A fractional chromatic number of a graph $G=(V,E)$, is $min~\sum_{I}y_{I}$, and for every vertex $v\in V$ we have $\sum_{\{I:v \in I\}}y_{I} \ge1$(the condition).
I'm just a little confused on notation. Let's consider this graph
There are $3$ independent sets: $\{V_{3}\},\{V_{2}\},\{V_{1},V_{4}\}$ So, is a fractional chromatic number of this graph $\le 3$ (sum of all vertices in each independent graph)? Am I supposed to verify every independent graph if the sum isn't more than $1$, to check the condition ? f.e. for the first set ($\{V_{3}\}$) the sum of all vertices is $1$, and $1 \le 1$, so the condition is met, similarly the next set ($\{V_{4}\}$, and for the last set ($\{V_{1},V_{4}\}$) the sum of all verticies is $\frac{1}{2}+\frac{1}{2}= 1$ and $1 \le 1$ so the condition is met for this and any other set, so indeed, $\chi^{*} \le 3$ ? Is it correct ?
There is a variable $y_I$ for each independent set $I$; note that $I$ need not be maximal. Let's call the variables $y_1, y_2, y_3, y_4$, and $y_{14}$. This gives the following optimization problem:
$$\textrm{minimize} \sum_I y_I\\ \begin{align} \textrm{s.t.}\quad &y_1 + y_{14} &\geq 1\\ & y_2 &\geq 1\\ & y_3 &\geq 1\\ & y_4 + y_{14} &\geq 1\\ \end{align}$$
Clearly $y_2 = y_3 = 1$. For the remaining variables, we'll get the smallest sum when we let $y_{14}$ do all of the work: $y_1=y_4=0$, and $y_{14}=1$. The interpretation of this is that we have no reason to assign different colours to $v_1$ and $v_4$.
So the sum is $3$, and all variables are integer. This corresponds to a $3-$colouring of the graph, which should make sense. (Choose three colours for $\{v_1,v_4\}, v_2$, and $v_3$.) There is no benefit to a fractional colouring for this graph.