User:GFauxPas/Sandbox

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Munkres Supplementary Exercises to Chapter 1

Theorem
The Well-Ordering Theorem implies the Hausdorff Maximal Principle.

Theorem
Let the Well-Ordering Theorem hold.

Then the Hausdorff Maximal Principle holds.

Proof
Let $X$ be a non-empty set.

Let $X$ contain at least two elements; otherwise, any non-empty ordering on $X$ is trivially a maximal chain.

By the Well-Ordering Theorem, $X$ can be well-ordered. Fix such a well-ordering.

Let $\le$ be any ordering on $X$.

Let $P\left({a,Y}\right)$ be the predicate:

''$a$ is $\le$-comparable to every $y \in Y$.

(Here $Y$ is a bound variable.)

That is, $P\left({a,Y}\right)$ holds if, for every $y \in Y$, $a \le y$ or $y \le a$.

Define the mapping:


 * $\rho\left({f: S_x \to \mathcal P(X) }\right) = \begin{cases}

f[S_x] \cup \{ x \} & P(x,S_x) \\ f[S_x] & \text{ otherwise } \end{cases}$

where $S_x$ is the initial segment in $X$ defined by $x$.

Also see

 * Hausdorff Maximal Principle Implies Well-Ordering Theorem