Topology
Selected Solutions to Homework 1
Instructor: W. D. Gillam
Section 10 Exercise 2: (a) The issue is to show that every non-maximal element x of a
well-ordered set I has a(n immediate) successor. Since x is non-maximal,
S := {y ∈ I : x < y}
is non-empty, so, since I is well-ordered, it has a minimal element z which is easily seen
to be the successor of x.
(b) Z is probably the most obvious example
Supplementary Exercise 1: We first prove, by transfinite induction, that for each
β ∈ J, there exists a unique function hβ : Sβ → C such that hβ (α) = ρ(hβ |Sα ) for every
α < β (i.e. for every α ∈ Sβ ).
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Case 1: β = γ + 1 is a successor. Then Sβ = Sγ {γ} and we know by induction that
there is a unique hγ : Sγ → C such that hγ (α) = ρ(h|Sα ) for each α ∈ Sγ . Define
hβ : Sβ → C by setting hβ (α) := hγ (α) for α ∈ Sγ ⊂ Sβ and hβ (γ) := ρ(hγ : Sγ → C).
Using the property of hγ you easily check that hβ has the requisite property and using
the uniqueness of hβ you can easily check the uniqueness of hβ (we had no choice in our
definition of hβ (γ)... you don’t even have to think!).
Case 2: β is a limit, so Sβ = ∪γ<β Sγ . By induction, we have, for each γ < β a unique
hγ : Sγ → C satisfying the usual property. If we pick any γ, γ 0 with γ ≤ γ 0 < β, then it
is easy to check that hγ 0 |Sγ = hγ (just note that the function on the left hand side has
the usual property and appeal to the uniqueness of hγ ), so the functions hγ for γ < β
determine a function hβ := ∪γ<β hγ : Sβ → C. You can easily check that this function is
the unique function with the desired property.
To complete the proof, one uses the existence and uniqueness of the hβ from above to
prove the existence and uniqueness of the desired h, arguing exactly as in Case 1 if P has
a maximal element, or exactly as in Case 2 if not.
Supplementary Exercise 2: (a) I’ll just do “(i) implies (ii)” and leave the other implication to you. Suppose, toward a contradiction, that h satisfies (i) but there is some
β for which h(β) is not given by the formula in (ii). First note that h(β) cannot be in
h(Sβ ), for then h would not even be injective (Why?), let alone order-preserving as we
assume in (i). (Using the book’s definition of “order-preserving” on Page 25 (you have
to look in the Definition of order-type), it is clear that any order-preserving map out of a
totally-ordered set has to be injective. This isn’t completely standard since some people
use “order-preserving” to mean that the relation ≤ is preserved, in which case a constant
map is order-preserving.) So h(β) must be in E \ h(Sβ ), but is not the minimum element
of E \ h(Sβ ). This means that there is some α ∈ E \ h(Sβ ) with α < h(β). But since h is
an order-preserving we must then have
α < h(β) ≤ h(γ)
for every γ ∈ J with β ≤ γ. Together with the fact that α ∈
/ h(Sβ ), this means that α
can’t be in the image of h (it isn’t the image of any element less than β or the image of
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any element greater than or equal to β). This contradicts the assumption in (i) that h(J)
is a segment of E because h(J) contains h(β), but not the strictly smaller element α.
(b) I’ll just show that no section Sβ of a well-ordered set E has the order-type of E and
leave the rest to you. Suppose we have an order preserving bijection h : Sβ → E. Note
that such an h satisfies (i) from part (a), so it must also satisfy (ii) from part (a) by what
we just proved. We can arrive at a contradiction by proving that h(α) = α for all α ∈ Sβ
(for then h can’t be surjective since β won’t be in its image). This last claim is proved
by contradiction: If it fails, then since Sβ is well-ordered, there is a minimal α for which
h(α) 6= α. On the other hand, the minimality of α and the formula for h(α) from (a)(ii)
make it clear that h(α) = α.
Supplementary Exercise 3: The hint given in the textbook is enough. All I will say is
that many of you mistakenly tried to prove that k itself is an isomorphism onto a segment
of E, which is not necessarily the case (take J = E = N, k(n) := n + 1, for example).
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Supplementary Exercise 4: Define an ordering ≤ on P := A B using the orderings ≤
on A and B by declaring p ≤ q iff one of the following occurs: p ∈ A ⊆ P and q ∈ B ⊆ P ,
p, q ∈ A ⊆ P and p ≤ q in A, or p, q ∈ B ⊆ P and p ≤ q in B. Obviously this makes
P into a well-ordered set such that the inclusions A, B ⊆ P are order-preserving. By the
previous exercise this implies that A (resp. B) is isomorphic to a segment Sα (resp. Sβ ) of
P . Now it is clear that at least one of the three indicated possibilities occurs, according
(respectively) to the cases α = β, α < β, β < α. The fact that no two of these possibilities
can occur simultaneously follows pretty easily from 2(b).
Supplementary Exercise 5: (a) is completely routine: you just check the axioms
(b) Suppose S is a non-empty subset of (B 0 , <0 ). We need to show that S has a minimal
element. Pick any b ∈ S ⊆ B 0 . By definition of B 0 , there is some (B, <) ∈ B with
b ∈ S ∩ B. (Note also that B ⊆ B 0 by definition of B 0 .) Since S ∩ B is a non-empty subset
of B and (B, <) is well-ordered, there is a <-minimum element c ∈ S ∩ B. I claim that
c is the <0 -minimum element of S. Consider an arbitrary element t ∈ S, not equal to c,
and let us show that c <0 t. By definition of B 0 , there is some (C, <) ∈ B with t ∈ S ∩ C.
Since B is simply ordered by the relation described in (a), one of the following occurs:
Case 1: (C, <) is a segment of (B, <) or is equal to (B, <). Then S ∩ C ⊆ S ∩ B, so our t
is an element of S ∩ B and hence c < t by minimality of c and then c <0 t by definition of
<0 .
Case 2: (B, <) is a segment of (C, <). One possibility is that t ∈ S ∩ C is actually in the
subset S ∩ B, in which case we argue exactly as in the previous case. The other possibility
is that t is not in the segment (B, <) of (C, <), but this means that any element of B is
< t, so in particular c < t, and hence c <0 t.
Supplementary Exercise 6: We first prove that the maximum principle implies the
well-ordering principle. Let X be any set. We want to prove that X can be well-ordered,
assuming the maximum principle. Define the poset A as in 5(a). By the maximum
principle, this poset has a maximal chain B. Define (B 0 , <0 ) as in 5(b). That exercise
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says that (B 0 , <0 ) is well-ordered, so the issue is to show that the inclusion
` B ⊆ X is an
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equality. Suppose not. Then there is some x ∈ X \ B . Let B := B {x}. Define an
ordering < on B 00 in the obvious way so that < |B 0 =<0 and x is the maximum element of
(B 00 , <). Then (B 00 , <) is well-ordered and (B 0 , <0 ) is the segment Sx of it. Using this you
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can check that (B 00 , <) ∈ A is greater than (according to the order on A) every element of
B, but it isn’t in B (if it were, then x would be in B 0 by definition of B 0 ), so maximality
of B is contradicted.
I already gave you a pretty good hint for the converse: You want to show that an arbitrary poset (P, ≤) has a maximal chain (assuming the well-ordering principle). Obviously
we want to make use of the well-ordering principle, so choose a well-ordering of the set P .
For α ∈ P , write Sα for the section of P with respect to our chosen well-ordering (this has
nothing to do with the other partial order ≤). Apply the Principle of Recursive Definition
(as formulated in Supplementary Exercise 1) with J = P (with our chosen well-ordering),
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C = {0, 1}, and ρ defined by setting ρ(f : Sα → {0, 1}) := 0 iff the subset f −1 (0) {α} of
P is a chain with respect to the ordering ≤ (i.e. is simply ordered by ≤). That Principle
furnishes a function h : P → {0, 1} with a certain nice property—use this nice property
to show that h−1 (0) is a maximal chain in (P, ≤).