Alphabeta Math
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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20 results · all verified · 0 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 20 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Graphs, Walks and Connectivity

1 · Prerequisites

2 · Summary

Finite cardinality supplies graph order, size, degree counts and the enumeration of two-element subsets, while Double counting: xXRx=R=yYRy\sum_{x \in X}\lvert R_x\rvert = \lvert R\rvert = \sum_{y \in Y}\lvert R^y\rvert for a relation between finite sets supplies the incidence identity behind the handshake lemma. The general definition Equivalence relation, equivalence class, and the quotient set A/A/{\sim}, from the required Relations, Functions, and Quotients development, is applied to reachability. The counts A finite set with nn elements has exactly (n2)\binom{n}{2} two-element subsets, and 2(n2)=n(n1)2\binom{n}{2} = n(n-1) and P(A)=2A\lvert\mathcal{P}(A)\rvert = 2^{\lvert A\rvert} for finite AA license the complete-graph and labelled-graph enumerations.

A graph is fixed as finite, simple and undirected before adjacency, subgraphs, walks, paths, cycles and connected components are defined. Removing repetitions turns walks into paths; reachability then partitions the vertices, shortest-path distance satisfies the metric axioms componentwise, and parity characterizes bipartite graphs. Standard graph constructions lead to the handshake lemma and its parity consequence. Vertex and edge cuts yield Whitney's inequalities, while longest-path and extremal edge arguments give cycle and connectivity criteria.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

A finite simple graph is a finite vertex set together with a set of two-element vertex subsets

Definition

A finite simple graph is an ordered pair G=(V,E)G=(V,E) in which VV is a finite set and

E[V]2:={{u,v}V:uv}.E\subseteq [V]^2:=\{\,\{u,v\}\subseteq V: u\ne v\,\}.

The elements of V=V(G)V=V(G) are the vertices and the elements of E=E(G)E=E(G) are the edges. Thus every edge has two distinct endpoints, there are no loops, and an unordered pair of vertices occurs as at most one edge. Finiteness and the notation V|V| and E|E| are those of The cardinality A\lvert A\rvert of a finite set.

When no qualifier is present, the word graph has this meaning.

RemarkRemark: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Unless stated otherwise, graph means finite, simple and undirected; orders, sizes and empty-set conventions are fixed here

For a graph G=(V,E)G=(V,E) (A finite simple graph is a finite vertex set together with a set of two-element vertex subsets), its order is V|V| and its size is E|E| (The cardinality A\lvert A\rvert of a finite set). The null graph has V=V=\varnothing and E=E=\varnothing. An edgeless graph has E=E=\varnothing but may have vertices. Thus the null graph is the unique graph with no vertices, while an edgeless graph need not be null.

All sums indexed by VV or EE use their ordinary empty values. A minimum or maximum taken over the vertex set is used only when VV\ne\varnothing; in particular, minimum and maximum degree are not assigned values for the null graph. Connectivity conventions for the null graph and the one-vertex graph are stated with the definition of connectivity.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Multigraphs, loops and directed graphs as variants distinct from the default finite simple graph

Definition

A finite undirected multigraph consists of a finite vertex set VV, a finite edge set MM, and an endpoint map assigning to each eMe\in M either a singleton {v}\{v\} or a two-element subset {u,v}\{u,v\} of VV. An edge assigned {v}\{v\} is a loop at vv. Distinct members of MM assigned the same endpoint set are parallel edges.

A finite directed graph, or digraph, consists of a finite vertex set VV and a set AV×VA\subseteq V\times V of arcs. An arc (u,v)(u,v) is directed from uu to vv; an arc (v,v)(v,v) is a directed loop.

These objects are variants, not instances of the default graph convention in A finite simple graph is a finite vertex set together with a set of two-element vertex subsets and Unless stated otherwise, graph means finite, simple and undirected; orders, sizes and empty-set conventions are fixed here. A theorem stated merely for a graph therefore neither permits loops or parallel edges nor supplies directions to its edges.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree

Definition

Let G=(V,E)G=(V,E) be a graph (A finite simple graph is a finite vertex set together with a set of two-element vertex subsets).

Distinct vertices u,vu,v are adjacent, or neighbours, when {u,v}E\{u,v\}\in E. A vertex vv and an edge ee are incident when vev\in e. The open neighbourhood and closed neighbourhood of vv are

NG(v):={uV:{u,v}E},NG[v]:=NG(v){v}.N_G(v):=\{\,u\in V:\{u,v\}\in E\,\},\qquad N_G[v]:=N_G(v)\cup\{v\}.

The degree of vv is degG(v):=NG(v)\deg_G(v):=|N_G(v)|, equivalently the number of edges incident with vv. A graph is rr-regular when every vertex has degree rr; it is cubic when it is 33-regular. The multiset of the vertex degrees, usually written in nonincreasing order, is the degree sequence.

When VV\ne\varnothing, the minimum degree and maximum degree are

δ(G):=minvVdegG(v),Δ(G):=maxvVdegG(v).\delta(G):=\min_{v\in V}\deg_G(v),\qquad \Delta(G):=\max_{v\in V}\deg_G(v).

They are defined because VV is a nonempty finite set (The cardinality A\lvert A\rvert of a finite set). Neither δ(G)\delta(G) nor Δ(G)\Delta(G) is defined for the null graph.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Subgraphs, induced subgraphs and spanning subgraphs

Definition

Let G=(V,E)G=(V,E) be a graph. A graph H=(W,F)H=(W,F) is a subgraph of GG when WVW\subseteq V and FE[W]2F\subseteq E\cap[W]^2.

For WVW\subseteq V, the subgraph induced by WW is

G[W]:=(W, E[W]2).G[W]:=(W,\ E\cap[W]^2).

Thus it retains every edge of GG whose two endpoints lie in WW. A subgraph H=(W,F)H=(W,F) is induced when H=G[W]H=G[W], and it is spanning when W=VW=V. These constructions remain finite simple graphs by A finite simple graph is a finite vertex set together with a set of two-element vertex subsets.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges

Definition

Let G=(V,E)G=(V,E) be a graph. A walk of length \ell is a finite vertex list

W=(v0,v1,,v)W=(v_0,v_1,\ldots,v_\ell)

such that {vi1,vi}E\{v_{i-1},v_i\}\in E for every 1i1\le i\le\ell. Its endpoints are v0v_0 and vv_\ell. The length is the number \ell of traversed edges, not the number of displayed vertices. A walk is closed when v0=vv_0=v_\ell.

A trail is a walk in which the traversed edges are distinct. A path is a walk in which the vertices are distinct. In particular, (v)(v) is a path of length 00 from vv to itself.

A cycle is a closed walk (v0,,v1,v=v0)(v_0,\ldots,v_{\ell-1},v_\ell=v_0) of length 3\ell\ge3 in which v0,,v1v_0,\ldots,v_{\ell-1} are distinct. A graph is acyclic when it contains no cycle. Reversing any walk, trail, path, or cycle preserves its type and length.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31Open item page →

Every walk between two vertices contains a path between the same endpoints

Statement

Let W=(v0,,v)W=(v_0,\ldots,v_\ell) be a walk in a graph. By deleting zero or more closed segments from its vertex list, one obtains a path from v0v_0 to vv_\ell of length at most \ell. If WW repeats a vertex, the resulting path can be chosen to have length strictly less than \ell.

Facts & Assumptions

Given: A walk W=(v0,,v)W=(v_0,\ldots,v_\ell) in a finite simple graph.

[F2]

If vi=vjv_i=v_j with i<ji<j, deleting vi+1,,vjv_{i+1},\ldots,v_j leaves the list (v0,,vi,vj+1,,v)(v_0,\ldots,v_i,v_{j+1},\ldots,v_\ell), which is a walk with the same endpoints and length (ji)<\ell-(j-i)<\ell.

[F3]

A finite vertex list has only finitely many lists obtainable by deleting segments, so any nonempty family of their lengths has a least member (The cardinality A\lvert A\rvert of a finite set).

Proof

technique · constructive
1.1

Let W\mathcal W be the nonempty finite family of walks obtained from WW by deleting zero or more closed segments, and choose PWP\in\mathcal W of least length.

givenF2F3chooseconstruct
2.1

The walk PP has no repeated vertex: if two positions carried the same vertex, [F2] would delete the intervening closed segment and produce a shorter member of W\mathcal W, contrary to the choice of PP.

step 1.1F2
3.1

By [F1], PP is a path with the same endpoints as WW, and its length is at most \ell because WWW\in\mathcal W. If WW repeats a vertex, applying [F2] once before minimizing gives a member of W\mathcal W shorter than WW, so PP is strictly shorter than WW.

step 1.1step 2.1F1F2discharge-construct
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Connected graphs and connected components defined by the existence of vertex paths

Definition

Vertices uu and vv of a graph GG are connected, or reachable from one another, when a path in GG has endpoints uu and vv. Equivalently, a walk joins them, because Every walk between two vertices contains a path between the same endpoints turns such a walk into a path.

For vV(G)v\in V(G), its connected component is the induced subgraph on

CG(v):={uV(G):u is reachable from v}.C_G(v):=\{\,u\in V(G):u\text{ is reachable from }v\,\}.

A graph is connected when its vertex set is nonempty and every two vertices are reachable from one another. Thus the null graph is not connected. The one-vertex graph is connected, since its vertex is joined to itself by the path of length 00 from Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Being joined by a walk or path is the same equivalence relation on the vertex set

Statement

On the vertex set of a graph, define uvu\sim v when a walk joins uu to vv. Then \sim is an equivalence relation (Equivalence relation, equivalence class, and the quotient set A/A/{\sim}), and uvu\sim v holds exactly when a path joins uu to vv.

Facts & Assumptions

Given: A graph G=(V,E)G=(V,E) and the relation uvu\sim v defined by the existence of a walk from uu to vv.

[F1]

An equivalence relation is reflexive, symmetric and transitive (Equivalence relation, equivalence class, and the quotient set A/A/{\sim}).

[L1]

Every walk between two vertices contains a path between the same endpoints (Every walk between two vertices contains a path between the same endpoints).

[F2]

A one-vertex list is a path of length 00, reversing a walk gives a walk with reversed endpoints, and concatenating a uu to vv walk with a vv to ww walk gives a uu to ww walk (Connected graphs and connected components defined by the existence of vertex paths, Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

Proof

technique · direct
1.1

For every uVu\in V, the length-00 walk (u)(u) joins uu to itself, so uuu\sim u.

F2
1.2

If uvu\sim v, reversing a witnessing walk gives a walk from vv to uu, so vuv\sim u.

givenF2
1.3

If uvu\sim v and vwv\sim w, concatenating witnessing walks gives a walk from uu to ww, so uwu\sim w.

givenF2
2.1

Steps 1.1, 1.2 and 1.3 prove reflexivity, symmetry and transitivity, hence \sim is an equivalence relation by [F1].

step 1.1step 1.2step 1.3F1
3.1

Every path is a walk, while [L1] turns every witnessing walk into a path with the same endpoints; therefore walk-reachability and path-reachability coincide.

L1F2
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

The connected components of a graph partition its vertex set and are its maximal connected subgraphs

Statement

The vertex sets CG(v)C_G(v) of the connected components of a graph GG are nonempty, cover V(G)V(G), and any two are equal or disjoint. Each induced graph G[CG(v)]G[C_G(v)] is connected and contains every connected subgraph of GG that contains vv. Consequently the connected components are precisely the maximal connected subgraphs of GG.

Facts & Assumptions

Given: A graph G=(V,E)G=(V,E) and the reachability classes CG(v)C_G(v).

[L1]
[F1]

CG(v)C_G(v) is the set of vertices reachable from vv, and the connected component at vv is the induced graph G[CG(v)]G[C_G(v)] (Connected graphs and connected components defined by the existence of vertex paths, Subgraphs, induced subgraphs and spanning subgraphs).

Proof

technique · direct
1.1

Reflexivity in [L1] gives vCG(v)v\in C_G(v), so every class is nonempty and the classes cover VV.

L1F1
1.2

If CG(u)C_G(u) and CG(v)C_G(v) meet at ww, then uu reaches ww and ww reaches vv by symmetry; transitivity gives that uu reaches vv, and another use of symmetry and transitivity shows a vertex reaches uu exactly when it reaches vv. Thus CG(u)=CG(v)C_G(u)=C_G(v), so distinct classes are disjoint.

L1F1
1.3

If x,yCG(v)x,y\in C_G(v), then xx reaches vv and vv reaches yy; concatenating these paths gives a walk from xx to yy, and deleting any repeated segments gives a path all of whose vertices remain reachable from vv. Hence G[CG(v)]G[C_G(v)] is connected.

L1F1
1.4

If a connected subgraph HH contains vv, every vertex xx of HH is joined to vv by a path in HH, hence also in GG, so xCG(v)x\in C_G(v). Thus HH is a subgraph of G[CG(v)]G[C_G(v)].

F1
2.1

Steps 1.1 and 1.2 give the partition assertion, and steps 1.3 and 1.4 say exactly that its induced subgraphs are connected and maximal among connected subgraphs.

step 1.1step 1.2step 1.3step 1.4
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Graph distance within a component, eccentricity, diameter and girth, including the acyclic convention

Definition

Let uu and vv lie in the same connected component of a graph GG. Their distance is

dG(u,v):=min{N:some path from u to v has length }.d_G(u,v):=\min\{\,\ell\in\mathbb N:\text{some path from }u\text{ to }v\text{ has length }\ell\,\}.

The set being minimized is nonempty by connectivity and finite because a path has no repeated vertex and V(G)V(G) is finite (The cardinality A\lvert A\rvert of a finite set). No ordinary natural-number distance is assigned to vertices in different components.

For a vertex vv, its eccentricity is the maximum of dG(v,u)d_G(v,u) over the vertices uu in its component. The diameter of a connected graph is the maximum of dG(u,v)d_G(u,v) over all pairs of its vertices. Thus every component has its own diameter, while a disconnected graph has no single graph diameter under this convention.

The girth g(G)g(G) is the minimum length of a cycle in GG. If GG is acyclic, set g(G)=g(G)=\infty by convention. This symbol records absence of a cycle and is not a distance between vertices.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Shortest-path distance is a metric on every connected component

Statement

Let CC be the vertex set of a connected component of a graph GG. The function dG:C×CNd_G:C\times C\to\mathbb N satisfies, for all u,v,wCu,v,w\in C,

  1. dG(u,v)0d_G(u,v)\ge0, with dG(u,v)=0d_G(u,v)=0 if and only if u=vu=v;
  2. dG(u,v)=dG(v,u)d_G(u,v)=d_G(v,u);
  3. dG(u,w)dG(u,v)+dG(v,w)d_G(u,w)\le d_G(u,v)+d_G(v,w).

These are the metric axioms, so shortest-path distance is a metric on CC.

Facts & Assumptions

Given: Vertices u,v,wu,v,w in one connected component CC of GG.

[F1]

dG(x,y)d_G(x,y) is the least length of an xx to yy path, and the length is a natural number (Graph distance within a component, eccentricity, diameter and girth, including the acyclic convention).

[L1]

Every walk contains a path between the same endpoints whose length is no greater (Every walk between two vertices contains a path between the same endpoints).

[F2]

Reversing a path preserves its length, and concatenating an xx to yy path of length aa with a yy to zz path of length bb gives an xx to zz walk of length a+ba+b.

Proof

technique · direct
1.1

Every path length is a natural number, hence dG(u,v)0d_G(u,v)\ge0. The length-00 path (u)(u) gives dG(u,u)=0d_G(u,u)=0. Conversely a path of length 00 has one vertex, so dG(u,v)=0d_G(u,v)=0 forces u=vu=v.

F1
1.2

Reversing a shortest uu to vv path gives a vv to uu path of the same length, so dG(v,u)dG(u,v)d_G(v,u)\le d_G(u,v); reversing the roles gives the opposite inequality, hence equality.

F1F2
1.3

Concatenate shortest paths from uu to vv and from vv to ww. This is a uu to ww walk of length dG(u,v)+dG(v,w)d_G(u,v)+d_G(v,w), and [L1] supplies a uu to ww path of no greater length. Minimality in [F1] gives dG(u,w)dG(u,v)+dG(v,w)d_G(u,w)\le d_G(u,v)+d_G(v,w).

F1F2L1
2.1

Steps 1.1, 1.2 and 1.3 establish separation, symmetry and the triangle inequality directly on CC, so dGd_G is a metric there.

step 1.1step 1.2step 1.3
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

A bipartite graph and a proper two-colouring of its vertices

Definition

A graph G=(V,E)G=(V,E) is bipartite when there are disjoint subsets A,BVA,B\subseteq V with V=ABV=A\cup B such that every edge has one endpoint in AA and the other in BB. The ordered pair (A,B)(A,B) is a bipartition.

Equivalently, a proper two-colouring is a map c:V{0,1}c:V\to\{0,1\} satisfying c(u)c(v)c(u)\ne c(v) whenever {u,v}E\{u,v\}\in E: take A=c1[{0}]A=c^{-1}[\{0\}] and B=c1[{1}]B=c^{-1}[\{1\}], or define cc from a bipartition. Either part may be empty, so every edgeless graph, including the null graph, is bipartite.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-07-31Open item page →

Every closed walk of odd length contains a cycle of odd length

Statement

Every closed walk of odd length in a graph contains, among its traversed edges, a cycle of odd length.

Facts & Assumptions

Given: A finite simple graph and its closed walks.

[F1]

A simple graph has no loop, so an odd closed walk has length at least 33; a closed walk of length m3m\ge3 whose vertices before the repeated endpoint are distinct is a cycle (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

[L1]

Mathematical induction: a property true at 00 and inherited from mm to m+1m+1 holds for every natural mm (The principle of mathematical induction).

[F2]

If a closed walk (v0,,vr=v0)(v_0,\ldots,v_r=v_0) has vi=vjv_i=v_j with 0i<j<r0\le i<j<r, it splits into closed walks of lengths jij-i and r(ji)r-(j-i); when rr is odd, exactly one of these lengths is odd.

Proof

technique · induction
1.1

Let P(m)P(m) say that every odd closed walk of length at most mm contains an odd cycle. At m=0m=0 there is no odd positive length at most 00, so P(0)P(0) holds vacuously.

base
1.2

Fix mm and assume P(m)P(m). Let W=(v0,,vr=v0)W=(v_0,\ldots,v_r=v_0) be an odd closed walk with rm+1r\le m+1.

ihgiven
2.1

If rmr\le m, the inductive hypothesis supplies an odd cycle in WW.

step 1.2
2.2

Suppose instead that r=m+1r=m+1. If v0,,vr1v_0,\ldots,v_{r-1} are distinct, then WW is itself an odd cycle by [F1].

step 1.2F1
2.3

If those vertices are not distinct, choose 0i<j<r0\le i<j<r with vi=vjv_i=v_j. By [F2], WW splits into two shorter closed walks, and exactly one has odd length. Its length is at most r1=mr-1=m, so P(m)P(m) supplies an odd cycle in that subwalk and hence in WW.

step 1.2F2choose
3.1

Steps 2.1, 2.2 and 2.3 cover every odd closed walk of length at most m+1m+1, so P(m+1)P(m+1) holds.

step 2.1step 2.2step 2.3
4.1

By [L1], P(m)P(m) holds for every natural mm. Applying it at the length of any given odd closed walk yields an odd cycle contained in that walk.

step 1.1step 3.1L1discharge-induction
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-07-31Open item page →

A finite graph is bipartite if and only if it has no odd cycle

Statement

A finite simple graph is bipartite if and only if it contains no cycle of odd length.

Facts & Assumptions

Given: A finite simple graph G=(V,E)G=(V,E).

[F1]

A bipartition is an ordered pair (A,B)(A,B) of disjoint subsets with V=ABV=A\cup B and every edge running between its two parts (A bipartite graph and a proper two-colouring of its vertices).

[L1]

Every odd closed walk contains an odd cycle (Every closed walk of odd length contains a cycle of odd length).

[F2]

Path-reachability is an equivalence relation, and its equivalence classes are the connected components; hence two vertices lie in the same component exactly when a path joins them. Every such path has a length and therefore a parity (Being joined by a walk or path is the same equivalence relation on the vertex set, Connected graphs and connected components defined by the existence of vertex paths, Graph distance within a component, eccentricity, diameter and girth, including the acyclic convention).

[L2]

The connected components are nonempty, cover the vertex set, and are pairwise equal or disjoint (The connected components of a graph partition its vertex set and are its maximal connected subgraphs).

Proof

technique · direct
1.1

Suppose GG has a bipartition (A,B)(A,B). Along any walk the successive vertices alternate between AA and BB, so a closed walk returns to its initial part only after an even number of edges. In particular, every cycle has even length.

F1
1.2

Conversely, suppose GG has no odd cycle. Then by [L1] it has no odd closed walk.

givenL1
2.1

Let CC be an arbitrary connected component. It is nonempty by [L2], so fix a root rCr\in C. Put ACA_C equal to the vertices joined to rr by an even-length path and BCB_C equal to those joined to rr by an odd-length path. Every vertex of CC lies in at least one of these sets by [F2]. This construction is vacuous when the graph has no components.

step 1.2F2L2choose
3.1

The sets ACA_C and BCB_C are disjoint: an even rr to vv path followed by the reverse of an odd rr to vv path would be an odd closed walk, contrary to step 1.2.

step 1.2step 2.1
4.1

Every edge {u,v}\{u,v\} inside CC runs between ACA_C and BCB_C. Otherwise choose paths from rr to uu and from rr to vv having the common parity of their class. Traversing the first path, the edge {u,v}\{u,v\}, and the reverse of the second path gives an odd closed walk, contrary to step 1.2.

step 1.2step 2.1step 3.1
5.1

By [L2], the connected components partition the finite vertex set and therefore form a finite family. Repeating this construction for each component and taking AA and BB to be the unions of their respective parts gives disjoint sets with V=ABV=A\cup B. The endpoints of an edge are joined by that one-edge path and hence lie in one component by [F2], so every edge runs between the two global parts. Thus (A,B)(A,B) is a bipartition.

step 2.1step 3.1step 4.1F1F2L2
6.1

Step 1.1 proves that bipartite graphs have no odd cycle, and steps 1.2 to 5.1 prove the converse.

step 1.1step 5.1
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Empty and complete graphs, complete bipartite graphs, and the convention that PnP_n and CnC_n have nn vertices

Definition

On a finite vertex set VV, the empty graph has edge set \varnothing and the complete graph KVK_V has edge set [V]2[V]^2. When VV is an nn-element labelled set, these are also denoted Kn\overline K_n and KnK_n.

For disjoint finite sets AA and BB, the complete bipartite graph KA,BK_{A,B} has vertex set ABA\cup B and edge set

{{a,b}:aA, bB}.\{\,\{a,b\}:a\in A,\ b\in B\,\}.

If A=m|A|=m and B=n|B|=n, it is denoted Km,nK_{m,n}.

The path graph PnP_n has nn vertices. For n1n\ge1, take vertices 0,,n10,\ldots,n-1 and edges {i,i+1}\{i,i+1\} for 0i<n10\le i<n-1; it has length n1n-1. Set P0P_0 equal to the null graph. The cycle graph CnC_n is defined only for n3n\ge3: it has vertices 0,,n10,\ldots,n-1, the consecutive edges {i,i+1}\{i,i+1\} for 0i<n10\le i<n-1, and the closing edge {n1,0}\{n-1,0\}. Thus in this notation the subscript records the number of vertices, not the length of a path.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

The Petersen graph on the two-element subsets of a five-element set, adjacent when disjoint

Definition

Fix a five-element set Ω\Omega. The Petersen graph has vertex set [Ω]2[\Omega]^2, the set of two-element subsets of Ω\Omega, and edge set

{{A,B}[Ω]2:{A,B}=2 and AB=}.\{\,\{A,B\}\subseteq [\Omega]^2:|\{A,B\}|=2\text{ and }A\cap B=\varnothing\,\}.

In words, two vertices are adjacent exactly when the corresponding two-element subsets are disjoint. This is a finite simple graph, and its vertex set has (52)=10\binom52=10 elements by A finite set with nn elements has exactly (n2)\binom{n}{2} two-element subsets, and 2(n2)=n(n1)2\binom{n}{2} = n(n-1). The definition is independent of the names assigned to the five elements, because a relabelling preserves disjointness.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Graph isomorphisms, automorphisms and graph complements

Definition

Let G=(V,E)G=(V,E) and H=(W,F)H=(W,F) be graphs. A graph isomorphism is a bijection φ:VW\varphi:V\to W (Injection, surjection, bijection) such that, for all distinct u,vVu,v\in V,

{u,v}E{φ(u),φ(v)}F.\{u,v\}\in E\quad\Longleftrightarrow\quad\{\varphi(u),\varphi(v)\}\in F.

Graphs are isomorphic, written GHG\cong H, when such a map exists. An automorphism of GG is an isomorphism from GG to itself.

The complement of GG is the graph

G:=(V,[V]2E).\overline G:=(V,[V]^2\setminus E).

Thus exactly one of GG and G\overline G contains any given pair of distinct vertices as an edge, and G=G\overline{\overline G}=G.

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Vertex and edge deletion, edge contraction, graph minors, subdivisions and topological minors

Definition

Let G=(V,E)G=(V,E) be a graph. For vVv\in V, vertex deletion gives Gv:=G[V{v}]G-v:=G[V\setminus\{v\}]. For eEe\in E, edge deletion gives Ge:=(V,E{e})G-e:=(V,E\setminus\{e\}). The same notation is used for deleting a set of vertices or edges.

For an edge e={u,v}e=\{u,v\}, contracting ee identifies uu and vv. More precisely, take a new vertex xex_e, put V=(V{u,v}){xe}V'=(V\setminus\{u,v\})\cup\{x_e\}, and let q:VVq:V\to V' send u,vu,v to xex_e and fix every other vertex. Then

G/e:=(V, {q[f]:fE{e}, q[f]=2}).G/e:=\bigl(V',\ \{\,q[f]:f\in E\setminus\{e\},\ |q[f]|=2\,\}\bigr).

The condition q[f]=2|q[f]|=2 deletes loops, while writing the edge family as a set merges parallel edges. Thus contraction returns a simple graph.

A graph HH is a minor of GG when it can be obtained by vertex deletions, edge deletions and edge contractions. Subdividing an edge {u,v}\{u,v\} replaces it by {u,x}\{u,x\} and {x,v}\{x,v\} for a new vertex xx. A subdivision repeats this operation zero or more times. The graph HH is a topological minor of GG when some subdivision of HH is isomorphic to a subgraph of GG (Subgraphs, induced subgraphs and spanning subgraphs).

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

The line graph whose vertices are the edges of the original graph and whose adjacency records a shared endpoint

Definition

For a graph G=(V,E)G=(V,E), its line graph L(G)L(G) has vertex set EE. Distinct vertices e,fEe,f\in E are adjacent in L(G)L(G) exactly when the edges ee and ff of GG share an endpoint, that is, when efe\cap f\ne\varnothing.

Because EE is finite and adjacency is recorded as a set of two-element subsets of EE, the line graph is again a finite simple graph (A finite simple graph is a finite vertex set together with a set of two-element vertex subsets). Incidence in GG has become adjacency in L(G)L(G).

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Handshake lemma: the sum of the vertex degrees is twice the number of edges

Statement

For every finite simple graph G=(V,E)G=(V,E),

vVdegG(v)=2E\sum_{v\in V}\deg_G(v)=2|E|

as an identity of natural numbers. It remains valid for the null graph, when both sides are 00.

Facts & Assumptions

Given: A finite simple graph G=(V,E)G=(V,E).

[L1]

For a relation RX×YR\subseteq X\times Y between finite sets, the sum of the row-fibre cardinalities and the sum of the column-fibre cardinalities both equal R|R| (Double counting: xXRx=R=yYRy\sum_{x \in X}\lvert R_x\rvert = \lvert R\rvert = \sum_{y \in Y}\lvert R^y\rvert for a relation between finite sets).

[F1]

The row and column fibres of a finite incidence relation are finite and are defined by fixing the first or second coordinate (A relation RX×YR \subseteq X \times Y between finite sets, its row fibres RxR_x and its column fibres RyR^y).

[F2]

degG(v)\deg_G(v) is the number of edges incident with vv (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

Proof

technique · direct
1.1

Define the incidence relation R:={(v,e)V×E:ve}R:=\{\,(v,e)\in V\times E:v\in e\,\}. Its row fibre at vv is the set of edges incident with vv, so it has cardinality degG(v)\deg_G(v) by [F2].

givenF1F2
1.2

Every edge eEe\in E is a two-element subset of VV, so its column fibre has cardinality 22.

givenF1
2.1

Applying [L1] to RR and using steps 1.1 and 1.2 gives vVdegG(v)=R=eE2=2E\sum_{v\in V}\deg_G(v)=|R|=\sum_{e\in E}2=2|E|.

step 1.1step 1.2L1algebra
CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Every finite graph has an even number of vertices of odd degree

Statement

In every finite simple graph, the set of vertices having odd degree has even cardinality.

Facts & Assumptions

Given: A finite simple graph G=(V,E)G=(V,E) and the finite set O:={vV:degG(v) is odd}O:=\{v\in V:\deg_G(v)\text{ is odd}\}.

[L1]

The handshake lemma gives vVdegG(v)=2E\sum_{v\in V}\deg_G(v)=2|E|, an even natural number (Handshake lemma: the sum of the vertex degrees is twice the number of edges).

[F1]

The cardinality O|O| is a natural number because OO is a subset of the finite set VV (The cardinality A\lvert A\rvert of a finite set).

Proof

technique · direct
1.1

Split the degree sum into the vertices in OO and those outside OO. The latter summands are even, so their sum is even.

given
1.2

A finite sum of kk odd natural numbers has the same parity as kk: writing each summand as 2qi+12q_i+1 makes the sum 2iqi+k2\sum_iq_i+k. Hence vOdegG(v)\sum_{v\in O}\deg_G(v) has the same parity as O|O|.

F1algebra
2.1

The total degree sum is even by [L1], and subtracting the even contribution from VOV\setminus O leaves an even sum over OO. Step 1.2 therefore makes O|O| even.

step 1.1step 1.2L1
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

The complete graph on an nn-element vertex set has (n2)\binom{n}{2} edges

Statement

If VV is an nn-element set, then the complete graph KVK_V has exactly (n2)\binom n2 edges.

Facts & Assumptions

Given: A finite set VV with V=n|V|=n.

[F1]

The edge set of KVK_V is [V]2[V]^2, the set of all two-element subsets of VV (Empty and complete graphs, complete bipartite graphs, and the convention that PnP_n and CnC_n have nn vertices).

Proof

technique · direct
1.1

By [F1], E(KV)=[V]2E(K_V)=[V]^2.

F1
2.1

By [L1], [V]2=(n2)|[V]^2|=\binom n2, so step 1.1 gives E(KV)=(n2)|E(K_V)|=\binom n2.

step 1.1L1
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

There are 2(n2)2^{\binom{n}{2}} simple graphs on a fixed labelled nn-element vertex set

Statement

On a fixed labelled nn-element vertex set VV, there are exactly 2(n2)2^{\binom n2} finite simple graphs.

Facts & Assumptions

Given: A fixed finite set VV with V=n|V|=n.

[F1]

A simple graph on VV is uniquely specified by choosing an edge set E[V]2E\subseteq[V]^2 (A finite simple graph is a finite vertex set together with a set of two-element vertex subsets).

[L2]

A finite set with mm elements has a power set with 2m2^m elements (P(A)=2A\lvert\mathcal{P}(A)\rvert = 2^{\lvert A\rvert} for finite AA).

Proof

technique · direct
1.1

Sending a graph (V,E)(V,E) to its edge set is a bijection from the simple graphs on the fixed labelled set VV to the power set P([V]2)\mathcal P([V]^2), by [F1].

F1
2.1

By [L1] and [L2], P([V]2)=2[V]2=2(n2)|\mathcal P([V]^2)|=2^{|[V]^2|}=2^{\binom n2}. Combining with step 1.1 gives the stated count.

step 1.1L1L2
DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-07-31Open item page →

Vertex cuts, edge cuts, vertex connectivity κ(G)\kappa(G) and edge connectivity λ(G)\lambda(G), with conventions for complete and one-vertex graphs

Definition

Let G=(V,E)G=(V,E) be connected with n=V2n=|V|\ge2. A vertex cut is a set SVS\subseteq V for which GSG-S is disconnected or has at most one vertex. An edge cut is a set FEF\subseteq E for which GFG-F is disconnected. The vertex connectivity and edge connectivity are

κ(G):=min{S:S is a vertex cut},λ(G):=min{F:F is an edge cut}.\kappa(G):=\min\{\,|S|:S\text{ is a vertex cut}\,\},\qquad\lambda(G):=\min\{\,|F|:F\text{ is an edge cut}\,\}.

The finite families being minimized are nonempty: deleting all but one vertex is a vertex cut, and deleting all edges is an edge cut. For the complete graph KnK_n, no deletion leaving at least two vertices disconnects it, so the first definition gives the standard exceptional convention κ(Kn)=n1\kappa(K_n)=n-1.

For the one-vertex graph K1K_1, set κ(K1)=λ(K1)=0\kappa(K_1)=\lambda(K_1)=0. The null graph and every disconnected graph are outside the domain of these two connectivity parameters under this convention. Thus the inequalities involving δ(G)\delta(G) are stated only for nontrivial connected graphs, where the vertex set is nonempty and δ(G)\delta(G) is defined.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

For every nontrivial connected graph, λ(G)δ(G)\lambda(G)\le\delta(G)

Statement

For every connected finite simple graph GG with at least two vertices, λ(G)δ(G)\lambda(G)\le\delta(G).

Facts & Assumptions

Given: A connected graph G=(V,E)G=(V,E) with V2|V|\ge2.

[F2]

A vertex vv of minimum degree has exactly δ(G)\delta(G) incident edges (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

Proof

technique · direct
1.1

Choose a vertex vv with degG(v)=δ(G)\deg_G(v)=\delta(G) and let FF be the set of all edges incident with vv. Then F=δ(G)|F|=\delta(G).

givenF2choose
2.1

In GFG-F, the vertex vv is isolated while at least one other vertex remains, so GFG-F is disconnected. Thus FF is an edge cut.

step 1.1F1
3.1

Minimality in [F1] gives λ(G)F=δ(G)\lambda(G)\le|F|=\delta(G).

step 1.1step 2.1F1
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)Open item page →

For every nontrivial connected graph, κ(G)λ(G)\kappa(G)\le\lambda(G)

Statement

For every connected finite simple graph GG with at least two vertices, κ(G)λ(G)\kappa(G)\le\lambda(G).

Facts & Assumptions

Given: A connected graph G=(V,E)G=(V,E) with n=V2n=|V|\ge2.

[F1]

κ(G)\kappa(G) is the least size of a vertex set whose deletion disconnects GG or leaves at most one vertex, and λ(G)\lambda(G) is the least size of an edge cut (Vertex cuts, edge cuts, vertex connectivity κ(G)\kappa(G) and edge connectivity λ(G)\lambda(G), with conventions for complete and one-vertex graphs).

Proof

technique · constructive
1.1

Choose an edge cut FF of size λ(G)\lambda(G). Let AA be the vertex set of one component of GFG-F and put B=VAB=V\setminus A. Then AA and BB are nonempty, and every edge of GG between AA and BB belongs to FF.

givenF1F2chooseconstruct
2.1

If Fn1|F|\ge n-1, deleting all but one vertex gives a vertex cut of size n1Fn-1\le|F|, so κ(G)λ(G)\kappa(G)\le\lambda(G).

step 1.1F1
2.2

Suppose F<n1|F|<n-1. The bipartite set of all pairs with one endpoint in AA and one in BB has ABA+B1=n1|A||B|\ge|A|+|B|-1=n-1 members. Hence not every such pair is an edge of GG, for all cross edges lie in FF. Choose nonadjacent aAa\in A and bBb\in B.

step 1.1algebrachoose
3.1

For each edge eFe\in F, choose one endpoint s(e)s(e) different from both aa and bb; this is possible because e{a,b}e\ne\{a,b\}. Put S={s(e):eF}S=\{s(e):e\in F\}. Then a,bSa,b\notin S, every edge of FF meets SS, and SF|S|\le|F|.

step 2.2construct
4.1

Any path from aAa\in A to bBb\in B must use an edge between AA and BB, hence an edge of FF by step 1.1, and therefore a vertex of SS by step 3.1. Thus GSG-S has no aa to bb path and is disconnected, so SS is a vertex cut.

step 1.1step 3.1F1F2
5.1

In the case F<n1|F|<n-1, step 4.1 gives κ(G)SF=λ(G)\kappa(G)\le|S|\le|F|=\lambda(G); step 2.1 gives the same inequality in the remaining case.

step 2.1step 3.1step 4.1F1discharge-construct
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Whitney's inequalities: κ(G)λ(G)δ(G)\kappa(G)\le\lambda(G)\le\delta(G) for every nontrivial connected graph

Statement

For every connected finite simple graph GG with at least two vertices,

κ(G)λ(G)δ(G).\kappa(G)\le\lambda(G)\le\delta(G).

This includes complete graphs under the convention κ(Kn)=n1\kappa(K_n)=n-1: for KnK_n with n2n\ge2, all three quantities equal n1n-1.

Facts & Assumptions

Given: A connected finite simple graph GG with at least two vertices.

[L1]

κ(G)λ(G)\kappa(G)\le\lambda(G) for every nontrivial connected graph (For every nontrivial connected graph, κ(G)λ(G)\kappa(G)\le\lambda(G)).

[L2]

λ(G)δ(G)\lambda(G)\le\delta(G) for every nontrivial connected graph (For every nontrivial connected graph, λ(G)δ(G)\lambda(G)\le\delta(G)).

Proof

technique · direct
1.1

Applying [L1] and [L2] to GG gives κ(G)λ(G)δ(G)\kappa(G)\le\lambda(G)\le\delta(G).

L1L2
2.1

For KnK_n, deleting fewer than n1n-1 vertices leaves a nonempty complete graph and deleting n1n-1 leaves one vertex, so κ(Kn)=n1\kappa(K_n)=n-1; every vertex has degree n1n-1, and deleting all n1n-1 edges incident with one vertex is an edge cut, while any smaller edge deletion leaves every pair joined through a remaining direct edge or a two-edge path. Hence λ(Kn)=δ(Kn)=n1\lambda(K_n)=\delta(K_n)=n-1, as stated.

L1L2algebra
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Every nonempty finite simple graph contains a path with at least δ(G)\delta(G) edges

Statement

Every nonempty finite simple graph GG contains a path having at least δ(G)\delta(G) edges.

Facts & Assumptions

Given: A nonempty finite simple graph G=(V,E)G=(V,E).

[F2]

A path has no repeated vertex, and appending a neighbour not already on it produces a longer path (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

[F3]

A finite graph has only finitely many paths, so a path of greatest length exists (The cardinality A\lvert A\rvert of a finite set).

Proof

technique · constructive
1.1

Choose a path P=(v0,v1,,v)P=(v_0,v_1,\ldots,v_\ell) of greatest length in GG.

givenF3chooseconstruct
2.1

Every neighbour of v0v_0 lies among v1,,vv_1,\ldots,v_\ell: a neighbour outside this list could be prepended to PP, giving a longer path and contradicting step 1.1.

step 1.1F2
3.1

The degG(v0)\deg_G(v_0) distinct neighbours of v0v_0 therefore occupy distinct positions among the \ell vertices v1,,vv_1,\ldots,v_\ell, so degG(v0)δ(G)\ell\ge\deg_G(v_0)\ge\delta(G) by [F1]. Thus PP has at least δ(G)\delta(G) edges.

step 2.1F1discharge-construct
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31Open item page →

Every finite simple graph of minimum degree at least two contains a cycle

Statement

If GG is a finite simple graph with δ(G)2\delta(G)\ge2, then GG contains a cycle.

Facts & Assumptions

Given: A finite simple graph GG with δ(G)2\delta(G)\ge2.

[L1]

GG contains a path with at least δ(G)\delta(G) edges (Every nonempty finite simple graph contains a path with at least δ(G)\delta(G) edges).

[F1]

A closed walk of length at least 33 whose vertices before the final repeated endpoint are distinct is a cycle (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

Proof

technique · direct
1.1

By [L1], GG has a path of length at least 22. Among all paths choose one of greatest length, say P=(v0,,v)P=(v_0,\ldots,v_\ell), so 2\ell\ge2.

givenL1choose
2.1

Every neighbour of v0v_0 lies on PP, since a neighbour outside PP could be prepended to make a longer path. Since degG(v0)2\deg_G(v_0)\ge2, besides v1v_1 there is a neighbour vjv_j with j2j\ge2.

step 1.1given
3.1

The list (v0,v1,,vj,v0)(v_0,v_1,\ldots,v_j,v_0) is a closed walk of length j+13j+1\ge3 with distinct vertices before its final v0v_0, so it is a cycle by [F1].

step 1.1step 2.1F1
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)Open item page →

An nn-vertex simple graph with more than (n12)\binom{n-1}{2} edges is connected

Statement

Let GG be a finite simple graph with n1n\ge1 vertices. If E(G)>(n12)|E(G)|>\binom{n-1}{2}, then GG is connected.

Facts & Assumptions

Proof

technique · contradiction
1.1

Suppose, for contradiction, that GG is disconnected. Choose one component with vertex set AA, put B=VAB=V\setminus A, and write k=Ak=|A|. Then 1kn11\le k\le n-1, B=nk|B|=n-k, and no edge joins AA to BB.

assume-contraF1choose
2.1

By [L1], the edges with both endpoints in AA number at most (k2)\binom k2, and those with both endpoints in BB number at most (nk2)\binom{n-k}2. Step 1.1 accounts for every edge, so E(k2)+(nk2)|E|\le\binom k2+\binom{n-k}2.

step 1.1L1
2.2

For 1kn11\le k\le n-1, elementary binomial algebra gives (k2)+(nk2)=(n12)(k1)(nk1)(n12)\binom k2+\binom{n-k}2=\binom{n-1}2-(k-1)(n-k-1)\le\binom{n-1}2.

step 1.1algebra
3.1

Steps 2.1 and 2.2 give E(n12)|E|\le\binom{n-1}{2}, contradicting the hypothesis. Therefore GG is connected.

step 2.1step 2.2givendischarge-contradiction

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

FALSE: a finite simple graph is determined up to isomorphism by its degree sequence

Statement

FALSE. A finite simple graph is determined up to isomorphism by its degree sequence.

Facts & Assumptions

Given: The cycle graph C6C_6 and the graph HH that is the disjoint union of two copies of C3C_3.

[F2]

A graph isomorphism is a bijection preserving adjacency in both directions, and therefore sends cycles to cycles of the same length (Graph isomorphisms, automorphisms and graph complements).

Refutation

technique · direct
1.1

Every vertex of C6C_6 has degree 22. Every vertex of HH lies on one of its two triangles and also has degree 22. Thus both degree sequences are (2,2,2,2,2,2)(2,2,2,2,2,2).

givenF1F3
1.2

The graph HH contains a cycle of length 33, namely either triangle, whereas C6C_6 contains no triangle because its only edges join consecutive vertices on its six-cycle.

givenF1
2.1

If C6HC_6\cong H, [F2] would carry a triangle of HH to a triangle of C6C_6, contradicting step 1.2. Hence the graphs have the same degree sequence but are not isomorphic, refuting the claim.

step 1.1step 1.2F2
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)Open item page →

FALSE: a connected simple graph in which every vertex has even degree must itself be a cycle

Statement

FALSE. Every connected finite simple graph in which every vertex has even degree is a cycle graph.

Facts & Assumptions

Given: The bowtie graph with vertices x,a,b,c,dx,a,b,c,d and edges xa,ab,bx,xc,cd,dxxa,ab,bx,xc,cd,dx, consisting of two triangles that share only xx.

[F2]

A connected graph has a path between every two vertices (Connected graphs and connected components defined by the existence of vertex paths).

[F3]

In the cycle graph CnC_n each vertex is joined by an edge exactly to its predecessor and its successor modulo nn (Empty and complete graphs, complete bipartite graphs, and the convention that PnP_n and CnC_n have nn vertices); with [F1] this gives every vertex of a cycle graph exactly two neighbours.

Refutation

technique · direct
1.1

Every two vertices of the bowtie are joined by a path, using xx when they lie in different triangles, so the graph is connected by [F2].

givenF2
1.2

The vertex xx has neighbours a,b,c,da,b,c,d and hence degree 44; each other vertex has degree 22. Thus all degrees are even by [F1].

givenF1
2.1

The bowtie is not a cycle graph because its vertex xx has degree 44, whereas [F3] gives degree 22 at every vertex of a cycle graph. Steps 1.1 and 1.2 therefore refute the claim.

step 1.1step 1.2F3
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

FALSE: vertex connectivity, edge connectivity and minimum degree are always equal

Statement

FALSE. Every nontrivial connected finite simple graph satisfies κ(G)=λ(G)=δ(G)\kappa(G)=\lambda(G)=\delta(G).

Facts & Assumptions

Given: The bowtie graph GG with two triangles xabxxabx and xcdxxcdx sharing only the vertex xx.

[L1]

Whitney's theorem guarantees only κ(G)λ(G)δ(G)\kappa(G)\le\lambda(G)\le\delta(G) (Whitney's inequalities: κ(G)λ(G)δ(G)\kappa(G)\le\lambda(G)\le\delta(G) for every nontrivial connected graph).

Refutation

technique · direct
1.1

Deleting xx separates the two edges abab and cdcd, while the connected graph has no vertex cut of size 00. Hence κ(G)=1\kappa(G)=1.

givenF1
1.2

Deleting the two edges xaxa and xbxb separates a,ba,b from the other triangle, so λ(G)2\lambda(G)\le2. No single edge disconnects GG, because every edge lies on one of the two triangles and the other two edges of that triangle give an alternate path between its endpoints. Hence λ(G)=2\lambda(G)=2.

givenF1
1.3

The four noncentral vertices have degree 22 and xx has degree 44, so δ(G)=2\delta(G)=2.

given
2.1

Thus κ(G)=1<2=λ(G)=δ(G)\kappa(G)=1<2=\lambda(G)=\delta(G), a strict instance of the first Whitney inequality and a counterexample to the asserted equality.

step 1.1step 1.2step 1.3L1
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31Open item page →

FALSE: a shortest walk between two vertices may repeat a vertex and fail to be a path

Statement

FALSE. A shortest walk between two vertices may repeat a vertex and fail to be a path.

Facts & Assumptions

Given: Vertices u,vu,v in one connected component and a uu to vv walk WW of minimum length.

[L1]

Every walk contains a path between the same endpoints of no greater length, and if the walk repeats a vertex the path can be chosen strictly shorter (Every walk between two vertices contains a path between the same endpoints).

[F1]

The distance dG(u,v)d_G(u,v) is the minimum length of a uu to vv path (Graph distance within a component, eccentricity, diameter and girth, including the acyclic convention).

Refutation

technique · contradiction
1.1

Suppose, for contradiction, that the minimum-length walk WW repeats a vertex and is not a path.

assume-contragiven
2.1

By [L1], deleting a closed segment from WW yields a uu to vv path of length strictly less than the length of WW. This path is also a walk, contradicting the minimality of WW.

step 1.1L1F1
3.1

Therefore every shortest walk is a path, and the claimed possibility cannot occur.

step 2.1discharge-contradiction

Sources