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From Wikipedia :

An independent set of $\sqrt{n}$ vertices in an $n$-vertex triangle-free graph is easy to find: either there is a vertex with more than $\sqrt{n}$ neighbors (in which case those neighbors are an independent set) or all vertices have less than $\sqrt{n}$ neighbors (in which case any maximal independent set must have at least $\sqrt{n}$ vertices)

Why does any maximal stable set have at least square root of $n$ vertices?

Thanks.

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The point is that the graph is triangle free, so if $j$ and $k$ are both neighbours of $i$, then the edge $\{j,k\}$ is not in the graph. This means that $N_i = \{j : \text{$i$ is a neighbour of $j$}\}$ is an independent set for any $i$.

The other fact being used is that a graph with $n$ vertices of maximum degree $\Delta$ has an independent set of size at least $\lfloor n/(\Delta+1)\rfloor$.

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  • $\begingroup$ Can you please provide more details? And "has a set" and "every maximal" is different, no? $\endgroup$
    – Eugene
    Aug 8, 2017 at 20:01
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    $\begingroup$ If the maximum degree is $\Delta$ than any independent set of size $t$ has at most $t\Delta$ neighbours. Unless $t(\Delta +1) \ge n$, it's not maximal. $\endgroup$
    – Louis
    Aug 8, 2017 at 20:55
  • $\begingroup$ Okay, so if maximum degree is less than $\sqrt{n}$, then we have the bound. so here the triangle free property isn't used right? It is used only when saying that if a vertex has big neighborhood then take it as independent set. Am I right ? $\endgroup$
    – Eugene
    Aug 8, 2017 at 21:57
  • $\begingroup$ Yes. That's right. $\endgroup$
    – Louis
    Aug 8, 2017 at 22:16

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