3
$\begingroup$

I am looking for a short proof of this fact. This is clearly true by drawing these trees, but I am having trouble putting it into writing. Somehow I need to select 3 of the 5 vertices and show that there must be a $3$-cycle or an independent set of size $3$. (Is this a Ramsey number or something like that?)

Is there an obvious/short argument here, or perhaps a lemma I could use to knock it out quickly?

  • 0
    OK, I've read the proof a bit more carefully and it seems solid with one exception. If there exists a vertex with maximum degree greater than or equal $\sqrt{n}$ then it's neighborhood is an independent set, so $\alpha \ge \sqrt{n}$. Otherwise, \Delta < \sqrt{n}. The next step in the proof implicitly uses Brooke's theorem. A triangle free graph is obviously not complete but it may be a cycle. So for all **non-cycle** triangle-free graphs, we have \chi \le \Delta \implies \alpha \ge \frac{n}{\chi} \ge \frac{n}{\Delta} > \sqrt{n}2012-11-24

1 Answers 1

5

In general let $F$ be a forest of order $n$ with $k$ connected components. It follows that every connected component is a tree, thus every component is bipartite (because trees do not have odd cycles). Thus the ith connected component contains an independent set of size greater than ceiling($n_i/2$) (Where $n_i$ is the order of the ith connected component) . Since the union of all these independent sets is again independent (because they come from different connected components), thus we have:

$\sum_{i=1}^k ceiling(n_i/2)\leq \alpha(F)$

A weaker version of this inequality would be: $n/2=\sum_{i=1}^k (n_i/2)\leq \alpha(F)$

Since the order of $F$ in your question is 5, we know that $5/2\leq \alpha(F)$ hence $3\leq \alpha(F)$

  • 0
    I might be missing something, but what is $\alpha(F)$ ? and how does $3 \leq \alpha(F)$ follow from $5/2 \leq \alpha(F) $ ? since $\alpha (F)$ could equal 2.52016-06-17