Download Problem Set 4 - MIT Mathematics

18.510 Introduction to Mathematical Logic and Set Theory
Problem Set #4 (due Thursday, October 8)
On problems 5 and 6, you don’t need to be super-picky about justifying every
assertion about finite or countable sets. Everything you say should be true, and
you should give a correct mathematical proof in the usual classroom/journal
style, but don’t worry about justifying things like the union of finite sets being
finite. These sorts of things aren’t hard once you are used to them, and they are
a distraction from the main point of these problems. By contrast, in problems
2 and 3, justifying these sorts of intuitively obvious facts is the point, and in
problem 4, you should make sure not to use the Axiom of Choice by accident.
1. Let T be a well-ordered set, and give S ⊆ T the induced well-ordering. (I.e.,
order the elements of S the same way as in T .) Prove that type(S) ≤ type(T )
as ordinals, where type(S) denotes the unique ordinal isomorphic to S.
2. Prove that a set S is infinite (i.e., not in bijection with an element of ω) if
and only if there is an injective map from ω to S.
3. Prove that a set S is infinite if and only if there is a bijection from S to one
of its proper subsets.
4. An antichain in a poset is a set of incomparable elements in the poset.
Without assuming the Axiom of Choice as an axiom, prove that the following
are equivalent:
1. The Axiom of Choice.
2. Every poset contains a maximal antichain (i.e., an antichain not contained
in any larger antichain) and every set can be totally ordered.
Extra credit: prove that the Axiom of Choice is equivalent to “every poset
contains a maximal antichain.” (This is much harder to prove, so it’s not worth
spending time on it unless you’ve finished everything else and are looking for a
challenge.)
5. A subset S of a poset T is called cofinal if for every t ∈ T , there exists s ∈ S
such that s ≥ t.
1. Prove that every totally ordered set T has a cofinal subset that is wellordered (in the ordering it inherits from T ).
2. Let α be the least uncountable ordinal, viewed as a poset under the ordering
∈. Prove that α has no countable cofinal subset. The word “least” is
important here; give an example of a larger ordinal than α that does have
a countable cofinal subset.
6. Let S be a set, and let F (standing for “forbidden”) be a subset of P(S). Say
a subset G ⊆ S is good if no element of F is a subset of G. Intuitively, a good
subset is one that doesn’t contain a forbidden subset.
1
1. Suppose every element of F is finite and ∅ 6∈ F . Prove that S contains a
maximal good subset (i.e., a good subset not contained in any larger good
subset).
2. Without the assumption that every element of F is finite, construct an
example in which S has no maximal good subset.
As motivation for those who have studied linear algebra, let S be the set of
vectors in a given vector space, and let F be the set of finite, linearly dependent
subsets of S. Then a subset of S is good if and only if it is linearly independent,
and a maximal good subset is a basis of the vector space. Thus, part 1 implies
that every vector space has a basis. However, no knowledge of linear algebra is
required to solve this problem.
2