Exercises and examples from Chapter 2

The axioms for a poset are universal or ${\forall }_1$, that is to say that each axiom is of the form $\forall x \forall y \dots \forall z \theta (x,y,\dots ,z)$ where $\theta$ is quantifier-free. (See Chapter 9 on first order logic.)

Given $<$ on $X$ let $x⩽y$ be defined by $x<y\vee x=y$. Then the axioms in Definition 2.4 hold. For example if $x⩽y\wedge y⩽x$ then $x=y$ since the other case, $x<y$ and $y<x$ is forbidden by irreflexivity, since by transitivity it implies $x<x$. Now let $x<\prime y$ be defined by $x⩽y\wedge x\ne y$. So $x<\prime y\iff (x<y\vee x=y)\wedge x\ne y\iff x<y$ so $<$ is recovered this way. For the other case, given $⩽$ on $X$ satisfying Definition 2.4 let $x<y\iff x⩽x\wedge x\ne y$ and observe that the axioms in Definition 2.1 hold. For example if $x<y$ and $y<z$ then $x⩽z$ by transitivity of $⩽$ and if $x=z$ then $x<y$ and $y<x$ so by the definition of $<$, $x⩽y$ and $y⩽x$ so by Definition 2.4 $x=y$, so $x<y$ is false, a contradiction. Therefore $x\ne z$ and so $x<z$. Now let $x⩽\prime y\iff x<y\vee x=y$. So $x⩽\prime y\iff x=y\vee (x⩽y\wedge x\ne y)\iff x⩽y$ since $x⩽y$ holds when $x=y$ by Definition 2.4. So $⩽$ is recovered.

$\sim$ is an equivalence relation as $x⩽x$ holds by axiom (ii) of a preorder so $\sim$ is reflexive; if $x⩽y$ and $y⩽x$ and also $y⩽z$ and $z⩽y$ then $x⩽z$ and $z⩽x$ by transitivity of $⩽$ twice; and clearly the definition $x\sim y$ is symmetric in $x,y$. Definte $\left[x\right]⩽\left[y\right]$ to mean $x⩽y$. This is well-defined for if $x⩽y$ and $x\prime \sim x$, $y\prime \sim y$ then $x⩽x\prime$, $x\prime ⩽x$, $y⩽y\prime$, $y\prime ⩽y$, so by transitivity (twice) $x\prime ⩽y\prime$. Thus the definition of $\left[x\right]⩽\left[y\right]$ does not depend on the choice of representatives of the equivalence classes $\left[x\right]$, $\left[y\right]$.

$⩽$ is a partial order on $X/\sim$. Transitivity and reflexivity are easy and follow from properties on $⩽$ on $X$. For the remaining axiom (iii), suppose $\left[x\right]⩽\left[y\right]$ and $\left[y\right]⩽\left[x\right]$. Then by the previous paragraph $x⩽y$ and $y⩽x$ so $x\sim y$ and hence $\left[y\right]=\left[x\right]$.

(Proof of Theorem 2.12)

If $(X,⩽)$ is directed and $u,v\in X$ are maximal then there is $w\in X$ with $u⩽w$ and $v⩽w$. But $uw$ and $vw$ since they are both maximal. So $u=w$ and $v=w$ hence $u=v$.

We can show there is a surjection $f:\mathbb{N}\to \mathbb{Q}$. It follows that $\mathbb{Q}$ is countable as $\mathbb{Q}=\left\{f\left(n\right):n\in \mathbb{N}\right\}$. Observe that there is a bijection $\mathbb{N}\times \mathbb{N}\to \mathbb{N}$ (see Figure 11.1, p163 for a hint) and both $\mathbb{Z}$ and $\mathbb{N}\setminus \left\{0\right\}$ are countable. Then composing functions we have a surjection

where the last arrow is $(n,k)\mapsto n/k$.

If $f:\mathbb{N}\times \mathbb{N}\to \bigcup \left\{X_i:i\in \mathbb{N}\right\}$ is given as shown then there is a surjection

obtained by composing with the bijection $\mathbb{N}\to \mathbb{N}\times \mathbb{N}$ of Figure 11.1.

Follow the hint given.

If $P=P\left(X\right)$ and $f:X\to P$, let $Y=\left\{x\in X:x\notin f\left(x\right)\right\}$. If $f$ is onto then $Y=f\left(y\right)$ for some $y\in X$. If $y\in f\left(y\right)$ then $y\notin Y$ by the definition of $Y$ hence $y\notin f\left(y\right)$, and if $y\notin f\left(y\right)$ then $y\in Y$ so $y\in f\left(y\right)$. This gives the contradiction.

Then if the condition to the right of the $:$ here is false for some $T\subseteq G$ there are $t_1\ne t_2\in T$ and $h_1,h_2\in H$ with $h_1{t}_1=h_2{t}_2$. In other words every $T\subseteq G$ containing $t_1,t_2$ failes to be in $X$, which shows the condition in Proposition 2.20 holds.

A set $U\subseteq V$ is linearly independent if whenever $n\in \mathbb{N}$ and $u_1,\dots ,u_n\in U$ all distinct and ${\lambda }_1,\dots ,{\lambda }_n\in F$ with ${\lambda }_1{u}_1+\dots +{\lambda }_n{u}_n=0$ then ${\lambda }_1=\dots ={\lambda }_n=0$. (Note that there is in general no way of adding infinitely many vectors from $V$.) A basis is a maximal linear independent set. The collection $X$ of linearly independent sets has the Zorn property bu Proposition 2.20 since if $U$ is dependent there is a finite subset $\left\{u_1,\dots ,u_n\right\}\subseteq U$ which is dependent and any $U\prime$ containing this is also dependent. So a maximal independent set exists by Zorn's lemma. If $B\subseteq V$ is such a set and $v\in V$ then there are $n\in \mathbb{N}$ and $u_1,\dots ,u_n\in B$, ${\lambda }_1,\dots ,{\lambda }_n\in F$ such that ${\lambda }_1{u}_1+\dots +{\lambda }_n{u}_n=x$. For if not one cna show that $B\cup \left\{x\right\}$ is independent, contradicting maximality of $B$. For if $u_1,\dots ,u_n\in B$ are distinct with ${\lambda }_1{u}_1+\dots +{\lambda }_n{u}_n+\mu x=0$ then $\mu \ne 0$ and $x={\mu }^{-1}{\lambda }_1{u}_1+\dots +{\mu }^{-1}{\lambda }_n{u}_n$.

Given a bijection $f:B\to C$ define

for each ${\lambda }_1,\dots ,{\lambda }_n\in F$ and $u_1,\dots ,u_n\in B$. This is well-defined and linearly maps $V\to W$ since each $v\in V$ is ${\lambda }_1{u}_1+\dots +{\lambda }_n{u}_n$ for some unique choice of non-zero ${\lambda }_1,\dots ,{\lambda }_n\in F$ and $u_1,\dots ,u_n\in B$, and $F$ has an inverse map

As you can check. (There are a number of detailed checks omitted that the reader must do here.)

The set of ideals containing $I$ is a poset with the Zorn property by Proposition 2.20. So a maximal ideal $M\supseteq I$ exists. $R/M$ is a field since if $x\in R$ and there is no $y$ with $xy\in M$ then $M\cup \left\{x\right\}$ generates an ideal properly extending $M$. The details are left to the reader.