hcju/proofwikiqa huggingface dataset:
Let $p \in C$. Let $\VV$ be the set of all subsets of $A$ containing $p$ that are clopen relative to $A$.. By Quasicomponents and Components are Equal in Compact Hausdorff Space and Quasicomponent is Intersection of Clopen Sets: :$C$ is the intersection of $\VV$. {{AimForCont}}: :$C \cap \partial A = \O$ By Boundary of Set is Closed, $K \cap \partial A$ is closed for each $K \in \VV$. Thus by Compact Space satisfies Finite Intersection Axiom, there exists a finite set $\VV' \subseteq \VV$ such that $\partial A \cap \bigcap \VV' = \O$. But then: :$\displaystyle K = \bigcap \VV' \in \VV$ {{explain|Where from?}} Therefore there exists a $K \in \VV$ such that $K \cap \partial A = \O$. Since $A$ is closed in $S$, and $K$ is clopen in $A$, $K$ is closed in $S$. {{explain|Where from?}} We have that: :$\partial A = A^- \setminus \map {\operatorname {Int} } A$ where $A^-$ is the closure of $A$ and $\map {\operatorname {Int} } A$ is the interior of $A$ {{explain|Where from?}} Hence as $K \subseteq A$, it follows that $K \subseteq \map {\operatorname {Int} } A$. Since $K$ is open relative to $A$, it is open relative to $\map {\operatorname {Int} } A$. {{explain|"open relative"}} {{explain|Where from?}} We have that $\map {\operatorname {Int} } A$ is open in $S$. {{explain|Where from?}} Therefore $K$ is open in $S$. Thus $K$ is clopen in $S$. We have that $p \in K \subseteq A \subsetneqq S$. {{explain|Where from?}} Therefore $S$ is not connected. {{explain|Where from?}} From this contradiction it follows that: :$C \cap \partial A \ne \O$ {{qed}} Category:Sierpiński's Theorem 93ep903c30ci3yyz9dihxkqlg0fqh5i
Let $p \in C$. Aiming for a contradiction, suppose that $C$ is not connected. Therefore, by definition of connected, there exist disjoint closed sets $A, B$ of $\left({X, \tau}\right)$ such that $C = A \cup B$. By Compact Hausdorff Space is T4, there exist disjoint open sets $U, V$ of $\left({X, \tau}\right)$ such that $U \supseteq A$ and $V \supseteq B$. By Quasicomponent is Intersection of Clopen Sets, $C$ is the intersection of all clopen sets of $\left({X, \tau}\right)$ containing $p$. Since $U$ and $V$ are open, $X \setminus \left({U \cup V}\right)$ is closed. Hence $X \setminus \left({U \cup V}\right)$ is compact. {{explain|Where from?}} Let $S$ be the set of clopen sets of $\left({X, \tau}\right)$ containing $p$. Let $S'$ be the set of complements relative to $\left({X, \tau}\right)$ of elements of $S$. Then $S'$ is an open cover of $X \setminus \left({U \cup V}\right)$. {{explain|Needs explaining as to why - might be worth elaborating the fact that $X \setminus \left({U \cup V}\right) \subseteq X \setminus C$ in order for the above to hold.}} Thus by compactness has a finite subcover $T'$. {{explain|specifically, what has, and why - compactness of what? Above sentence is incomplete.}} Let $T$ be the set of complements of elements of $T'$. {{explain|complements relative to what?}} Then $\displaystyle C \subseteq \bigcap T \subseteq U \cup V$. {{explain|why?}} Furthermore, since $T$ is a finite set of clopen sets of $\left({X, \tau}\right)$]], $\bigcap T$ is clopen. Let $\displaystyle U' = U \cap \bigcap T$ and let $\displaystyle V' = V \cap \bigcap T$. Then $C \subseteq U' \cup V' = T$. {{explain|Where from?}} Since $T$ is clopen, so is $U'$. But $C$ contains points in $U'$ and points not in $U'$, contradicting the fact that $C$ is a quasicomponent of $\left({X, \tau}\right)$. {{explain|What specifically is the nature of the relation between $U'$ and $C$ which invalidates one or more criteria for $C$ to be a quasicomponent?}} It follows that $C$ is connected. {{qed}} Category:Compact Spaces Category:Connectedness Between Two Points ls4qkzrw8xoy4txtrq8dura9fta35pl
{{ProofWanted|Follows from Quasicomponent of Compact Hausdorff Space is Connected}} Category:Compact Spaces Category:Connectedness Between Two Points Category:Connected Spaces 6a14pwguev71d12haxtjgdhsa7tbvp0
Let $x \in S''$. Let $U$ be an open neighborhood of $x$. Then by the definition of derived set, $U$ contains an element $y$ of $S'$ such that $x \ne y$. Then $U \setminus \set x$ is an open neighborhood of $y$ by the definition of a $T_1$ space. The definition of derived set is applied once more to see that $U \setminus \set x$ contains an element $z$ of $S$. Thus $z \in U \cap S$ and $z \ne x$. As this holds for all open neighborhoods $U$ of $x$, it follows that $x \in S'$. This holds for all $x \in S''$. It follows by definition of subset that $S'' \subseteq S'$. Thus $S'$ is closed. {{qed}} Category:Topology 82loeeeh5jjq7gkk61eb3jdg5xzxktr
Let $\struct {S, \circ}$ be the algebraic structure defined as: :$\forall x, y \in S: x \circ y = x$ That is, $\circ$ is the left operation. From Element under Left Operation is Right Identity, every element serves as a right identity. Then given any $a \in S$, we have that $x \circ a = x$ and as $x$ is an identity, axiom $(3)$ is fulfilled as well. But from More than one Right Identity then no Left Identity, there is no left identity and therefore no identity element. Hence $\struct {S, \circ}$ is not a group. {{qed}}
First, suppose $k \perp m$. That is: :$\gcd \set {k, m} = 1$ By Bézout's Identity: :$\exists u, v \in \Z: u k + v m = 1$ Thus: :$\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$ Thus: :$\eqclass u m$ is an inverse of $\eqclass k m$ Suppose that: :$\exists u \in \Z: \eqclass u m \eqclass k m = \eqclass {u k} m = 1$ Then: :$u k \equiv 1 \pmod m$ and: :$\exists v \in \Z: u k + v m = 1$ Thus from Bézout's Identity, $k \perp m$. {{qed}}
By definition, a group is also a semigroup. Thus the result Element Commutes with Square in Semigroup can be applied. {{qed}}
{{begin-eqn}} {{eqn | l = x \circ \left({x \circ x}\right)       | r = \left({x \circ x}\right) \circ x       | c = by definition of semigroup: $\circ$ is associative }} {{end-eqn}} {{qed}} Category:Semigroups mi2ugs4sugszvqgjktx05jh5zremnba
=== Necessary Condition === {{begin-eqn}} {{eqn | l = \mathbf v       | r = \mathbf 0       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = -1_{\mathbb F} \circ \mathbf v       | r = -1_{\mathbb F} \circ \mathbf 0       | c = scaling both sides by the negative of the unity of $\mathbb F$ }} {{eqn | ll= \leadsto       | l = -\mathbf v       | r = \mathbf 0       | c = Vector Inverse is Negative Vector, Zero Vector Scaled is Zero Vector }} {{eqn | ll= \leadsto       | l = \mathbf v       | r = -\mathbf v }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Utilizing the vector space axioms: {{begin-eqn}} {{eqn | l = \mathbf v       | r = -\mathbf v       | c = by hypothesis }} {{eqn | ll =\leadsto       | l = \mathbf v - \mathbf v       | r = -\mathbf v - \mathbf v       | c = adding $-\mathbf v$ to both sides }} {{eqn | ll= \leadsto       | l = \mathbf 0       | r = \left({-1_{\mathbb F} \circ \mathbf v}\right) + \left({-1_{\mathbb F} \circ \mathbf v}\right)       | c = Vector Inverse is Negative Vector }} {{eqn | ll= \leadsto       | l = \mathbf 0       | r = \left({-2 \cdot 1_{\mathbb F} }\right)\circ \mathbf v }} {{end-eqn}} {{qed}} By hypothesis, $\mathbb F$ is infinite. By Characteristic of Division Ring is Zero or Prime: : $\operatorname{Char} \left({\mathbb F}\right) = 0$ so $-2 \cdot 1_{\mathbb F} \ne 0$. Thus from Vector Product is Zero only if Factor is Zero, $\mathbf v = \mathbf 0$. {{qed}}{{proofread}} Category:Vector Algebra ik5njc69b6vtfss13careo9tk67dcgu
=== Definition 1 implies Definition 2 === Let $S = T$ by Definition 1. Then: {{begin-eqn}} {{eqn | l = S       | r = T       | c =  }} {{eqn | ll= \leadsto       | l = \leftparen {x \in S}       | o = \iff       | r = \rightparen {x \in T}       | c = {{Defof|Set Equality}} }} {{eqn | ll= \leadsto       | l = \leftparen {x \in S}       | o = \implies       | r = \rightparen {x \in T}       | c = Biconditional Elimination }} {{eqn | ll= \leadsto       | l = S       | o = \subseteq       | r = T       | c = {{Defof|Subset}} }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = S       | r = T       | c =  }} {{eqn | ll= \leadsto       | l = \leftparen {x \in S}       | o = \iff       | r = \rightparen {x \in T}       | c = {{Defof|Set Equality}} }} {{eqn | ll= \leadsto       | l = \leftparen {x \in T}       | o = \implies       | r = \rightparen {x \in S}       | c = Biconditional Elimination }} {{eqn | ll= \leadsto       | l = T       | o = \subseteq       | r = S       | c = {{Defof|Subset}} }} {{end-eqn}} Thus by the Rule of Conjunction: :$S \subseteq T \land T \subseteq S$ and so $S$ and $T$ are equal by Definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $S = T$ by Definition 2: :$S \subseteq T \land T \subseteq S$ First: {{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = T       | c =  }} {{eqn | ll= \leadsto       | l = \leftparen {x \in S}       | o = \implies       | r = \rightparen {x \in T}       | c = {{Defof|Subset}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = T       | o = \subseteq       | r = S       | c =  }} {{eqn | ll= \leadsto       | l = \leftparen {x \in T}       | o = \implies       | r = \rightparen {x \in S}       | c = {{Defof|Subset}} }} {{end-eqn}} Thus by Biconditional Introduction: :$\forall x: \paren {x \in S \iff x \in T}$ and so $S$ and $T$ are equal by Definition 1. {{qed}}
Let $S \cap T = \O$. Let $S' \subseteq S$ and $T' \subseteq T$. {{AimForCont}} $S' \cap T' \ne \O$. Then: {{begin-eqn}} {{eqn | l = \exists x       | o = \in       | r = S' \cap T'       | c =  }} {{eqn | ll= \leadsto       | l = x        | o = \in       | r = S'       | c = {{Defof|Set Intersection}} }} {{eqn | lo= \land       | l = x        | o = \in       | r = T'       | c =  }} {{eqn | ll= \leadsto       | l = x        | o = \in       | r = S       | c = {{Defof|Subset}} }} {{eqn | lo= \land       | l = x        | o = \in       | r = T       | c =  }} {{eqn | ll= \leadsto       | l = x        | o = \in       | r = S \cap T       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = S \cap T       | o = \ne       | r = \O       | c = {{Defof|Set Intersection}} }} {{end-eqn}} From this contradiction: :$S' \cap T' = \O$ Hence the result by definition of disjoint sets. {{qed}}
=== Definition 1 implies Definition 2 === Let $\RR$ be a relation which fulfils the condition: :$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$ Let $\tuple {x, y} \in \RR$ such that $x \ne y$. {{AimForCont}} that $\tuple {y, x} \in \RR$. Then $\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR$. By hypothesis, this implies that $x = y$. From this contradiction it is concluded that $\tuple {y, x} \notin \RR$. It follows that the condition: :$\tuple {x, y} \in \RR \land x \ne y \implies \tuple {y, x} \notin \RR$ holds for $\RR$. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\RR$ be a relation which fulfils the condition: :$\tuple {x, y} \in \RR \land x \ne y \implies \tuple {y, x} \notin \RR$ Let $\tuple {x, y} \in \RR$ such that $\tuple {y, x} \in \RR$ also. {{AimForCont}} that $x \ne y$. By hypothesis, this implies that $\tuple {y, x} \notin \RR$. From this contradiction it is concluded that $\tuple {y, x} \notin \RR$. It follows that the condition: :$\tuple {x, y} \in \RR \land \tuple {y, x} \in \RR \implies x = y$ holds for $\RR$. {{qed}}
=== Network as Multigraph === Let $N = \struct {V, E, w}$ be a network, either directed or undirected. {{WLOG}}, suppose $N$ is undirected, and in the following argument allow the term edge to include arcs. Let $e \in E$ be an edge of the underlying graph $\struct {V, E}$ of $N$. Let $e = u v$, where $u, v \in V$ are vertices of the underlying graph $\struct {V, E}$ of $N$. Let $n = \map w e$ be the weight of $e$. By hypothesis, $n$ is a strictly positive integer. Let $n$ edges be identified between $u$ and $v$. Repeat this process for all edges of $N$. The resulting object is a multigraph. === Multigraph as Network === Let $G = \struct {V, E}$ be a multigraph, where each element of $E$ may occur multiple times. Let $e = u v$ where $e \in E$ and $u, v \in V$. Let $\map m e$ be the multiplicity of $e$. Then $m$ can be identified with a weight function from $w: E \to \Z_{>0}$. Thus the network $N = \struct {V, E, w}$ can be created. {{qed}}
By definition of complete graph, $K_p$ has $p$ vertices. Also by definition of complete graph, each vertex of $K_p$ is adjacent to all the other $p - 1$ vertices of $K_p$. As $K_p$ is a simple graph, there can be only one edge joining any pair of vertices of $K_p$. So each vertex of $K_p$ has $p - 1$ edges to which it is incident. So, by definition, $K_p$ is $p-1$-regular. {{qed}}
Recall that a perfect graph is one where each vertex is of different degree. We note in passing that the simple graph consisting of one vertex trivially fulfils the condition for perfection. {{AimForCont}} $G$ is a simple graph of order $n$ where $n \ge 2$ such that $G$ is perfect. First, suppose that $G$ has no isolated vertices. By the Pigeonhole Principle, for all vertices to have different degrees, one of them must be of degree at least $n$. That means it must connect to at least $n$ other vertices. But there are only $n - 1$ other vertices to connect to. Therefore $G$ cannot be perfect. Now suppose $G$ has an isolated vertex. There can be only one, otherwise there would be two vertices of degree zero, and so $G$ would not be perfect. Again by the Pigeonhole Principle, for all vertices to have different degrees, one of them must be of degree at least $n - 1$. But of the remaining $n - 1$ vertices, one of them is of degree zero. So it cannot be adjacent to any vertex. So there are only $n - 2$ other vertices to connect to. Therefore $G$ cannot be perfect. {{qed}}
Let $\NN \subseteq \FF$ be a nest. We will show that $\bigcup \NN \in \FF$. Let $F$ be a finite subset of $\bigcup \NN$. By the definitions of subset and of union, each element of $F$ is an element of at least one element of $\NN$. By the Principle of Finite Choice, there is a mapping $c: F \to \NN$ such that: :$\forall x \in F: x \in \map c x$ Then $c \sqbrk F$ is a finite subset of $\NN$. From Finite Totally Ordered Set is Well-Ordered, $f \sqbrk F$ has a greatest element $P \in \NN \subseteq \FF$. Then: :$F$ is a finite subset of $P$ and: :$P \in \FF$ Since $\FF$ has finite character: :$F \in \FF$ We have thus shown that every finite subset of $\bigcup \NN$ is in $\FF$. Since $\FF$ is of finite character: :$\bigcup \NN \in \FF$ Thus by Zorn's Lemma, $\FF$ has a maximal element. {{qed}} {{AoC|Zorn's Lemma}} {{Namedfor|John Wilder Tukey|name2 = Oswald Teichmüller|cat = Tukey|cat2 = Teichmüller}}
Suppose that the set of positive real numbers $\R_{\ge 0}$ has the Continuum Property. Aiming for a contradiction, let $T \subseteq \N$ be a subset of $\N$ which has no smallest element. Then: : $\forall t \in T: \exists u \in T: u < t$ Let $A \in \R_{\ge 0}$. For every $t \in T$, let $a_t = \dfrac A {2^t}$. We have that $t \in \N$, so $t \ge 0$. Therefore $2^t \ge 1$. This $a_t$ can never be greater than $A$. Therefore $\left\{{a_t: t \in T}\right\}$ has an upper bound. As $\R_{\ge 0}$ has the Continuum Property, it follows that $\left\{{a_t: t \in T}\right\}$ has a supremum $B$. We have that $\dfrac {7 B} 8 < B$. Thus $\dfrac {7 B} 8$ is not an upper bound of $\left\{{a_t: t \in T}\right\}$. Therefore there exists $m \in T$ such that: :$a_m > \dfrac {7B} 8$ Then there is $n \in T$ such that $n < m$.  Let $n = m - p$.  Then: {{begin-eqn}} {{eqn | l = a_n       | r = \dfrac A {2^n} }} {{eqn | r = \dfrac A {2^{m - p} } }} {{eqn | r = \dfrac {A \cdot 2^p} {2^m} }} {{eqn | o = \ge       | r = \dfrac {A \cdot 2} {2^m} }} {{eqn | r = 2 \cdot a_m }} {{eqn | o = >       | r = 2 \cdot \dfrac {7B} 8 }} {{eqn | r = \dfrac {7B} 4 }} {{eqn | o = >       | r = B }} {{end-eqn}} Therefore $a_n > B$. This contradicts the supposition that $B$ is the supremum of $\left\{{a_t: t \in T}\right\}$ Thus our initial supposition that $T \subseteq \N$ has no smallest element was false. As $T$ is arbitrary, it follows that each $T \subseteq \N$ has a smallest element. Thus the Well-Ordering Principle holds for $\N$, as was to be proved. {{qed}}
By definition, a circuit contains at least one edge. That is, a single vertex does not comprise a trivial degenerate circuit with no edges. Suppose $C$ contains only one edge $e$. As $G$ is a simple graph it contains no loops. Therefore $e$ starts and ends at two distinct vertices. Therefore $C$, consisting entirely of $e$ can not be a circuit. Suppose $C$ contains exactly $2$ edges $e$ and $f$. Let $e = u v$ where $v$ and $v$ are vertices of $G$. Then for $C$ to be a circuit it must follow that $f = v u$. Thus there are two edges between the same vertices $v$ and $u$. As $G$ is an undirected graph, it must be that $e$ and $f$ are part of a multiple edge between $u$ and $v$. But $G$ is not a multigraph, and so such a circuit can not exist. A circuit containing $3$ edges definitely exists, though. Let $G = \struct {V, E}$ where $V = \set {u, v, w}$ and $E = \set {\set {u, v}, \set {u, w}, \set {v, w} }$. Then $\tuple {u, w, v, u}$ is a $3$-edge circuit. {{qed}}
By definition of cut-vertex, $G - v$ contains at least $2$ components. That it can contain more components than $2$ is best proved by illustration: :500px {{qed}}
Let $G$ be a connected graph and $e = u v$ be a bridge of $G$. By definition of bridge, $G - e$ has to be of at least $2$ components. Aiming for a contradiction, suppose that $G - e$ were of more than $2$ components. Let $G_1, G_2, G_3$ be $3$ of those components such that $u \in G_1$ and $v \in G_2$. Note that $u$ and $v$ cannot both be in the same component or $e$ would not be a bridge. Now let $e$ be replaced so as to connect $G_1$ and $G_2$ again. Then $G_3$ would still be disconnected from the rest of $G$, and $G$ itself would therefore be a disconnected graph. From this contradiction it follows that $G - e$ can only be of two components. {{qed}}
Let the vertices of $G$ all be even. Then by Characteristics of Eulerian Graph, $G$ is Eulerian. By definition of Eulerian, $G$ therefore contains a Eulerian circuit. Thus every edge of $G$ lies on a circuit of $G$. From Condition for Edge to be Bridge, if an edge $e$ of $G$ is a bridge, then it does not lie on a circuit. Hence no edge of $G$ is a bridge. {{qed}}
Proof by demonstration: :400px A Hamiltonian cycle is indicated in <span style="color:blue">blue</span>.
Let $G$ be a simple graph. === Definition 1 implies Definition 2 === Let $G$ have no circuits. As by definition a cycle is a circuit, it follows that $G$ has no cycles. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $G$ have no cycles. Aiming for a contradiction, suppose $G$ contains a circuit $C$. From Circuit contains Cycle as Subgraph it follows that $G$ contains a cycle. From that contradiction it follows that $G$ can have no circuits. {{qed}} Category:Tree Theory eqq1nj4m8pyjgobeuug04c8p4xiajhc
Let $T$ be a tree. {{AimForCont}} there exists a pair of 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.
Let $T$ be such that between any two vertices there is exactly one path. Then for a start $T$ is by definition connected. Suppose $T$ had a circuit, say $\left({u, u_1, u_2, \ldots, u_n, v, u}\right)$. Then there are two paths from $u$ to $v$: : $\left({u, u_1, u_2, \ldots, u_n, v}\right)$ and : $\left({u, v}\right)$. Hence, by Modus Tollendo Tollens, $T$ can have no circuits. That is, by definition, $T$ is a tree. {{qed}}
Suppose $T$ is a tree with $n$ nodes. We need to show that $T$ has $n - 1$ edges. {{questionable|This is actually a Proof by Complete Induction}} Proof by induction: Let $T_n$ be a tree with $n$ nodes. For all $n \in \N_{>0}$, let $\map P n$ be the proposition that a tree with $n$ nodes has $n-1$ edges. ==== Basis for the Induction ==== $\map P 1$ says that a tree with $1$ vertex has no edges. It is clear that $T_1$ is $N_1$, the edgeless graph, which has $1$ node and no edges. So $\map P 1$ is (trivially) true. This is our basis for the induction. ==== Induction Hypothesis ==== Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :Any tree with $k$ nodes has $k - 1$ edges. Then we need to show: :Any tree with $k + 1$ nodes has $k$ edges. ==== Induction Step: Proof 1 ==== {{:Size of Tree is One Less than Order/Necessary Condition/Induction Step/Proof 1}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. {{qed|lemma}} ==== Induction Step: Proof 2 ==== {{:Size of Tree is One Less than Order/Necessary Condition/Induction Step/Proof 2}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Suppose $T$ is a connected simple graph of order $n$ with $n - 1$ edges. We need to show that $T$ is a tree. {{AimForCont}} $T$ is not a tree. Then it contains a circuit. It follows from Condition for Edge to be Bridge that there is at least one edge in $T$ which is not a bridge. So we can remove this edge and obtain a graph $T'$ which is connected and has $n$ nodes and $n - 2$ edges. Let us try and construct a connected graph with $n$ nodes and $n - 2$ edges. We start with the edgeless graph $N_n$, and add edges till the graph is connected. We pick any two vertices of $N_n$, label them $u_1$ and $u_2$ for convenience, and use one edge to connect them, labelling that edge $e_1$. We pick any other vertex, label it $u_3$, and use one edge to connect it to either $u_1$ or $u_2$, labelling that edge $e_2$. We pick any other vertex, label it $u_4$, and use one edge to connect it to either $u_1, u_2$ or $u_3$, labelling that edge $e_3$. We continue in this way, until we pick a vertex, label it $u_{n - 1}$, and use one edge to connect it to either $u_1, u_2, \ldots, u_{n - 2}$, labelling that edge $e_{n - 2}$. That was the last of our edges, and the last vertex still has not been connected. Therefore a graph with $n$ vertices and $n-2$ edges that such a graph ''cannot'' be connected. Therefore we cannot remove any edge from $T$ without leaving it disconnected. Therefore all the edges in $T$ are bridges. Hence $T$ can contain no circuits. Hence, by Proof by Contradiction, $T$ must be a tree. {{qed}}
Let $T_{k + 1}$ be an arbitrary tree with $k + 1$ nodes. Take any node $v$ of $T_{k+1}$ of degree $1$. Such a node exists from Finite Tree has Leaf Nodes. Consider $T_k$, the subgraph of $T_{k + 1}$ created by removing $v$ and the edge connecting it to the rest of the tree. By Connected Subgraph of Tree is Tree, $T_k$ is itself a tree. The order of $T_k$ is $k$, and it has one less edge than $T_{k + 1}$ by definition. By hypothesis, $T_k$ has $k - 1$ edges. So $T_{k + 1}$ must have $k$ edges.
Let $T_{k + 1}$ be any tree with $k + 1$ nodes. Remove any edge $e$ of $T_{k + 1}$. By definition of tree $T_{k + 1}$ has no circuits Therefore from Condition for Edge to be Bridge it follows that $e$ must be a bridge. So removing $e$ disconnects $T_{k + 1}$ into two trees $T_1$ and $T_2$, with $k_1$ and $k_2$ nodes, where $k_1 + k_2 = k + 1$. By hypothesis, $T_1$ and $T_2$ have $k_1 - 1$ and $k_2 - 1$ edges. Putting the edge $e$ back again, it can be seen that $T_{k + 1}$ has $\paren {k_1 - 1} + \paren {k_2 - 1} + 1 = k$ edges. Therefore a tree of order $k+1$  is of size $k$.