Let $P = \left \{\,\,(\,k \,\,,\, k+1): k \in \mathbb{N} \cup \{0\}\,\,\right\}$
Alternatively, $P =\left\{\,\,(0 \,\,,\, 1), \,\,(1 \,\,,\, 2), \,\,(2 \,\,,\, 3), \,\,(3 \,\,,\, 4), \,\,(4 \,\,,\, 5), \,\,(5 \,\,,\, 6), \,\,(7 \,\,,\, 8)\,\cdots\, \right\}$
Suppose you add one new edge to $P$ consistent with the usual $<$ operator.
Now introduce a new edge
and add another edge
and another edge $\cdots$
Using this procedure, can you eventually build up the less-than ($<$) operator on the natural numbers?
Let $\mathtt{ONION}$ be the set of all ordered pairs $(A_1, A_2^{\prime})$ such that all of the following are true:
- $\forall \alpha \in \{A_1, A_2\}$
- $\forall n, m \in \mathbb{N}, (n, m) \in \alpha \implies (n < m)$
- $\forall k \in \mathbb{N} \cup \{0\}, (k,\,\, k+1) \in \alpha$
- $\exists \text{ unique pair } x, y \in \mathbb{N} \cup \{0\} \text{ such that } $
- $(x, y) \not\in A_1$
- $(x, y) \in A_2$
- $x < y$
We conjecture that we can begin with $P = \left \{\,\,(\,k \,\,,\, k+1): k \in \mathbb{N} \cup \{0\}\,\,\right\}$ and eventually get to $(<)$ if and only if $\exists$ a $Q \subset \mathtt{ONION}$ such that $Q$ is sufficiently like a path.
Weakly Path-ish
Definition of Weakly Path-ish
Suppose that $VS$ is a non-finite set of vertices.
Also suppose that $ES \subseteq VS \times VS$ such that:
- ${\forall W \in VS} \quad \exists \text{ unique } V \in VS \text{ such that } (V, W) \in ES$
- ${\forall W \in VS} \quad \exists \text{ unique } X \in VS \text{ such that } (W, X) \in ES$
We say that $E$ is a weakly path-ish Relation.
One thing about this definition I do not like is that we might want a start or end to the path.
For example, ${\exists b \in VS} \text{ such that } \exists c \in VS \text{ such that } (b, c) \in ES$ but $\forall a \in VS, (a, b) \not\in ES$
Example of a Weakly Path-ish Relation
An example of weakly path-ish set is $E^{\prime} = \{(x, x+2): x \in \mathbb{Z}\}$
$E^{\prime}$ is basically a path through the even numbers unioned with a path through the odd numbers.
$$E^{\prime} = \left\{\cdots (-3, -1), (-1, 1), (1, 3), \cdots \right\} \cup \left\{\cdots (-4, -2), (-2, 0), (0, 2), (2, 4), \cdots \right\}$$
However, the two paths are separate, which is not ideal.
Definition of Strongly Path-ish
for any weakly-path-ish relation $R$,
we say that $R$ is strongly path-ish
if and only if
there does not exist a weakly path-ish relation $R^{\prime}$ on the same domain as $R$ such that
for all $x, y, z$ in the domain,
$(x, y) \in R^{\prime}$ and $(y, z) \in R^{\prime}$ if and only if $(x, z) \in R$
Conjectured Example of a Strongly Path-ish Relation
We conjecture that $\{(x, x+1): x \in \mathbb{Z}\}$ is an example of a strongly path-ish relation.
Example of a set which is NOT strongly path-ish
Let $ES = \bigg\{\left(\frac{1}{1}, \frac{1}{2}\right), \left(\frac{1}{2}, \frac{1}{4}\right), \left(\frac{1}{4}, \frac{1}{8}\right), \left(\frac{1}{8}, \frac{1}{16}\right), \cdots, \left(\frac{1}{2^{99}}, \frac{1}{2^{100}}\right), \cdots, \left(\frac{-1}{2^{99}}, \frac{-1}{2^{100}}\right), \cdots, \left(\frac{-1}{1}, \frac{-1}{2}\right) \bigg)$.
Or more formally, $ES = \left\{\left(2^{-k}, 2^{-(k+1)}\right): k \in \mathbb{N} \cup \{0\} \right\} \cup \left\{\left(-2^{-(k+1)}, -2^{-k}\right): k \in \mathbb{N} \cup \{0\}\right\}$
Let us put aside the fact that $ES$ has at least one start/end, and an earlier definition required weakly path-ish objects to have no beginning or end.
Let $ES^{\prime} = \bigg\{\left(\frac{+1}{1}, \frac{-1}{1}\right), \left(\frac{-1}{1}, \frac{+1}{2}\right), \left(\frac{+1}{2}, \frac{-1}{2}\right), \left(\frac{-1}{2}, \frac{+1}{4}\right), \left(\frac{+1}{4}, \frac{-1}{4}\right), \left(\frac{-1}{4}, \frac{+1}{8}\right), , \left(\frac{+1}{8}, \frac{-1}{8}\right), \cdots \bigg)$.
Consider the following algorithm: For each vertex $v$ in $P$ in increasing order, we append the edges $(v-2, v), (v-3, v), ...,(0, v)$. For any $v$, this takes only a finite number of steps, so to do it for all of $\mathbb{N}$ takes only countably-many steps.
Given $a < b$, the edge $(a, b)$ is clearly added when this algorithm reaches vertex $b$ (which it must do in a finite number of steps).