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Applying Probability to Ramsey Theory

The following theorem, due to P. Erdős, is a true classic, and is presented here in a manner that is faithful to how it was first published.

Applying Probability to Ramsey Theory

The following theorem, due to P. Erdős, is a true classic, and is presented here in a manner that is faithful to how it was first published. As we shall see later, it was subsequently recastbut that's getting the cart ahead of the horse.

Theorem

If\(n\) is a positive integer. Then \[\begin{aligned}\end{aligned}\]

Proof

Let \(t\) be an integer with \(t>n\) and consider the set \(\cgF\) of all labeled graphs with vertex set \(\{1,2,\dots,t\}\). Clearly, there are \(2^{C(t,2)}\) graphs in this family. Let \(\cgF_1\) denote the subfamily consisting of those graphs which contain a complete subgraph of size\(n\). It is easy to see that \[\begin{aligned}\end{aligned}\] Similarly, let \(\cgF_2\) denote the subfamily consisting of those graphs which contain an independent set of size\(n\). It follows that \[\begin{aligned}\end{aligned}\]

We want to take the integer \(t\) as large as we can while still guaranteeing that \(|\cgF_1|+|\cgF_2|\le |\cgF|\). This will imply that there is a graph \(G\) in \(\cgF\) which does not contain a complete subgraph of size \(n\) or an independent set of size\(n\). So consider the following inequality: \[\begin{aligned}\end{aligned}\] Now we ask how large can \(t\) be without violating inequality? To answer this, we use the trivial inequality \(\binom{t}{n}\le t^n/n!\) and the use the Stirling approximation for \(n!\). After some algebra and taking the \(n^{\text{th} }\) root of both sides, we see that we need only guarantee that \[\begin{aligned}\end{aligned}\]

Now let's take a second look at the proof of . We consider a probability space \((S,P)\) where the outcomes are graphs with vertex set \(\{1,2,\dots,t\}\). For each \(i\) and \(j\) with \(1\le i \lt j\le t\), edge \(ij\) is present in the graph with probability \(1/2\). Furthermore, the events for distinct pairs are independent.

Let \(X_1\) denote the random variable which counts the number of \(n\)-element subsets of \(\{1,2,\dots,t\}\) for which all \(\binom{n}{2}\) pairs are edges in the graph. Similarly, \(X_2\) is the random variable which counts the number of \(n\)-element independent subsets of \(\{1,2,\dots,t\}\). Then set \(X=X_1+X_2\).

By linearity of expectation, \(E(X)=E(X_1)+E(X_2)\) while \[\begin{aligned}\end{aligned}\] If \(E(X)\lt 1\), then there must exist a graph with vertex set \(\{1,2,\dots,t\}\) without a \(K_n\) or an \(I_n\). And the question of how large \(t\) can be while maintaining \(E(X)\lt 1\) leads to exactly the same calculation we had before.

Condensed: the full section is in Keller & Trotter, Applied Combinatorics.

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Partea acestei pagini este adaptată de la Keller & Trotter, Applied Combinatorics (CC BY-SA 4.0). Condensată şi reexplicată aici; erorile sunt ale noastre.

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