I am now reading Nash's 1951 paper Non-cooperative games and I am trying to understand the intuition behind solution concepts.
Solutions(Nash 1951)
A game is solvable if its set $S$ of equilibrium points satisfies the condition for all $i$'s $$(t;r_{i})\in S\,\,and\,\,s\in S \implies (s;r_{i})\in S\tag{1}$$ Here $t$ and $s$ are $n$-tuple strategy profile. Additionally, if $t=(t_{1},t_{2},\cdots,t_{n})$, then $(t;r_{i})=(t_{1},\cdots,t_{i-1},r_{i},t_{i+1},\cdots)$. I.e $(t;r_{i})$ is the strategy profile with player $i$'s strategy now being $r_{i}$.
The solution of a solvable game is its set $S$ of equilibrium points.
A game is strongly solvable if it has a solution $S$ s.t for all $i$'s $$s\in S\,\,and\,\, p_{i}(s;r_{i})=p_{i}(s)\implies (s;r_{i})\in S\tag{2}$$ Here $p_{i}(s)=\max_{r_{i}}p_{i}(s;r_{i})$, where $p_{i}$ is the payoff for player $i$. i.e $p_{i}(s)$ is the maximal payoff for player $i$ given the opponents playing $s_{-i}$.
Sub-solutions: If $S$ is a subset of the set of equilibrium points of a solvable game and for every $s \in S$, the condition $p_{i}(s)=max_{r_{i}}p_{i}(s;r_{i})$ is satisfied, and if $S$ is maximal relative to this property, we call $S$ a sub-solution.
Factor sets: For any sub-solution $S$ we define the $i$-th factor set $S_{i}$ as the set of all $s_{i}$ s.t $(t;s_{i})\in S$ for some $t$,
Theorem 3: A sub-solution $S$ is the set of all $n$-tuples $(s_{1},\dots,s_{n})$ s.t each $s_{i}\in S_{i}$ where $S_{i}$ is the $i^{th}$ factor set of $S$. Geometrically, $S$ is the product of its factor sets.
Here are my questions.
1. Why do we need these solution concepts?
I don't quite get the intuition behind these solution concepts. Moreover, since a solution may not exist in a game, it seems that the concept of equilibrium points can be applied to a broader class of games than these solution concepts. Additionally, it has been mentioned in the paper that if a solution exists, then it is unique. Then it seems that we don't need to differentiate between solutions and strong solutions because there can be only one solution (strong solution if it exists) and those conditions (1) and (2) will be trivially satisfied.
2. The difference between sub-solution and equilibrium points
I am a little confused about the difference between sub-solution and equilibrium points. From the definition of sub-solution, it seems that it is exactly the set which contains all the equilibrium points. However, based on Theorem 3, it is the Cartesian product of its factor sets. For example, suppose in a game where the set of all equilibrium points is $S=\{(U,L),(D,R)\}$. Then based on the definition, the sub-solution should also be $S$. But by theorem $3$, the sub-solution should be $\{(U,L),(U,R),(D,L),(D,R)\}$.
I really appreciate it if someone can give me some help on these. Thanks in advance!