This web page provides answers, or hints and clues, to exercises
from Chapter 2 of
The Mathematics of Logic and also some additional
information on some of the examples. As elsewhere in the book
means "and" and means "or".
The axioms for a poset are universal or , that is to say that each axiom is of the form where is quantifier-free. (See Chapter 9 on first order logic.)
Given on let be defined by . Then the axioms in Definition 2.4 hold. For example if then since the other case, and is forbidden by irreflexivity, since by transitivity it implies . Now let be defined by . So so is recovered this way. For the other case, given on satisfying Definition 2.4 let and observe that the axioms in Definition 2.1 hold. For example if and then by transitivity of and if then and so by the definition of , and so by Definition 2.4 , so is false, a contradiction. Therefore and so . Now let . So since holds when by Definition 2.4. So is recovered.
is an equivalence relation as holds by axiom (ii) of a preorder so is reflexive; if and and also and then and by transitivity of twice; and clearly the definition is symmetric in . Definte to mean . This is well-defined for if and , then , , , , so by transitivity (twice) . Thus the definition of does not depend on the choice of representatives of the equivalence classes , .
is a partial order on . Transitivity and reflexivity are easy and follow from properties on on . For the remaining axiom (iii), suppose and . Then by the previous paragraph and so and hence .
(Proof of Theorem 2.12)
If is directed and are maximal then there is with and . But and since they are both maximal. So and hence .
We can show there is a surjection . It follows that is countable as . Observe that there is a bijection (see Figure 11.1, p163 for a hint) and both and are countable. Then composing functions we have a surjection
where the last arrow is .
If is given as shown then there is a surjection
obtained by composing with the bijection of Figure 11.1.
The result in Exercise 2.27 requires care: it shows that under certain conditions
a countable union of countable sets is easily seen to be countable. (The conditions
required are that each is countable by a single function
Follow the hint given.
If and , let . If is onto then for some . If then by the definition of hence , and if then so . This gives the contradiction.
Then if the condition to the right of the here is false for some there are and with . In other words every containing failes to be in , which shows the condition in Proposition 2.20 holds.
A set is linearly independent if whenever
and all distinct and
then . (Note that there is
in general no way of adding
infinitely many vectors from .)
A basis is a maximal linear independent set. The collection of
linearly independent sets has the Zorn property bu Proposition 2.20 since if
is dependent there is a finite subset
which is dependent and any containing this is also dependent. So a maximal
independent set exists by Zorn's lemma. If
is such a set and then there are and
. For if not
one cna show that is independent, contradicting maximality
of . For if are distinct with
Given a bijection define
for each and . This is well-defined and linearly maps since each is for some unique choice of non-zero and , and 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 is a poset with the Zorn property by Proposition 2.20. So a maximal ideal exists. is a field since if and there is no with then generates an ideal properly extending . The details are left to the reader.