### 数学代写|图论作业代写Graph Theory代考|MATH7331

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## 数学代写|图论作业代写Graph Theory代考|The topological end space

In this last section we shall develop a deeper understanding of the global structure of infinite graphs, especially locally finite ones, that can be attained only by studying their ends. This structure is intrinsically topological, but no more than the most basic concepts of point-set topology will be needed.

Our starting point will be to make precise the intuitive idea that the ends of a graph are the ‘points at infinity’ to which its rays converge. To do so, we shall define a topological space $|G|$ associated with a graph $G=(V, E, \Omega)$ and its ends. ${ }^8$ By considering topological versions of paths, cycles and spanning trees in this space, we shall then be able to extend to infinite graphs some parts of finite graph theory that would not otherwise have infinite counterparts (see the notes for more examples). Thus, the ends of an infinite graph turn out to be more than a curious new phenomenon: they form an integral part of the picture, without which it cannot be properly understood.

To build the space $|G|$ formally, we start with the set $V \cup \Omega$. For every edge $e=u v$ we add a set $\dot{e}=(u, v)$ of continuum many points, making these sets $\ddot{e}$ disjoint from each other and from $V \cup \Omega$. We then choose for each $e$ some fixed bijection between $\dot{e}$ and the real interval $(0,1)$, and extend this bijection to one between $[u, v]:={u} \cup \grave{e} \cup{v}$ and $[0,1]$. This bijection defines a metric on $[u, v]$; we call $[u, v]$ a topological edge with inner points $x \in \dot{e}$. Given any $F \subseteq E$ we write $\stackrel{\circ}{F}:=\bigcup{\dot{e} \mid e \in F}$.

When we speak of a ‘graph’ $H \subseteq G$, we shall often also mean its corresponding point set $V(H) \cup \tilde{E}(H)$.

Having thus defined the point set of $|G|$, let us choose a basis of open sets to define its topology. For every edge $u v$, declare as open all subsets of $(u, v)$ that correspond, by our fixed bijection between $(u, v)$ and $(0,1)$, to an open set in $(0,1)$. For every vertex $u$ and $\epsilon>0$, declare as open the ‘open star around $u$ of radius $\epsilon$ ‘, that is, the set of all points on edges $[u, v]$ at distance less than $\epsilon$ from $u$, measured individually for each edge in its metric inherited from $[0,1]$. Finally, for every end $\omega$ and every finite set $S \subseteq V$, there is a unique component $C(S, \omega)$ of $G-S$ that contains a ray from $\omega$. Let $\Omega(S, \omega):=\left{\omega^{\prime} \in \Omega \mid C\left(S, \omega^{\prime}\right)=C(S, \omega)\right}$. For every $\epsilon>0$, write $E_\epsilon(S, \omega)$ for the set of all inner points of $S$ $C(S, \omega)$ edges at distance less than $\epsilon$ from their endpoint in $C(S, \omega)$. Then declare as open all sets of the form
$$\hat{C}\epsilon(S, \omega):=C(S, \omega) \cup \Omega(S, \omega) \cup \dot{E}\epsilon(S, \omega) .$$

## 数学代写|图论作业代写Graph Theory代考|Ramsey’s original theorems

In its simplest version, Ramsey’s theorem says that, given an integer $r \geqslant 0$, every large enough graph $G$ contains either $K^r$ or $\overline{K^r}$ as an induced subgraph. At first glance, this may seem surprising: after all, we need about $(r-2) /(r-1)$ of all possible edges to force a $K^r$ subgraph in $G$ (Corollary 7.1 .3 ), but neither $G$ nor $\bar{G}$ can be expected to have more than half of all possible edges. However, as the Turán graphs illustrate well, squeezing many edges into $G$ without creating a $K^r$ imposes additional structure on $G$, which may help us find an induced $\overline{K^r}$.

So how could we go about proving Ramsey’s theorem? Let us try to build a $K^r$ or $\overline{K^r}$ in $G$ inductively, starting with an arbitrary vertex $v_1 \in V_1:=V(G)$. If $|G|$ is large, there will be a large set $V_2 \subseteq V_1 \backslash\left{v_1\right}$ of vertices that are either all adjacent to $v_1$ or all non-adjacent to $v_1$. Accordingly, we may think of $v_1$ as the first vertex of a $K^r$ or $\overline{K^r}$ whose other vertices all lie in $V_2$. Let us then choose another vertex $v_2 \in V_2$ for our $K^r$ or $\overline{K^r}$. Since $V_2$ is large, it will have a subset $V_3$, still fairly large, of vertices that are all ‘of the same type’ with respect to $v_2$ as well: either all adjacent or all non-adjacent to it. We then continue our search for vertices inside $V_3$, and so on (Fig. 9.1.1).

How long can we go on in this way? This depends on the size of our initial set $V_1$ : each set $V_i$ has at least half the size of its predecessor $V_{i-1}$, so we shall be able to complete $s$ construction steps if $G$ has order about $2^s$. As the following proof shows, the choice of $s=2 r-3$ vertices $v_i$ suffices to find among them the vertices of a $K^r$ or $\overline{K^r}$.
Theorem 9.1.1. (Ramsey 1930)
For every $r \in \mathbb{N}$ there exists an $n \in \mathbb{N}$ such that every graph of order at least $n$ contains either $K^r$ or $\overline{K^r}$ as an induced subgraph.

Proof. The assertion is trivial for $r \leqslant 1$; we assume that $r \geqslant 2$. Let $n:=2^{2 r-3}$, and let $G$ be a graph of order at least $n$. We shall define a sequence $V_1, \ldots, V_{2 r-2}$ of sets and choose vertices $v_i \in V_i$ with the following properties:
(i) $\left|V_i\right|=2^{2 r-2-i} \quad(i=1, \ldots, 2 r-2)$;

(ii) $V_i \subseteq V_{i-1} \backslash\left{v_{i-1}\right} \quad(i=2, \ldots, 2 r-2)$;
(iii) $v_{i-1}$ is adjacent either to all vertices in $V_i$ or to no vertex in $V_i$ $(i=2, \ldots, 2 r-2)$.

# 图论代考

## Matlab代写

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