ are open sets.
== Proof ==
Let $S = \ds \bigcup \set {U \subseteq S : U \in \tau \text { and } U \text { is path-connected} }$.
Let $C$ be a path component of $T$.
=== Lemma ===
Then:
 
Hence $C$ is the union of open sets.
By definition of a topology then $C$ is an open set.
The result follows.
Category:Path Components are Open iff Union of Open Path-Connected Sets
Path Graph from Cycle Graph

== Theorem ==
Removing one edge from a cycle graph leaves a path graph.
== Proof ==
Let $G_n$ be the graph obtained by removing any edge from the cycle graph $C_n$.
As each of those edges lies on a cycle, by Condition for Edge to be Bridge, none of them is a bridge.
So removing any edge from $C_n$ leaves the resulting subgraph of $C_n$ connected.
From Size of Cycle Graph equals Order, the cycle graph $C_n$ has $n$ edges.
So $G_n$ has $n$ vertices and $n - 1$ edges, and is connected.
From Finite Connected Simple Graph is Tree iff Size is One Less than Order, $G_n$ is a tree.
But $G_n$ has two vertices of degree $1$, and all the others are of degree $2$.
It follows that $G_n$ is traversable, and hence by definition is a path graph.
Category:Cycle Graphs
Category:Path Graphs
Path Homotopy is Equivalence Relation

== Theorem ==
Let $X$ be a topological space.
Let $p, q \in X$.
Then path homotopy on the set of all paths in $X$ from $p$ to $q$ is an equivalence relation.
== Proof ==
Suppose $X$ is a topological space.
Let $p, q \in X$.
Let $\sim$ denote the path homotopy on the set of all paths in $X$ from $p$ to $q$.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Thus $\sim$ is seen to be reflexive.
=== Symmetric ===
Thus $\sim$ is seen to be symmetric.
=== Transitive ===
Thus $\sim$ is seen to be transitive.
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence the result.
== Sources ==
* : $\S 7$: Homotopy and the Fundamental Group. Homotopy
Category:Homotopy Theory
Category:Examples of Equivalence Relations
Path as Parameterization of Contour

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $\gamma: \closedint a b \to \C$ be a path.
Let there exist $n \in \N$ and a subdivision $\set {a_0, a_1, \ldots, a_n}$ of $\closedint a b$ such that:
:$\gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } }$ is a smooth path for all $k \in \set {1, \ldots, n}$
where $\gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } }$ denotes the restriction of $\gamma$ to $\closedint {a_{k - 1} } {a_k}$.
Then there exists a contour $C$ with parameterization $\gamma$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Put $\gamma_k = \gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } } : \closedint {a_{k - 1} } {a_k} \to \C$.
By definition, it follows that there exists a directed smooth curve $C_k$ with parameterization $\gamma_k$.
For all $k \in \set {1, \ldots, n - 1}$, we have:
:$\map {\gamma_k} {a_k} = \map {\gamma_{k + 1} } {a_k}$
Define the contour $C$ as the concatenation $C_1 \cup C_2 \cup \ldots \cup C_n$.
Then, it follows by definition that $\gamma$ is a possible parameterization of $C$.
Category:Path as Parameterization of Contour
Category:Contour Integrals
Category:Complex Contour Integrals
Path as Parameterization of Contour/Corollary 1

== Theorem ==
If $\gamma$ is a closed path, then $C$ is a closed contour.
== Proof ==
By definition of closed path, we have
:$\map \gamma a = \map {\gamma_1} {a_0} = \map {\gamma_n} {a_n}$
so:
:$C_1$ has start point $\map \gamma a$
and:
:$C_n$ has end point $\map \gamma a$.
By definition, it follows that $C$ is a closed contour.
Category:Path as Parameterization of Contour
Path as Parameterization of Contour/Corollary 2

== Theorem ==
If $\gamma$ is a Jordan arc, then $C$ is a simple contour, and if $\gamma$ is a Jordan curve, then $C$ is a simple closed contour.
== Proof ==
Let $k_1, k_2 \in \set {1, \ldots, n}$, and $t_1 \in \hointr {a_{k_1 - 1} } {a_{k_1} }, t_2 \in \hointr {a_{k_2 - 1} } {a_{k_2} }$.
Then by the definition of Jordan arc, or Jordan curve:
:$\map \gamma {t_1} \ne \map \gamma {t_2}$
so:
:$\map {\gamma_{k_1} } {t_1} \ne \map {\gamma_{k_2} } {t_2}$
Let instead:
:$k \in \set {1, \ldots, n}$
and:
:$t \in \hointr {a_{k - 1} } {a_k}$ with $t \ne a_1$.
Then by the definition of Jordan arc, or Jordan curve:
:$\map \gamma t \ne \map \gamma {a_n}$
so:
:$\map {\gamma_k} t \ne \map {\gamma_n} {a_n}$
By definition, it follows that $C$ is a simple contour.
A Jordan curve is a closed path by its definition.
Path as Parameterization of Contour/Corollary 1 shows that if $\gamma$ is a Jordan curve, then $C$ will be closed, so $C$ is a simple closed contour.
Category:Path as Parameterization of Contour
Path in Digraph/Examples/Arbitrary Example 1

== Example of Path in Digraph ==
Consider the following digraph:
:
The sequence of vertices $1 \to 2 \to 3 \to 4$ highlighted in is a path from $1$ to $4$.
== Sources ==
* : Chapter $1$: Preliminaries: $1.2$ Graphs and Trees: Directed Graphs
Category:Examples of Paths in Digraphs
Path in Tree is Unique

== Theorem ==
Let $T$ be a graph.
Then $T$ is a tree there is exactly one path between any two distinct vertices.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : $\S 2.3.4.1$: Free Trees: Theorem $\mathrm A$
Category:Path in Tree is Unique
Category:Paths (Graph Theory)
Category:Tree Theory
Path in Tree is Unique/Necessary Condition/Proof 1

== Theorem ==
== Proof ==
Let $T$ be a tree.
 there exists a pair of distinct vertices $u$ and $v$ in $T$ such that there is not exactly one path between them.
If there is no path between $u$ and $v$, $T$ is not connected.
In this case, $T$ is certainly not a tree.
So, in keeping with our supposition, there is more than one path between $u$ and $v$.
Let two of these paths be:
:$P_1 = \tuple {u, u_1, \ldots, u_i, r_1, r_2, \ldots, r_{j - 1}, r_j, u_{i + 1}, \ldots, v}$
:$P_2 = \tuple {u, u_1, \ldots, u_i, s_1, s_2, \ldots, s_{k - 1}, s_k, u_{i + 1}, \ldots, v}$
Now consider the path:
:$P_3 = \tuple {u_i, r_1, r_2, \ldots, r_{j - 1}, r_j, u_{i + 1}, s_k, s_{k - 1}, \ldots, s_2, s_1, u_i}$
It can be seen that $P_3$ is a circuit.
Thus by definition $T$ can not be a tree.
From Proof by Contradiction it follows that there is exactly one path between any pair of vertices.
== Sources ==
* : $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees: Theorem $4.1$
* : $\S 2.3.4.1$: Free Trees: Theorem $\mathrm A$
Category:Path in Tree is Unique
Path in Tree is Unique/Sufficient Condition/Proof 1

== Theorem ==
== Proof ==
Let $T$ be such that between every pair of distinct vertices of $T$ there exists exactly one path.
Then $T$ is connected.
Suppose $T$ had a circuit, say $\tuple {u, u_1, u_2, \ldots, u_n, v, u}$.
Then there are two paths from $u$ to $v$:
:$\tuple {u, u_1, u_2, \ldots, u_n, v}$
and
:$\tuple {u, v}$
Hence, by Modus Tollendo Tollens, $T$ can have no circuits.
That is, by definition, $T$ is a tree.
== Sources ==
* : $\S 2.3.4.1$: Free Trees: Theorem $\mathrm A$
Category:Path in Tree is Unique