coloringgraph theory
Could you give an example or a proof for a graph that has no triangles but still has chromatic number $\chi(G) \ge n$, where $n \gt 3$ is the number of vertices? This is from coloring of a graph.
The only way I can think about it is in terms of complete graph $K_n$, but that has triangles and its chromatic number is $\chi(K_n) \ge n$. But can you give an example for a graph that is not complete and still has $\chi(G) \ge n$?
The notion of $\chi$-boundedness is a generalization of perfect graphs. A graph $G$ is perfect if $\chi(H)=\omega(H)$ for each induced subgraph $H$ of $G$.
A graph $G$ is $\chi$-bounded if there exists a function $f:\mathbb{N}\to \mathbb{N}$ such that $\chi(H) \leq f(\omega(H))$ for each induced subgraph $H$ of $G$.
If we let $f:\omega \mapsto \omega$ (the identity function), then $G$ is a perfect graph iff $G$ is $\chi$-bounded by $f$. Hence, the above is a direct generalization.
In that sense, a $\chi$-binding function is an upper bound for the chromatic number, but in form of a function from its clique number $\omega$ to a natural number.
For example, the class of even-hole-free graphs, that is the class of graphs that forbidds induced cycles of even length, is $\chi$-bounded with $\chi$-binding function $\chi(G) \leq 2\omega(G)-1$ as shown by Addario-Berry et.al.
Question: Are all graphs $\chi$-bounded by some function $f$? No. There are graphs that are triangle free, thus clique number strictly smaller than $3$, but with an arbitrary large chromatic number (consider the Mycielskian construction for an example).
Answering the question whether or not a given family of graphs is $\chi$-bounded or not is far from trivial and an open research question for many classes of graphs.
There is another interesting problem.
Problem: Given a class of graphs that is $\chi$-bounded by some function $f$. Is $f$ the smallest $\chi$-binding function for this class?
Answering the above is an open research question for many graph classes.
Your second questions asks what we can say about the induced subgraphs of a graph $G$ if we are provided with a $\chi$-binding function. For a concrete graph, these are the usual answers one could provide if we know the chromatic number of a graph (which is the same as saying we know the $\chi$-binding function of a specific graph that provides a tight bound). This is not interesting. It is more interesting to consider a whole family of graphs and its $\chi$-binding function, because it allows us to make certain statements about the structure of the whole family.
Is there a general answer about the structure of a class of graphs if we know its $\chi$-binding function? I am afraid that there is none, but I can't say this for sure.
If you want to dive deeper into this topic, I suggest the following two survey papers:
[1] "Schiermeyer, Ingo, and Bert Randerath. "Polynomial $\chi$-Binding Functions and Forbidden Induced Subgraphs: A Survey." Graphs and Combinatorics 35.1 (2019): 1-31."
[2] "Scott, A., & Seymour, P. (2020). A survey of χ-boundedness." (I have only found it on arXiv)
Though we can think of the graph in question,
as $5$ tetrahedrons glued together, it is more profitable to think of it as $C_5 + K_2$, where $+$ denotes graph join: we take disjoint copies of $C_5$ and $K_2$, and draw all the edges between them. In the diagram, $C_5$ is the $5$ outside vertices, and $K_2$ is the two center vertices.
Similarly, the graph $C_5 + K_n$ has chromatic number $n+3$ (you need $3$ colors to color $C_5$, and $n$ separate colors to color $K_n$) and clique number $n+2$ (at most two vertices from $C_5$, plus at most $n$ vertices from $K_n$, can be part of a clique). So $C_5 + K_{k-3}$ is a graph with the property you want.
In fact, it is the smallest such graph. $C_5 + K_{k-3}$ has $k+2$ total vertices, so there's only a few cases to check:
The argument is basically the same as for $C_5 + K_2$.
Best Answer
An upper bound for chromatic number is $\chi(G) \le \Delta(G)+1$ (see Greedy coloring). And we know that for any simple graph $G$, we have $\Delta(G) \le n-1$ (Note that if $G$ is not simple, we can consider its condensation by removing parallel edges as they don't affect chromatic number). Thus, we have $\chi(G) \le n$. So, it is not hard to see that $\chi(G) = n$ is possible only when we have $G = K_n$. But, as you noticed, this is not a triangle-free graph for $n \ge 3$.
However, if you are not restricting yourself with number of vertices, I recommend you to see Mycielski Graph as they can generate a graph with any chromatic number you want and they are always triangle-free.