Errata to "The Mathematics of Logic"
- Page 1, line 3 from bottom: remove the word
means. (The fact
go downwards does not rule out loops or cycles, at least
for some idea of
graph but a tree is not allowed to have loops. Note
that my definition of tree is not the same as some others you will find, for
example in graph theory. In particular my trees are rooted trees, meaning
they have a special designated node as root.)
- Page 5, definition 1.5:
A tree is a non-empty set of sequences of finite length such
that... (Add the words
of finite length.)
- Page 8, line 5:
does not appear should be
was an unintentional double negative: a sequence is -free
if it has no subsequence of form .)
- Page 14, line 6: should read . (Upper-case X.)
- Page 15, last line: should read
. (Upper-case X.)
- Page 17, line 8 (first line of proof):
for ... (Correcting an obvious typo.)
- Page 20, Exercise 2.34. The ring should be taken to
have an element .
- Page 21, proof of Zorn's lemma: in the proof, the function
needs to return a strict bound of the chain , and
the proof originally given that any two chains in one is an
initial segment of the other can be clarified. Replace the second
paragraph of this proof with:
We first apply the Axiom of Choice. Considering
as a non-empty set, and the set of all non-empty subsets
of , by the Axiom of Choice there is a function
such that for all . Now let
be a chain. By the Zorn property there is
some upper bound, , for . Since has no maximum element
there is with . In other words, the
is non-empty and hence in . Thus
is a strict upper bound for . Composing functions
we obtain a function
such that is a strict upper bound of
whenever is a chain.
- Page 22, replace the first complete paragraph on page 22 with:
Suppose to start with that there is which is
not in . Then there is a least such
is an initial segment of as required,
so suppose not. Let be least in
Observe that if with then
, and as is a chain we must have
since is impossible.
Thus and it follows that contradicting our assumption that is not in .
So is an initial segment of .
If there is which is not in then
a similar argument shows is an initial segment
of , and if neither of these applies then
- Page 59, line 2 of the proof of Proposition 5.17:
least such should read
- Page 70, line 4 from bottom: add the word
just before .