Chapter 14, The axiom of choice Video Solutions, Set theory and its philosophy: a critical introduction

Problem 1

Do the 'easy exercise' mentioned in the proof of proposition 14.1.1.

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Problem 1

Show that the axiom of choice is equivalent to the assertion that if $\left(B_i\right)_{i \in I}$ is a family of non-empty sets, then $\prod_{i \in I} B_i$ is non-empty.

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02:03

Problem 1

Show that the set $\mathfrak{P}(A, B)$ of functions $f$ such that $\operatorname{dom}[f] \subseteq A$ and $\operatorname{im}[f] \subseteq B$ is inductively ordered by inclusion.

James Chok

Numerade Educator

02:27

Assuming the axiom of countable dependent choice, show that if $(A, \leqslant)$ is an infinite partially ordered set, then $A$ has either an infinite totally ordered subset or an infinite totally unordered subset. [Suppose that all totally unordered subsets of $A$ are finite: show that every infinite subset $B$ of $A$ has a maximal totally unordered subset and therefore has an element $b$ which is comparable with infinitely many elements of B.]

Angelo Rendina

Numerade Educator

Problem 2

If $(A, \leqslant)$ is inductively ordered and $a \in A$, show that $\{x \in A: x \leqslant a\}$ is also inductively ordered.

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Problem 3

Show that the axiom of countable dependent choice is equivalent to the assertion that if $(A, \leqslant)$ is a partially ordered set and $\left(D_n\right)$ is a sequence of cofinal subsets of $\mathrm{A}$, then there is an increasing sequence $\left(x_n\right)$ in $A$ such that $\left\{x_n: n \in \boldsymbol{\omega}\right)$ intersects every $D_n$.

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06:04

Problem 3

Is $\mathfrak{F}(\boldsymbol{\omega})$ inductively ordered by inclusion?

Mengchun Cai

Numerade Educator

Problem 4

Let Well $(A)$ be the set of all relations on $A$ which are well-orderings. If $r, r^{\prime} \in$ Well $(A)$, define $r \leqslant r^{\prime}$ iff $r \subseteq r^{\prime}$ and dom $[r]$ is an initial subset of the well-ordered set $\left(\operatorname{dom}\left[r^{\prime}\right], r^{\prime}\right)$. Show that $(\mathrm{Well}(A), \leqslant)$ is an inductively ordered set. Hence deduce the well-ordering property directly from Zorn's lemma.

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01:19

Problem 5

Show by an example that a function of the kind referred to in lemma 14.5 .1 need not have a least fixed point.
${ }^2$ For the details see appendix A.

Ankit Gupta

Numerade Educator

Problem 6

Show that the axiom of choice is equivalent to the assertion that every partially ordered set has a maximal totally ordered subset. [Necessity. Use the Teichmüller/ Tukey property. Sufficiency. Prove Zorn's lemma.]

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03:38

Problem 7

Assuming the axiom of choice, prove that every partially ordered set $(A, \leqslant)$ has a maximal totally unordered subset. [First method. Use the well-ordering principle. Second method. Show that the set of all totally unordered subsets of $\mathrm{A}$ is of finite character and then use the Teichmuller/Tukey property.]

Prathan Jarupoonphol

Numerade Educator

08:12

Problem 8

Assuming the axiom of choice, prove that a relation is a partial ordering [resp. partial well-ordering] on the set $A$ iff it is the intersection of a set of total orderings [resp. well-orderings] on $A$.

Prathan Jarupoonphol

Numerade Educator

Problem 9

Assuming the axiom of choice, prove that every partially ordered set $(A, \leqslant)$ has a cofinal partially well-ordered subset. [Let $\mathcal{A}$ be the set of partially well-ordered subsets of $A$. Apply Zorn's lemma to $\mathcal{A}$ with the partial ordering 'is an initial subset of.]

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