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Exercise (from T.Tao's Analysis 1 textbook):

Let $X$ be a set, and let $\Omega$ be the space of all pairs $(Y,\leq)$, where $Y$ is a subset of $X$ and $\leq$ is a well-ordering of $Y$. If $(Y,\leq)$ and $(Y',\leq ')$ are elements of $\Omega$, we say that $(Y,\leq)$ is an \emph{initial segment} of $(Y',\leq ')$ if there exists an $x\in Y'$ such that $Y=\{y\in Y': y<'x\}$ (so in particular $Y\subset Y'$), and for any $y,y'\in Y$, $y\leq y'$ iff $y\leq ' y'$. Define a relation $\preceq$ on $\Omega$ by defining $(Y,\leq)\preceq (Y',\leq ')$ if either $(Y,\leq)=(Y',\leq ')$, or if $(Y,\leq)$ is an initial segment of $(Y',\leq ')$.

(a) Show that $\preceq$ is a partial ordering of $\Omega$.

(b) There is exactly one minimal element of $\Omega$; what is it?

(c) Show that the maximal elements of $\Omega$ are precisely the well-orderings $(X,\leq)$ of $X$.

(d) Using Zorn's lemma, conclude the well-ordering principle: every set $X$ has at least one well-ordering.

(e) Conversely, use the well-ordering principle to prove the axiom of choice, Axiom 8.1. (Hint: place a well-ordering $\leq$ on $\bigcup_{\alpha\in I}X_{\alpha}$, and then consider the minimal elements of each $X_{\alpha}$.)

I've managed to do (a), (b), (c), (e) but I'm stuck on (d).

I've thought about applying Zorn's lemma in the following way: let $Z$ be a totally ordered subset of $\Omega$ (which is $\neq\emptyset$ since $(\emptyset,\leq_{\emptyset})\in\Omega$) then we define $W:=\{V:(V,\leq)\in Z\ for\ some\ well-ordering\ of\ V \}$ and $U:=\{\leq : (T,\leq)\in Z\ for\ some\ T\in 2^X\}$; then $(S,\leq_{S})$, where $S:=\bigcup W$ and $\forall z,z'\in Z$ we say that $z\leq_S z'$ iff $\exists\leq\in U$ such that $z\leq z'$ could be a candidate upper bound for $Z$, but I'm not sure about this and in fact I haven't been able to prove that it is an upper bound for $Z$.

So I would appreciate any hint about how to show that Zorn's lemma holds for the set $\Omega$.

Best regards,

lorenzo.

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Suppose that $Z$ is a chain in $\Omega$, show that $Y=\bigcup_{(S,\leq_S)\in Z}S$ is a subset of $X$ and $\leq_Y$ defined by: $$y\leq_Yy'\iff\exists(S,\leq_S)\in Z: y,y'\in S, y\leq_S y'$$ is a well-ordering of $Y$, and that $(Y,\leq_Y)$ is an upper-bound for $Z$.

Then apply Zorn's lemma.

(You might notice that $\leq_Y$ is actually the union of $\leq_S$ from those $(S,\leq_S)\in T$, and you won't be wrong.)

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  • $\begingroup$ that's for sure, but my question in fact is: how to show that Zorn's lemma holds? $\endgroup$ – lorenzo Dec 5 '16 at 13:30
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    $\begingroup$ That's not your question. Read your question. $\endgroup$ – Asaf Karagila Dec 5 '16 at 13:32
  • $\begingroup$ ok, I clarified my question. $\endgroup$ – lorenzo Dec 5 '16 at 14:02

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