IS THERE A SET THAT IS NOT ALMOST OPEN BUT STILLMEASURABLE IN R
Di Wu
Submitted under the supervision of Professor Scot Adams to the UniversityHonors Program at the University of Minnesota-Twin Cities in partial fulfillment
of the requirements for the degree of Bachelor of Arts, summa cum laude inMathematics.
Department of MathematicsUniversity of Minnesota-Twin Cities
The United StatesMay, 9th , 2016
IS THERE A SET THAT IS NOT ALMOST OPEN BUT STILLMEASURABLE IN R?
Di Wu
A THESIS
in
Mathematics
Presented to the Faculties of the University of Minnesota-Twin Citiesin Partial Fulfillment of the Requirements for the Honors Degree ofBachelor of Arts, summa cum laude in Mathematics.
Spring, 2016
Supervisor of Dissertation
Scot Adams, Professor of Mathematics
Thesis Readers:Karel Prikry, Professor of MathematicsWayne Richter, Professor of Mathematics
Acknowledgments
First and foremost, I have to thank my thesis supervisors, Professor Scot Adams.
Without his assistance and dedicated involvement in every step throughout the
process, this paper would have never been accomplished. I would like to thank you
very much for your support and understanding over the past year.
I would also like to show gratitude to my committee, including Professor Karel
Prikry and Professor Wayne Richter. Without the help of Professor Karel Prikry,
we would not be able to figure out such an easier way to construct a not almost
open and measurable set.
The first time that I met Professor Scot Adams is through his lectures for GRE
math subject at the University of Minnesota. His teaching style and enthusiasm
for the topic made a strong impression on me and I have always carried positive
memories of his lectures with me. After several meetings, I decided to discuss the
possible Honors Thesis topics with Professor Scot Adams. He thought the possible
topics very carefully and raised many precious points in our discussion.
Getting through my thesis required more than academic support, and I have
ii
many, many people to thank for listening to and, at times, having to tolerate me
over the past three years. I cannot begin to express my gratitude and appreciation
for their friendship. Jun Li, Hanyu Feng, Eva Lian, Ronald Siegel, Carme Calderer,
Rina Ashkenazi and Andy Whitman have been unwavering in their personal and
professional support during the time I spent at the University. For many memorable
evenings out and in, I must thank everyone above as well as Sue Steinberg, and Matt
Hanson. I would also like to thank Eunice Lee and Christina Gee who opened both
their home and heart to me when I first arrived in the city.
Most importantly, none of this could have happened without my family. My
dad, who offered his encouragement during my hardest time. With his care, I
finally went through a really tough period of my life. To my parents it would be an
understatement to say that, as a family, we have experienced some ups and downs
in the past three years. Every time I was ready to quit, you did not let me and I
am forever grateful. This thesis stands as a testament to your unconditional love
and encouragement.
iii
ABSTRACT
IS THERE A SET THAT IS NOT ALMOST OPEN BUT STILL
MEASURABLE IN R
Di Wu
Professor Scot Adams, Supervisor
This paper demonstrates a simple way to construct a not almost open and
nonmeasurable set by taking the union of a Bernstein set and a meager and conull
set.
iv
Contents
1 Introduction 1
2 Not Almost Open and Measurable 2
2.1 Nowhere Dense and Almost Open . . . . . . . . . . . . . . . . . . . 2
2.2 Split and Continuum Cardinality . . . . . . . . . . . . . . . . . . . 10
2.3 A Not Almost Open but Measurable Set . . . . . . . . . . . . . . . 16
v
Chapter 1
Introduction
The motivations for this research comes from an analogy between measure theory
and the Baire category theory. In measure theory, to describe something“small”,
we have the concept of measure 0. While in the Baire category theory, to describe
“small” sets, we have the concept of meager sets. In addition, the idea of measurable
set is analogical to the concept of almost open in Baire category theory. However,
most of sets we deal with are often both measurable and almost open. Moreover,
when we try to construct a not almost open set, it is very easy to make this set non-
measurable simultaneously. So an interesting topic comes up: Could we construct
a not almost open but measure set?
1
Chapter 2
Not Almost Open and Measurable
2.1 Nowhere Dense and Almost Open
Definition 2.1. Let S ⊆ R. Then S is nowhere dense if (S)o=∅.
Notes:
1) Loosely speaking, nowhere dense set is a set whose elements are not tightly
clustered anywhere in the defined topological space.
2) This definition can also be interpreted as saying that S is nowhere dense in the
defined topological space T if S contains no nonempty open set of T .
3) Any subset of a nowhere dense set is nowhere dense, obviously from the definition
and the elementary properties of closure and interior.
3) Some examples:
i) ∅ is nowhere dense since (∅)o=∅.
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ii) The boundary of an open (closed) set is nowhere dense ( Lemma 2.1).
iii) The Cantor set is nowhere dense ( Lemma 2.2).
Lemma 2.1. Let U ⊆ R be open or closed. Then ∂U is closed and nowhere dense.
Proof. On one hand, ∂U= U \ (U o), where U is closed and U o is open. Since we
know a closed set \ a open set= a closed set, then ∂U is closed. On the other hand,
(∂U)o=(∂U)o=(U \ (U o))o= (U)o ∩ ((U o)c)o=(U)o ∩ ((U o))c=((U)o) \ (U o).
1). If U is open, then U o = U . Thus (∂U)o=((U)o) \ (U o)=((U)o) \ U=∅.
2). If U is closed, then U=U . Thus (∂U)o=((U)o) \ (U o)=(U o) \ (U o)=∅.
So ∂U is nowhere dense. Thus ∂U is closed and nowhere dense.
Lemma 2.2. The (standard or ternary) Cantor set C is nowhere dense and has
Lebesgue measure 0 (Steen,1995).
Proof. Construction of Cantor ternary set is created by deleting the open middle
third from each of a set of line segment repeatedly. I start with C0 = [0, 1].
Step 1. Delete the open middle third (13, 23) with Lebesgue measure (20 ∗ 1
31= 1
3)
from [0, 1]. The remainder is the union of two line segments [0, 13] ∪ [2
3, 1]. Call this
set C1.
Step 2. Delete the open middle third of each of these remaining segments, i.e.
(19, 29) ∪ (7
9, 89) with Lebesgue measure (21 ∗ 1
32= 2
9). The remainder is the union of
four line segments [0, 19] ∪ [2
9, 13] ∪ [2
3, 79] ∪ [8
9, 1]. Call this setC2.
...
3
Step n. Continue to delete the open middle third of each of these remaining seg-
ments. The removed length at this step is (2n−1 ∗ 13n
= 2n−1
3n) and the remainder is
Cn = Cn−1
3∪ (2
3+ Cn−1
3).
...
Continue this infinite process, the Cantor ternary set is defines as:
C := lim∞n=1Cn=∩∞n=1Cn =[0, 1] \ ∪∞n=1 ∪2n−1−1m=0 (3m+1
3n, 3m+2
3n).
The Cantor set is the set of points in [0, 1], which are not removed. By the construc-
tion, the Cantor set cannot contain any interval with positive length. Otherwise,
suppose it contains some ε > 0 length interval. The reason is ∀ integer n >= 0, Cn
contains the interval of length 1/3n. Since C = ∩∞n=1Cn and 1/3n → 0, as n → ∞,
C contains no interval of positive length.
Definition 2.2. Let S ⊆ R. Then S is meager means:
∃S1, S2, ... ⊆ S, s.t. S =S1 ∪ S2 ∪ ... and s.t. ∀j ∈ N, Sj is nowhere dense.
Fact: Any subset of meager set is meager.
Proof. Let S be a meager set. Then ∃S1, S2, ... ⊆ S, s.t. S =S1 ∪ S2 ∪ ... and s.t.
4
∀j ∈ N, Sj is nowhere dense. For any subset A of S, and Sj ∩ A is nowhere dense
since Sj ∩A is the subset of nowhere dense set Sj. Therefore, any subset of meager
set is meager.
Definition 2.3. Let S ⊆ R. Then S is called an almost open set if ∃ an open set U
⊆ R,∃ meager sets Z,Z ′ ⊆ R s.t. S=(U \ Z) ∪ Z ′.
Notes:
1) I will denote A =∗ B for convenience iff ∃ meager sets Z,Z ′, s.t.(A\Z)∪Z ′ = B.
Figure 2.1: A =∗ B
2) Fact: A =∗ B iff A \B and B \ A are meager.
Proof. On one hand, if A =∗ B, choose meager sets Z,Z ′ s.t. (A \ Z) ∪ Z ′ = B.
A \B =A \ [(A \ Z) ∪ Z ′] ⊆ A \ (A \ Z) ⊆ Z.
B \ A=[(A \ Z) ∪ Z ′] \ A ⊆ (A ∪ Z ′) \ A ⊆ Z ′.
Since A \B and B \ A are subsets of meager sets, A \B and B \ A are meager.
On the other hand, if A\B and B \A are meager, since [A\ (A\B)]∪ (B \A) = B,
we know A =∗ B.
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3) Based on Fact 2), obviously we have A =∗ B iff A4B is meager.
4) Fact: If A =∗ B, then (R \ A)=*(R \B).
Proof. Because A =∗ B, A \B and B \ A are meager.
Then (R \ A) \ (R \B)=Ac \Bc=Ac ∩ (Bc)c=B ∩ Ac= B \ A.
Similarly, (R \B) \ (R \ A)= A \B. So (R \ A)=*(R \B).
3) This definition can be easily applied to define almost closed by changing the open
set U to some closed set C from Definition 1.3.
Definition 2.4. Let S ⊆ R, S is almost closed if ∃ clsoed set C ⊆ R s.t. S =∗ C.
Lemma 2.3. Let S ⊆ R be almost open. Then R \ S is almost open.
Proof. 1) First I will prove that R \ S is almost closed.
Since S ⊆ R be almost open, ∃ an open set U ⊆ R,∃ meager sets Z,Z ′ ⊆ R, s.t.
S=(U \Z)∪Z ′, written as S=*U. By the previous fact, R \S=*R \U , where R \U
is a close set.
2) Then I will prove that almost closed also implies almost open.
Recall the lemma 1.1, ∂U is closed and nowhere dense.
Thus R \ S=*((R \U) \ ∂U)∪ ∂U=*(R \U)∪ ∂U=* R \U . Since R \U is an open
set, R \ S is almost open.
Lemma 2.4. Let S1, S2, ... ⊆ R all be almost open. Then S1∪S2∪ ... is almost open.
Proof. Since S1, S2, ... ⊆ R are all almost open, ∃ open sets U1, U2, ..., s.t. S1=*U1, S2=*U2....
Then for i = 1, 2, ..., Si \ Ui and Ui \ Si are meager. What we want to prove is
6
(∪∞i=1Si) \ (∪∞i=1Ui) and (∪∞i=1Ui) \ (∪∞i=1Si) are meager.
(∪∞i=1Si) \ (∪∞i=1Ui) = (∪∞i=1Si) ∩ (∪∞i=1Ui)c = (∪∞i=1Si) ∩ (∩∞i=1U
ci ) ⊆ ∪∞i=1(Si ∩ U c
i ).
By the definition of meager set, it is obvious that a countable union of meager sets
is still meager. Also by the symmetry of (∪∞i=1Si)\ (∪∞i=1Ui) and (∪∞i=1Ui)\ (∪∞i=1Si),
we get what we want. Therefore S1 ∪ S2 ∪ ... is almost open.
Definition 2.5. Let S ⊆ R. Then S is negligible means : ∀ε > 0,∃ intervals
I1, I2, ... ⊆ R, s.t. S ⊆∞⋃j=1
Ij and∞∑j=1
[(sup Ij)− (inf Ij)] < ε.
Fact: Let S1, S2, ... ⊆ R all be negligible. Then S1 ∪ S2 ∪ ... is negligible.
Proof. Let ε > 0 be given, we seek a covering of S1 ∩ S2 ∩ S3... by countable many
intervals, whose total lengths < ε. Since for any given i ∈ N, Si is negligible, for each
Si, choose intervals Ii1, Ii2... s.t. Si ⊆∞⋃j=1
Iij and s.t.∞∑j=1
[(sup Iij) − (inf Iij)] <ε2i
.
Then∞⋃i=1
Si ⊆∞⋃i=1
∞⋃j=1
Iij, and∞∑i=1
∞∑j=1
[(sup Iij) − (inf Iij)] <∞∑i=1
ε2i
= ε. So∞⋃i=1
Si is
negligible.
Intuitively, both meager and negligible are used to describe “small” sets. However,
essentially they are very different from each other. In addition, one can find a mea-
ger set whose complement is negligible. Two examples are given below. Before we
get into these examples, I would like to introduce a collection of important sets,
called “Fat Cantor set”, which can be viewed as an extension of the Cantor set.
The Fat Cantor Set (Smith-Volterra-Cantor set, wiki). The construction of the
Fat Cantor set is very similar to the construction of the Cantor set. The Fat Cantor
7
set is constructed by removing certain intervals from the unite interval [0,1]. For
example, we are able to make a Fat Cantor set of measure 12
through the following
process.
Step 1. Remove the middle 14
from the interval [0,1] and the remainder is
[0, 38] ∪ [5
8, 1], called C ′1.
Step 2. Remove subintervals of width 116
from the middle of each segment intervals,
and the remainder is [0, 532
] ∪ [ 732, 38] ∪ [5
8, 2532
] ∪ [2732, 1], called C ′2.
...
Step n. Continuing removing subintervals of width 122n
from the middle of each of
the 2n−1 remaining segments, called C ′n.
...
Continuing this infinite process, the Fat Cantor set is defined as: C ′ = ∩∞i=1C′i.
Each subsequent iteration removes proportionally less from the remainder while
the Cantor set removes proportion as a constant from each remainder. The total
measure removed is equal to∞∑n=0
2n
22n+2 = 12, so C ′ has measure 1
2, which is positive.
Also C ′ is “ topologically small” in that no open interval is contained in C ′.
In general, for any given p ∈ (0, 1), one can construct a Fat Cantor set with measure
p in a more flexible way. For example, one can construct a Fat Cantor set with mea-
sure 34
by using a decreasing sequence that converges to 1 like 54, 65, 76, 87, . . .. For the
construction, we also require that every term of sequence is < 43. This then gives a
8
decreasing sequence (34)(5
4), (3
4)(6
5), (3
4)(7
6), . . . that converge to 3
4. A Fat Cantor set
of measure 34
can be constructed in the following process:
Step 1. We want the first remainder to have measure (34)(5
4) = 15
16. So we can
remove an open interval of 1− 1516
= 116
from the middle of interval [0,1]and call the
remainder C ′′1 . Then C ′′1 has measure 1516
, and is a union of two disjoint compact
intervals.
Step 2. We want the second remainder to have measure (34)(6
5) = 9
10. So we want
to remove a total measure of 1516− 9
10= 3
80from C ′′1 . This means we should remove
intervals of length 380/21 = 3
160from each of the two intervals in C ′′1 . Call the re-
minder C ′′2
...
Step n. Continuing, we remove subintervals of length (34) 1(n+2)(n+3)
/2n−1 from the
middle of each of the 2n−1 remaining intervals in C ′′n−1. to create C ′′n.
...
Continuing this infinite process, the Fat Cantor set is defined as: C ′′ = ∩∞i=1C′′i .
Then this Fat Cantor set C ′′ has measure 34.
Example 1. ∃P ⊆ [0, 1] s.t. P is meager, but [0, 1] \ P is negligible.
Solution: Construct a sequence of Fat Cantor sets P1, P2, . . . with measure 12, 23, 34. . ..
Pn. Let P =∞⋃n=1
Pn. Then P is meager but has measure 1. Thus P is meager but
[0, 1] \ P is negligible.
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Example 2. ∃Q ⊆ R s.t. Q is meager, but R \Q is negligible.
Solution: Let P be as in Example 1. Then let Q =∞⋃
k=−∞(P + k), where P + k =
{x+ k|x ∈ P} ⊆ [k, k + 1].
2.2 Split and Continuum Cardinality
Definition 2.6. Let A,C be sets, Then C splits A means: A∩C 6= ∅ and C \A 6= ∅.
Lemma 2.5. Let A1, A2, A3, ... ⊆ R. Assume A1, A2, . . . are all infinite. Then ∃ an
infinitely countable set C ⊆ R s.t., ∀j ∈ N, C splits Aj.
Proof. Step 1. Choose v1, w1 ∈ A1, s.t. v1 6= w1.
Step 2. Choose v2, w2 ∈ A2 \ {v1, w1}, s.t. v2 6= w2.
...
Step n. Choose vn, wn ∈ An \ {v1, w1, v2, w2, ...vn−1, wn−1}, s.t. vn 6= wn.
...
Continue this process countably many times .
Let B := {v1, v2, ...} and C := {w1, w2, ...}.
Then B ∩ C = ∅ and, for any given j ∈ N, we have , Aj ∩B 6= ∅ and Aj ∩ C 6= ∅.
∀j ∈ N, Aj ∩B ⊆ Aj \ C, then Aj \ C 6= ∅. Thus C splits Aj.
The following lemma is just another way to express the previous Lemma 2.5, but it
is helpful for understanding more generalized theorems later.
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Lemma 2.6. Let J := N, R := R, and let A : J → 2R. Assume ∀j ∈ J,∃ an injection
J ↪→ Aj. Then ∃h : {1, 2} × J ↪→ R, s.t. ∀j ∈ J, h1j, h2j ∈ A.
Proof. Step 1. Choose v1, w1 ∈ A1, s.t. v1 6= w1.
Step 2. Choose v2, w2 ∈ A2 \ {v1, w1}, s.t. v2 6= w2.
...
Step n. Choose vn, wn ∈ An \ {v1, w1, v2, w2, ...vn−1, wn−1}, s.t. vn 6= wn.
...
Define h : {1, 2} × J → R by hkj =
vj if k = 1
wj if k = 2
Corollary 2.2.1. Let J := N, R := R, and A ⊆ 2R. Assume ∃ a surjection J � A
and ∀A ∈ A,∃ an injection J ↪→ A. Then ∃ an infinitely countable set C ⊆ R s.t.
∀A ∈ A, C splits A, and ∃ a bijection J → C.
Proof. Choose a surjection A : J � A. ∀j ∈ J,∃ an injection J ↪→ Aj s.t. Aj is
infinite. For every j ∈ J, construct h1j, h2j in the previous way as in how we chose
vj, wj in Lemma 1.6.
This yields h : {1, 2} × J ↪→ R s.t. ∀j ∈ J, h1j, h2j ∈ Aj.
Define C := {h2j|j ∈ J}. Then j 7→ h2j is a bijection J → C.
∀j ∈ J , h2j ∈ Aj ∩ C so Aj ∩ C 6= ∅.
∀j ∈ J , h1j ∈ Aj \ C so Aj \ C 6= ∅.
Thus ∀j ∈ J , C splits Aj. Therefore C splits every set in A.
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Define A0 := {(a, b) ⊆ R|a, b ∈ Q, a < b}. Note that A0 is a countable set.
Fact: ∃ an infinitely countable set C ⊂ R s.t. ∀A ∈ A0 , C splits A. Proof: Just
apply the previous Corrollary 2.2.1 replacing A with A0.
Define B0 := { non empty open subsets of R}.
Now we want to consider the question whether ∃ an infinite countable set C ⊆ R,
s.t. ∀B ∈ B0, C splits B or not? The answer is yes. The main reason is that every
non-empty open set in R contains an interval with rational end points.
Remark 2.2.2. ∀B ∈ B0, ∃A ∈ A0 s.t. A ⊆ B.
Remark 2.2.3. Let A,B,C be sets. Assume C splits A and A ⊆ B. Then C splits
B.
Proof. ∅ 6= A ∩ C ⊆ B ∩ C and ∅ 6= A \ C ⊆ B \ C.
Remark 2.2.4. Denote J := N, I := {I ⊆ J | ∃j0 ∈ J s.t. I = {j ∈ J | j < j0}}∪{J}.
Then I = {∅, {1}, {1, 2}, ...} ∩ {J}.
Now Let us prove Lemma 2.6 in a different but more fancy way, which can lead
us to notice that J can be any infinite set and R can be any set. Before we go to
the details of the proof, let us review Zorn’s Lemma and Well-ordering Theorem.
Zorn’s Lemma : Suppose a partially ordered set P has the property that every
chain has an upper bound in P . Then set P contains at least one maximal element.
Well-ordering Theorem : Every set can be well-ordered.
12
Definition 2.7. Let S be a set with cardinal well-ordering <. We say < is a cardinal
well-ordering if ∀s ∈ S, {t ∈ S|t < s} is not bijective with S.
Remark 2.2.5. Every set has a cardinal well-ordering.
Lemma 2.7. Let J := N, R := R, and let A : J → 2R. Assume ∀j ∈ J,∃ an injection
J ↪→ Aj. Then ∃h : {1, 2} × J ↪→ R, s.t. ∀j ∈ J, h1j, h2j ∈ A.
Proof. J := N, I := {I ⊆ J | ∃j0 ∈ J s.t. I = {j ∈ J | j < j0}} ∪ {J} ⊆ 2J .
∀I ∈ I, FI := {f : {1, 2} × I ↪→ R s.t. ∀i ∈ I, f1i, f2i ∈ Ai}
What we want now: FJ 6= ∅.
F := ∪I∈IFI .
∀f, g ∈ F, f � g means dom[f ] ⊆ dom[g] and g|(dom[f]) = f . By Zorn’s Lemma,
we can choose h ∈ F s.t. h is �-maximal.
Choose I ∈ I s.t. h ∈ FI . To prove the lemma, we want to show : I = J .
Proof by contradiction: Suppose I 6= J , then I $ J .
Pick j1 =min (J \ I).
Define I1 := {j ∈ J |j ≤ j1}, i.e, I1 = I ∪ {j1}.
Choose v, w ∈ Aj1 \ (im[h]) s.t. v 6= w.
Define f := {1, 2} × I1 → R by fki =
hki if i ∈ I
v if i = j1 and k = 1
w if i = j1 and k = 2
Then f ∈ FI1 ⊆ F and h ≺ f (contradiction).
Theorem 2.2.6. Let J be an infinite set , R be a set, and let A : J → 2R. Assume
13
∀j ∈ J,∃ an injection J ↪→ Aj. Then ∃h : {1, 2} × J ↪→ R s.t. ∀j ∈ J, h1j, h2j ∈ A.
Proof. Let < be a cardinal well ordering on J . Let I := {I ⊆ J | ∃j0 ∈ J s.t.
I = {j ∈ J | j < j0}} ∪ {J} ⊆ 2J . (The following part of the proof is the same as
Lemma 2.7.)
∀I ∈ I, FI := {f : {1, 2} × I ↪→ R s.t. ∀i ∈ I, f1i, f2i ∈ Ai}
What we want now: FJ 6= ∅.
F := ∪I∈IFI .
∀f, g ∈ F, f � g means dom[f ] ⊆ dom[g] and g|(dom[f]) = f . By Zorn’s Lemma,
we can choose h ∈ F s.t. h is �-maximal.
Choose I ∈ I s.t. h ∈ FI . To prove the lemma, we want to show : I = J .
Proof by contradiction: Suppose I 6= J , then I $ J .
Pick j1 =min (J \ I).
Define I1 := {j ∈ J |j ≤ j1}, i.e, I1 = I ∪ {j1}.
Notice that card (im[h])=2(card(I)) < card(J) ≤ card (Aj).
Because < is a cardinal well-ordering, im[h] $ Aj.
Choose v, w ∈ Aj \ (im[h]) s.t. v 6= w.
Define f := {1, 2} × I1 → R by fki =
hki if i ∈ I
v if i = j1 and k = 1
w if i = j1 and k = 2
Then f ∈ FI1 ⊆ F and h ≺ f (contradiction).
Corollary 2.2.7. Let J be an infinite set, R be a set, and A ⊆ 2R. Assume ∃
14
a surjection J � A and ∀A ∈ A,∃ an injection J ↪→ A. Then ∃ an infinitely
countable set C ⊆ R s.t. ∀A ∈ A, C splits A, and ∃ a bijection J → C.
Proof. (The following proof is very similar to Corollary 2.2.1.) Choose a surjection
A : J � A, ∀j ∈ J,∃ an injection J ↪→ Aj.
By Theorem 2.2.5, choose h : {1, 2} × J ↪→ R s.t. ∀j ∈ J, h1j, h2j ∈ Aj.
Define C := {h2j|j ∈ J}. Then j 7→ h2j is a bijection J → C.
∀j ∈ J , h2j ∈ Aj ∩ C so Aj ∩ C 6= ∅.
∀j ∈ J , h1j ∈ Aj \ C so Aj \ C 6= ∅.
Thus ∀j ∈ J , C splits Aj. Therefore C simultaneously splits A.
Now let us move to discuss Continuum Cardinality.
Definition 2.8. Let A be a set. Then A has continuum cardinality means ∃ a
bijection: 2N → A.
Remark 2.2.8. R has continuum cardinality.
Define A1 := { closed subsets of R with continuum cardinality}.
Fact:
1) ∃ a surjection: 2N � {open subsets of R}.
Proof. Let B := {(a, b)| a, b ∈ Q, a < b}. Claim ∃ a surjection B : N � B.
Define f : 2N →{open subsets of R} by f(S) = ∪i∈SBi. Since B is a basis for the
topology on R, f is onto.
15
2) ∃ a bijection :{open subsets of R} → { closed subsets of R}. Take the complement
mapping which maps every open set to its complement.
3) ∃ a surjection: { closed subsets of R}� A1. This is obvious since A1 ⊂ { closed
subsets of R}.
Thus by 1)- 3), we know ∃ a surjection :2N � A1.
Remark 2.2.9. ∀A ∈ A1,∃ a bijection :2N → A.
Fact: ∃C ⊆ R s.t. ∀A ∈ A1, C splits A ( by Corollary 2.2.7 with J := 2N and
R := R ).
2.3 A Not Almost Open but Measurable Set
Define: B1 := {S ⊆ R|S is almost open and S is nonmeager}.
Remark 2.3.1. ∀B ∈ B1, ∃A ∈ A1 s.t.A ⊆ B.
Proof. This follows from Proposition 8.23 (i) ⇐⇒ (ii) in Kechris.
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Theorem 2.3.2. ∃C ⊆ R s.t. ∀B ∈ B1 , C splits B.
Proof. By the previous Fact, choose ∃C ⊆ R s.t. ∀A ∈ A1, C splits A. Let B ∈ B1
for given. By Remark 2.3.1, choose A ∈ A1 s.t.A ⊆ B. Since C splits A and A ⊆ B,
it follows by Remark 2.2.3 that C splits B.
Theorem 2.3.3. ∃C ⊆ R s.t. C is not almost open.
Proof. We follows Example 8.24 from Kechris.
Choose C ⊆ R s.t. ∀B ∈ B1, C splits B.
Proof by contradiction: Assume C is almost open.
∀B ∈ B1, C splits B. So C 6= B.
This implies C /∈ B1. So C is meager.
Since C is almost open, R \ C is also almost open.
Also R \ C is comeager and thus nonmeager. So R \ C ∈ B1.
Then C splits R \ C.
C splits R \ C implies C ∩ (R \ C) 6= ∅, but C ∩ (R \ C) = ∅ (contradiction).
Now we are able to construct a not almost open but measurable set.
Take C is not almost open (Theorem 2.3.3) and Q ⊆ R s.t. Q is meager, R \ Q is
negligible (Fat Cantor set Example 2). Let C1: =C ∪Q.
Claim: C1 is not almost open but measurable.
Proof. Suppose C1=C ∪ Q is almost open. Since Q is meager, this implies C is
almost open (contradiction). So C1 is not almost open. Take the complement of
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C1, Cc1 = Cc ∩ Qc ⊆ Qc. Since Qc has measure 0, Cc
1 also has measure 0, so Cc1 is
measurable. Then C1 is measurable.
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Bibliography
[1] Kechris, A. S. “Polish Spaces.” Classical Descriptive Set Theory. New York:
Springer-Verlag, 1995. 47-48. Print.
[2] Smith-Volterra-Cantor set. (n.d.). In Wikipedia. Retrieved May 5, 2016, from
https : //en.wikipedia.org/wiki/Smith− V olterra− Cantor − set
[3] Steen, Lynn Arthur; Seebach, J. Arthur Jr.(1995)[1978]. Counterexamples in
Topology, Berlin, New York: Springer-Verlag, ISBN 978-0-486-68735-3.
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