A TASTE OF SET THEORY: EXERCISE NUMBER 6
BOAZ TSABAN
1.
S Assume that λ < cf(κ) and for each α < λ, |Xα | < κ. Show that |
α<λ Xα 6= κ).
[Hint. Define f : λ → κ by f (α) = |Xα |. f is bounded.]
S
α<λ
Xα | < κ (in particular,
2. A coloring of (the elements of) a set A with λ many colors (λ may be finite or infinite) is a function
f : A → λ (so that each element a ∈ A is given the “color” f (a)). A subset C ⊆ A is monochromatic if f is
constant on C (i.e., all elements in C were given the same color).
Consider the following Ramsey-theoretic assertion concerning A, λ, and another cardinal κ, which may
be true or false:
A → (κ)1λ : For each coloring f of A with λ colors, there is a monochromatic C ⊆ A such that |C| = κ.
Prove the following assertions:
(1) ω → (ω)1n for all n ∈ N. (Prove this assertion directly.)
(2) A cardinal κ is regular if, and only if, κ → (κ)1λ for all λ < κ.
(3) Explain why (1) follows from (2).
[Hint. Using the results from class, this interesting exercise is very easy.]
3. Prove the following assertions:
(1) For each limit ordinal α, cf(ℵα ) = cf(α).
[Hint. It suffices to show that there is an increasing unbounded f : α → ℵα .]
(2) Compute: cf(ℵℵ1 ), cf(ℵℵℵ1 ).
4. Consider the following assertions:
S
AC: For set F whose members are nonempty sets, there is a choice function on F. (i.e., h : F → F
such that h(A) ∈ A for all A ∈ F.)
AC1 : For each A, B, if there is an onto f : A → B, then there is g : B → A such that f ◦ g is the identity
function.
You already proved that AC implies AC1 . Show that AC1 implies AC.
S
[Hint. Define f = {((a, A), A) : a ∈ A ∈ F}. f is a set. f is a function, from ( F) × F to F. Let g be as
in AC1 for f . From g we can define a choice function on F. (Thanks to Micki for the trick.)]
Good luck!
Department of Applied Mathematics and Computer Science, The Weizmann Institute of Science, Rehovot 76100,
Israel
E-mail address: [email protected]
URL: http://www.cs.biu.ac.il/~tsaban
Date: 11 Apr 05 ce.