In Tao Analysis I, Lemma 8.5.14 says
Let X be a partially ordered set with ordering
relation $\leq$ , and let $x_0$ be an element of $X$. Then there is a well-ordered subset $Y$ of $X$ which has $x_0$ as its minimal element, and which has no strict upper bound. For a given $Y$, Strict Upper Bound, $s(Y) \in X$, is defined by $\forall y \in Y( s(Y) > y )$.
Tao proves the Lemma by Contradiction where he assumes that every well-ordered subset $Y$ of $X$ which has $x_0$ as its minimal element at least one strict upper bound. This is done using Axiom of Choice to assign an $s(Y)$ for every $Y \subset X$ that contains $x_0$.
He then defines a special class of good subsets $Y$ of $X$, where Y is good $\leftrightarrow Y$ is well-ordered $\wedge$ $x_0 \in Y \wedge$ Y obeys property where $\forall x \in Y \setminus \{x_0\}[x = s(\{y \in Y: y<x\})]$.
Let $\Omega := \{Y \subseteq X: Y$ is good $ \}$.
Then came the confusing part:
We make the following important observation: if $Y$ and $Y'$ are
two good subsets of $X$, then every element of $Y'\setminus Y$ is a strict upper
bound for $Y$, and every element of $Y\setminus Y'$ is a strict upper bound
for $Y'$. (Exercise 8.5.13). In particular, given any two good sets
$Y$ and $Y'$, at least one of $Y'\setminus Y$ and $Y\setminus Y'$ must be empty (since
they are both strict upper bounds of each other). In other words,
$\Omega$ is totally ordered by set inclusion: given any two good sets $Y$
and $Y'$, either $Y \subseteq Y'$ or $Y' \subseteq Y$.
My main gripe (counterexample?) is if you take $\mathbb{N}$ for $X$, and let $x_0 = 7$ for the sake of example. Then, you take $Y := \{7, 8, 14, 15 \} $ and $Y' := \{7, 10, 11 \}$ both of which qualify as good subsets of $X$. Apparently, $8 \in Y \setminus Y'$ which cannot be strict upper bounds of $Y'$. So, I'm stuck here. Is there anything that I'm missing out on defining the good sets, $\Omega$, of $X$?
Best Answer
The property of being a good set depends on the function $s$ (the function that assigns every well-ordered set to a UNIQUE strict upper bound), the sets that you give in your example are not good sets, if you suppose that $Y$ and $Y'$ are good sets we get a contradiction: $8\in Y$ and $10\in Y'$ but since both sets are good sets then
$$8 = s(\{y\in Y : y < 8\}) = s(\{7\})$$ $$10 = s(\{y\in Y' : y < 10\}) = s(\{7\})$$
Here is the intuitive idea: The motivation behind a good set, is that are sets of the form:
$$X_1 = \{x_0\}$$
$$X_2 = \{x_0, s(X_1)\}$$
$$X_3 = \{x_0, s(X_1), s(X_2)\}$$
$$X_4 = \{x_0, s(X_1), s(X_2), s(X_3)\}$$
$$\vdots$$
So, if you take any two good sets
$$X_n=\{x_0, s(X_1),\dots, s(X_{n-1})\}$$
$$X_m=\{x_0, s(X_1),\dots, s(X_{m-1})\}$$
w.l.o.g suppose that $n < m$, then
$$X_m\backslash X_n = \{s(X_{n}),\dots, s(X_{m-1})\}$$
and clearly every $x\in X_m\backslash X_n$ is an strict upper bound for $X_n$.