Concentration of Chromatic Number in Four Values I am reviewing Theorem 7.3.3 from the The Probabilistic Method reproduced below:
Theorem 7.3.3: Let $p=n^{-\alpha}$ where $\alpha>\frac{5}{6}$ is fixed and let $G=G(n,p)$. Then there exists a $u=u(n,p)$ so that almost always
$$u\leq \chi(G)\leq u+3.$$
To prove this, the authors use the following lemma:
Lemma 7.3.4: Let $\alpha, c$ be fixed with $\alpha>\frac{5}{6}$. Let $p=n^{-\alpha}$. Then almost always every $c\sqrt{n}$ vertices of $G=G(n,p)$ may be 3-colored.



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*My first question is how would I justify that such a minimal set $T$ must have at least $3t/2$ edges?

*Naturally, since we want to show a property in a graph holds almost surely, the sum must be $o(1)$. But in the summation, would that not require $n\rightarrow\infty$? So, whenever a property almost always happens, are we referring to an infinite graph?

I am a bit confused about the last paragraph above in general and require some clarification. I do understand how the concentration inequalities on $Y$ were obtained. Since we have that the chromatic number is almost always concentrated on four values, does this property require that the graph be infinite?
 A: I only address your first question.
If $t$ vertices have degree $\geq 3$, then there are $\geq 3t$ edge incidences on vertices.  Since every edge is incident on $2$ vertices, the incidences overcount the edges by a factor of $2$.  That is, there are $\geq \frac{3t}{2}$ edges.
A: My answer is partial, because I am not familiar with this stuff.

how would I justify that such a minimal set $T$ must have at least $3t/2$ edges

Since each vertex $x\in T$ has at least 3 edges incident to other vertices of $T$. When we sum this over all $x\in T$ we obtain $3t$ and we have to divide the sum by $2$, since each edge was counted twice.

since we want to show a property in a graph holds almost surely, the sum must be $o(1)$. But in the summation, would that not require $n\rightarrow\infty$? So, whenever a property almost always happens, are we referring to an infinite graph?

I guess the considered graphs are finite, because we are calculating with their number of vertices. We don’t have to require  $n\rightarrow\infty$ in the summation. I guess it suffices to show that sum is less than some function $f(n)$, which is $o(1)$, that is $\lim_{n\to\infty} f(n)/n=0$.
