In mathematics, ordinals extend the concept of counting beyond finite numbers and incorporate infinite sequences. From a graph theory perspective, you can think of ordinals as a hierarchy of "layers" or "levels" that represent different sizes and types of infinity.
In this blog, we define ordinals as a (maybe infinite) competitive graph $$$\Gamma$$$ (A directed graph is a competitive graph, if exactly one of $$$(u, v)$$$ and $$$(v, u)$$$ is in the edge set, for every $$$u \ne v$$$.) with no loops and no reversed rays (that is, a sequence of vertices $$$v_0, v_1, v_2, \dots$$$, where there exists edges $$$(v_1, v_0), (v_2, v_1), (v_3, v_2), \dots$$$).
Finite ordinals are natural numbers. For example, $$$0$$$ is the empty graph, and $$$1$$$ is the only graph with one vertex. Further example is as below:
In fact, the set of natural number $$$\mathbb N$$$, usually denoted $$$\omega$$$, can be seen as an ordinal: $$$\Gamma_\omega = (V, E)$$$, where $$$V = \mathbb N$$$ and $$$E = \{(i, j) : i < j\}$$$, but $$$\mathbb Z$$$ cannot be seen as an ordinal in similar way, because there exists a reversed ray $$$v_i = -i$$$.
We define $$$\alpha \le \beta$$$ if $$$\beta$$$ can be seen as a subgraph of $$$\alpha$$$, and define $$$\alpha = \beta$$$, if $$$\alpha \le \beta$$$ and $$$\beta \le \alpha$$$. Like what we're familiar with, we have either $$$\alpha < \beta, \alpha = \beta$$$, or $$$\alpha > \beta$$$.
In fact, for every set $$$A$$$ of ordinals, we can have a ordinal $$$\min A$$$ in the set $$$A$$$. The proof is simply get the union of all the $$$S_\alpha$$$, where $$$\alpha \in A$$$ and $$$S$$$ is the set constructed before.
The successor of an ordinal $$$\alpha$$$ is a new graph $$$G$$$, where the vertices are the vertex of $$$\alpha$$$, together with a new vertex $$$v$$$, and link all the edges $$$(w, v)$$$, where $$$w \in \alpha$$$. We can denote the successor of $$$\alpha$$$ as $$$\alpha^+$$$. It can be seen that $$$\alpha < \alpha^+$$$. then, we can get the $$$n$$$-th successor of $$$\alpha$$$, denoted $$$\alpha + n$$$.
