Concerning a problem of K. Kuratowski
Except where reference is made to the work of others, the work described in this thesis is
my own or was done in collaboration with my advisory committee. This thesis does not
include proprietary or classifled information.
Van-Trinh Nguyen
Certiflcate of Approval:
Jack B. Brown
Professor
Department of Mathematics and
Statistics
Jo W. Heath, Chair
Professor
Department of Mathematics and
Statistics
Geraldo S. DeSouza
Professor
Department of Mathematics and
Statistics
Stephen L. McFarland
Graduate School Acting Dean
Concerning a problem of K. Kuratowski
Van-Trinh Nguyen
A Thesis
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulflllment of the
Requirements for the
Degree of
Master of Science
Auburn, Alabama
May 11, 2006
Concerning a problem of K. Kuratowski
Van-Trinh Nguyen
Permission is granted to Auburn University to make copies of this thesis at its
discretion, upon the request of individuals or institutions and at
their expense. The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
Van-Trinh Nguyen, daughter of Bach-Hac and Van Nguyen, was born in Saigon, Viet-
nam on June 13, 1977. In June 1997, she graduated from Trabuco Hills High School in
Mission Viejo, California. She attended University of California at Berkeley, California,
where she graduated with a Bachelor of Arts in Mathematics degree in May 2002.
iv
Thesis Abstract
Concerning a problem of K. Kuratowski
Van-Trinh Nguyen
Master of Science, May 11, 2006
(B.A., University of California, Berkeley, 2002)
30 Typed Pages
Directed by Jo W. Heath
Suppose (S;T) is a topological space and A is any subset of S. Then the functions f
and g from the power set of S, P(S), into P(S) are deflned as: f(A) is the closure of A
and g(A) is the complement of A. In this thesis, our goal is proving that there are at most
fourteen type of image from any subset of S by using flnite compositions of the closure
function f and the complement function g, including the null composition.
v
Acknowledgments
The author thanks Professor Jo W. Heath for the ideas and useful discussions con-
tributed to this research.
vi
Style manual or journal used Journal of Approximation Theory (together with the style
known as \aums"). Bibliograpy follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (speciflcally LATEX)
together with the departmental style-flle aums.sty.
vii
Table of Contents
1 INTRODUCTION 1
2 BASIC DEFINITIONS AND THEOREMS 3
3 SUPPORTING LEMMAS AND THEOREMS 6
4 THE CO-FINITE CASE 14
5 MAIN RESULTS 18
Bibliography 22
viii
Chapter 1
INTRODUCTION
Suppose (S;T) is a topological space and A is any subset of S, then the functions f
and g from the power set of S, P(S) into P(S), are deflned as: f(A) is the closure of A
and g(A) is the complement of A. There is referenced in Munkres "Topology" [1] a problem
created by K. Kuratowski. He asked how many images these two functions can generate
starting with a subset A of the real line. By using flnite compositions of the closure function
f and the complement function g, we can generate a fourteen difierent images of A. This
includes the image of the null composition, A itself. We also prove that no more than 14
images can ever be generated
For preparation of the proofs in chapters 4 and 5, in chapter 2, we want to introduce
to you the list of theorems that come from my class notes. For some of them, I did the
proofs and I presented them on the board in class. But these theorems are still not enough.
In chapter 3, we need to prove some more theorems that are necessary for my proofs in
chapters 4 and 5. Especially, theorems 3.5, 3.6a, and 3.6b; they help us to understand that
for any flnite composition with any exponent n on f or g (for any positive integer n), we
can reduce these exponents to 1 or 0.
Example: f4454(g3[f33(g4(A))]) = f(g[f(g0(A))]) = f(g[f(A)]).
In chapter 4, we do research on a co-flnite topological space: If (S;T) is a topological
space with the co-flnite topology and A is a subset of S, then the collection of all images
1
of A using flnite compositions of the closure function f and the complement function g,
including the null composition, has at most four elements.
Finally, in chapter 5, by using the information in the previous chapter, we want to
show you that for any topological space (S;T) with A is any subset of S, and by using
flnite compositions of the closure function f and the complement function g, we are always
able to generate a maximum of fourteen difierent images of A, including the image of the
null composition. Then, to demonstrate that this theorem cannot be improved, we build
an example on the real line that has fourteen difierent images.
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Chapter 2
BASIC DEFINITIONS AND THEOREMS
In this chapter, without proofs, we include the deflnitions, theorems and notations that
come from the class notes, and they are necessary for the proofs in the next chapters.
Deflnition 2.1: Suppose (S;T) is a topological space and A is a subset of S. If the set
of limit points of A is denoted A0, then A is deflned to be A[A0.
Deflnition 2.2: Suppose (S;T) is a topological space, and suppose M is subset of S.
The interior of M, denoted Int(M), is deflned to be the set of all points x in M such that
there is an open set O(x) satisfying x 2 O(x) M. The boundary of M, denoted Bd(M),
is deflned to the set of all points x such that every open set containing x contains a point
in M and a point not in M.
Deflnition 2.3: A subset M of a topological space (S;T) is said to be closed if the set
S nM is open.
Theorem 2.4-2.12. Suppose (S;T) is a topological space and M is a subset of S. From
the class notes we have the following reference theorems:
Theorem 2.4: M = M
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Theorem 2.5: M Int(M)[Bd(M)
Theorem 2.6: Int(M) is open.
Theorem 2.7: M is open ifi M = Int(M)
Theorem 2.8: M is closed ifi Bd(M) ? M
Theorem 2.9: S nM = S nInt(M)
Theorem 2.10: Bd(M) = M \S nM
Theorem 2.11: M = M [Bd(M)
Theorem 2.12: M is closed ifi M = M
Theorem 2.13: If (S;T) is a topological space, M is a closed set in S, and p is a limit
point of M, then p is an element of M.
Notation 2.14: N will denote the set of all positive integers.
Notation 2.15: Q will denote the set of all rational numbers.
4
Notation 2.16: I will denote the set of all irrational numbers.
Notation 2.17: R will denote the set of all real numbers.
Deflnition 2.18: Suppose S is any set and T consists of the empty set and any subset
of S whose complement is flnite. Then (S;T) is a co-flnite topological space.
Theorem 2.19: Suppose (S;T) is a topological space with the co-flnite topology. If A
is a flnite subset of S, then the closure of A is A, and if A is an inflnite subset, then the
closure of A is S.
Deflnition 2.20: Suppose (S;T) is the topological space. Then the functions f and g
from the power set of S, P(S), into P(S) are deflned as follows: if A is any subset of S,
then f(A) is the closure of A and g(A) is the complement of A.
5
Chapter 3
SUPPORTING LEMMAS AND THEOREMS
Beside the deflnitions and theorems that were introduced in chapter 2, in this chapter,
we will prove some more theorems to support the proofs of our results .
Lemma 3.1: If (S;T) is a topological space and B is a subset of S, then B is closed.
Proof: Since from theorem 2.4, B = B, and by theorem 2.12 with B substituted for M
and B for M, we will get B is closed.
Theorem 3.2-3.4: Suppose (S;T) is a topological space and A is a subset of S. Then
we will prove the following theorems.
Theorem 3.2: Int(A) = Int(Int(A))
Proof: First we will show that Int(A) Int(Int(A)).
By theorem 2.6, with A substituted for M, then Int(A) is open. And by theorem 2.7 with
Int(A) substituted for M, we have Int(A) = Int(Int(A)).
This implies Int(A) = Int(Int(A)).
Bytheorem2.5withInt(A)substitutedforM, wehaveInt(A) [Int(Int(A))[Bd(Int(A))].
Therefore, Int(Int(A)) Int[Int(Int(A))[Bd(Int(A))].
Since Int(Int(A)) = Int(A), we can remove one "Int" in Int(Int(A)).
So we get Int(A) Int[Int(A)[Bd(Int(A))].
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And by theorem 2.11 withInt(A) substituted for M, we have Int(A) = Int(A)[Bd[Int(A)].
This means that Int[Int(A)[Bd(Int(A))] = Int(Int(A)).
Hence, Int(A) Int(Int(A)) (Inclusion I).
Second we will show that Int(Int(A)) Int(A).
Let x 2 Int(Int(A)); then we will have 2 cases that can happen: x 2 Int(Int(A)) or
x =2 Int(Int(A)).
Case 1: If x 2 Int(Int(A)), then x 2 Int(A).
Case 2: If x =2 Int(Int(A)), then x must be a limit point of Int(Int(A)). This means
that x 2 [S nInt(Int(A))] and is a limit point of Int(Int(A)).
By theorem 2.9 with Int(A) substituted for M, we have x 2 S n(Int(A)).
Again, we have 2 cases that can happen: x 2 S nInt(A) or x =2 (S nInt(A)).
Case 2a: If x 2 S n Int(A), since x is a limit point of Int(Int(A)), then every open
set containing x contains at least one point in Int(Int(A)) which is difierent from x. But
[S nInt(A)] \ [Int(Int(A))] = `.
This gives us a contradiction since, by lemma 3.1 with Int(A) substituted for B, we have
Int(A) is closed, so S nInt(A) is open.
Case 2b: Therefore, x must be in [S nS nInt(A)].
This implies x 2 Int(A).
Therefore, Int(Int(A)) Int(A) (Inclusion II)
From inclusions I and II, we proved that Int(A) = Int(Int(A)).
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Theorem 3.3: S nInt(A) = S nInt(Int(A))
Proof: To start to show that S n Int(A) = S n Int(Int(A)), we will show that Int(A) =
Int(Int(A)).
First we will show that Int(A) Int(Int(A)).
By theorem 2.6, with A substituted for M, then Int(A) is open, and by theorem 2.7 with
Int(A) substituted for M, we have Int(A) = Int(Int(A)).
Bytheorem2.5withInt(A)substitutedforM, wehaveInt(A) Int(Int(A))[Bd(Int(A)).
Therefore, Int(Int(A)) Int[Int(Int(A))[Bd(Int(A))].
Since Int(Int(A)) = Int(A), we can remove one "Int" in Int(Int(A)).
This means that Int(A) Int[Int(A)[Bd(Int(A))].
But by theorem 2.11 withInt(A) substituted forM, we haveInt(A) = Int(A)[Bd(Int(A)).
This implies Int[Int(A)[Bd(Int(A))] = Int(Int(A)).
Hence, Int(A) Int(Int(A)) (Inclusion I)
Second we will show that Int(Int(A)) Int(A).
Let x be any limit point of Int(A). This means that x is a limit point of A.
By lemma 3.1 with A substituted for B, we have A is closed.
And by theorem 2.13 with A substituted for M, and x substituted for p, we will get x 2 A.
Hence (Int(A))0 A, and obviously, Int(A) A, too.
Notice that [Int(A)]0 [ Int(A) = Int(A).
Therefore Int(A) A.
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This implies Int(Int(A)) Int(A) (Inclusion II)
From inclusions I and II, we have Int(A) = Int(Int(A)).
This means that S nInt(A) = S nInt(Int(A)).
Theorem 3.4: S nInt(A) = S nInt(A)
Proof: By theorem 2.9 with Int(A) substituted for M, we have S nInt(A) = SnInt(Int(A)).
By theorem 3.3, we have S nInt(Int(A)) = S nInt(A).
Therefore, S nInt(A) = S nInt(A).
Theorem 3.5: Suppose (S;T) is a topological space and A is a subset of S. Then for
any positive integer n, the collection of all images of A using only flnite compositions of the
closure function f is fn(A) = f(A).
Proof: We will use induction to show that for any n 2 N, fn(A) = f(A).
Clearly, if n = 1, then fn(A) = f1(A) = f(A).
If n = 2, then fn(A) = f2(A) = f(f(A)).
But f(f(A)) is deflned to be (A) = A.
By theorem 2.4 with A substituted for M, we have A = A.
Therefore, f(f(A)) = A = f(A).
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Assume fk(A) = f(A) (for some k 2 N). Then we need to prove that fk+1(A) = f(A).
Since fk+1(A) = f(fk(A)), by substituting f(A) for fk(A), we will have fk+1(A) =
f(f(A)).
This means that fk+1(A) = f(A), as shown in the case n = 2.
Hence, fn(A) = f(A) (for any n 2 N).
Theorem 3.6a: Suppose (S;T) is a topological space and A is a subset of S. Then for
any non-negative integer n, the collection of all images of A using only odd flnite composi-
tions of the complement function g is g2n+1(A) = g(A).
Proof: By using induction:
If n = 0, then g2n+1(A) = g1(A) = g(A).
If n = 1, then g2n+1(A) = g3(A) = g(g[g(A)]) = g(g[S nA]) = g(S n[S nA]) = g(A).
Assume g2k+1(A) = g(A) (for some k 2 N). Then we need to prove that g2(k+1)+1(A) =
g(A).
Note that g2(k+1)+1(A) = g2k+3(A)) = g2k+1(g[g(A)]).
But g(g(A)) = g(S nA) = S n(S nA) = A.
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Therefore g2(k+1)+1(A) = g2k+1(g[g(A)]) = g2(k+1)(A) = g(A).
This means that g2(k+1)+1(A) = g(A).
Hence, g2n+1(A) = g(A) [for any n 2 (N [f0g)].
Theorem 3.6b: Suppose (S;T) is a topological space and A is a subset of S. Then for
any n is a positive integer, the collection of all images of A using only even flnite composi-
tions of the complement function g is g2n(A) = A.
Proof: By using induction:
If n = 1, then g2n(A) = g2(A) = g(g(A)) = g(S nA) = S n(S nA) = A.
Assume g2k(A) = A (for some k 2 N). Then we need to prove that g2(k+1)(A) = A.
Since g2(k+1)(A) = g2k+2(A) = g2k(g[g(A)]). By substituting A for g[g(A)], we will have
g2k(g[g(A)]) = g2k(A) = A.
This means that g2(k+1)(A) = A.
Hence, g2n(A) = A (for any n 2 N).
Corollary 3.7: The images of A generated by all flnite composition of f and g are also
generated by the compositions that alternate f and g; i.e. compositions of the form (fg)n,
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g(fg)n, (gf)n, or f(gf)n, for non-negative integers n.
Proof: We want to prove that any image of A is always generated by the flnite compo-
sitions that alternate f and g.
By theorem 3.5, we have all flnite compositions of f and g are also generated by the
flnite compositions that alternate f and the flnite compositions of the complement function
g.
We have 2 cases that can happen: when the flnite compositions of the complement
function g is odd or when the flnite compositions of the complement function g is even
Case 1: Where any of the flnite compositions of the complement function g is odd,
then by theorem 3.6a, we have the images of A are generated by the compositions that
alternate f and g.
Case 2: Where any of the flnite compositions of the complement function g is even.
Again, we have 2 cases:
? Case 2a: It is clear that if the function starts with g, then by theorem 3.6b, this
means that the function starts with f.
? Case 2b: If some flnite compositions of the complement function g are even, which
are in between two closure functions f. Then by theorem 3.6b, we have the images of A are
generated by some compositions that alternate ff and g, by theorem 3.5, this implies that
the images of A are generated by the compositions that alternate f and g.
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Therefore, the images of A generated by all flnite composition of f and g are also
generated by the compositions that alternate f and g.
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Chapter 4
THE CO-FINITE CASE
In this chapter, we will see that for any co-flnite topological space (S;T) with A a sub-
set of S, the collection of all images of A using flnite compositions of the closure function
f and the complement function g, including the null composition, always has at most four
elements, A, S nA, S, and `.
Example 4.1: Suppose (N;T) is the positive integers with the co-flnite topology, and A
is the flnite set f1;2;3g. By using the flnite compositions of the closure function f and the
complement function g, we will get :
f(A) = A = A = f1;2;3g (by theorem 2.19).
g(A) = N nA = N nf1;2;3g.
This implies that f(g(A)) = N (by theorem 2.19).
Hence, g(f(g(A))) = N nN = `.
From corollary 3.7, observe that for any composition of the closure f and the comple-
ment function g, an image of A is always either A or N nA or N or `.
Theorem 4.2: If (S;T) is a topological space with the co-flnite topology and A is a
subset of S, then the collection of all images of A using flnite compositions of the closure
function f and the complement function g, including the null composition, has at most four
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elements, A, S nA, S, and `.
Proof: By proving this theorem, we will get a maximum of four images from any subset A
of S, those images are: A, S nA, S, and `.
We have 2 cases that can happen: A is a flnite subset of S or A is an inflnite subset of S.
Case 1: If A is a flnite subset of S, then again, we have 2 cases: SnA is a flnite subset
of S or S nA is an inflnite subset of S.
Case 1a: If S nA is a flnite subset of S then:
We have f(A) = A = A (by theorem 2.19).
This implies that g(f(A)) = S nA.
Now, starting with g, g(A) = S nA.
Therefore f(g(A)) = S nA (by theorem 2.19).
Since A and S nA are flnite subsets of S, there are only two types of flnite composi-
tions and all of them alternate f and g. From corollary 3.7, this means that the images of
A generated by all flnite compositions of f and g will be either A or SnA (by theorem 2.19).
Case 1b: If S nA is an inflnite subset of S then:
We have f(A) = A = A (by theorem 2.19).
This implies that g(f(A)) = S nA and f(g(f(A))) = S.
Now, starting with g, we have g(A) = S nA.
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Hence, f(g(A)) = S nA = S (by theorem 2.19).
This implies that g(f(g(A))) = `.
By corollary 3.7, there are only four types of flnite compositions and all of them al-
ternate f and g. This means that the images we can generate will be either A, SnA, S or `.
Case 2: If A is an inflnite subset of S. Then again, we have 2 cases: S nA is a flnite
subset of S or S nA is an inflnite subset of S.
Case 2a: If S nA is a flnite subset of S then:
We have f(A) = A = S (by theorem 2.19).
Therefore g(f(A)) = `.
Now, starting with g, we get g(A) = S nA.
This implies that g(g(A)) = A.
By theorem 3.6b, it is clear to see that the flnite complements of A are g(A) = S nA
or null composition, A itself.
Hence, f(g(A)) = S nA (by theorem 2.19), and g(f(g(A))) = A.
From corollary 3.7, observe that there are only four types of flnite compositions and
all of them alternate f and g. This means that the images we can generate will be either
A, S nA, S or `.
Case 2b: If S nA is an inflnite subset of S then:
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We have f(A) = A = S (by theorem 2.19).
Therefore g(f(A)) = `.
Now, starting with g, we get g(A) = S nA.
This implies that g(g(A)) = A.
By theorem 3.6b, it is clear to see that the flnite complements of A are g(A) = S nA
or null composition, A itself.
Also, f(g(A)) = S nA = S (by theorem 2.19).
From corollary 3.7, we also observe that there are only four types of flnite compositions
and all of them alternate f and g, this means that the images those we can generate will
be either A, S nA, S or `.
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Chapter 5
MAIN RESULTS
Now, in this chapter, we want to show you that for any topological space (S;T), there
are at most fourteen type of image from any subset of S by using flnite compositions of the
closure function f and the complement function g, including the null composition.
We begin with an example of a subset of R that has fourteen difierent images.
Example 5.1: Let A = (0;1)[(1;2)[f3;4g[[5;6)[f7g[(8;9)[(Q\[9;10]), which is
in the reals with the usual topology. We will flnd seven difierent images using compositions
that start with g (steps 2 to 8) and another six difierent images using compositions that
start with f (steps 9 to 14). Then, counting A itself, we will have fourteen difierent images.
Facts: We will assume without proof that if a and b are two (unequal) real numbers then
the rational numbers, Q, and the irrational numbers, I, are both dense in the interval [a;b].
Therefore, by using these facts, we can get:
1. A = (0;1)[(1;2)[f3;4g[[5;6)[f7g[(8;9)[(Q\[9;10])
2. g(A) = (?1;0][f1g[[2;3)[(3;4)[(4;5)[[6;7)[(7;8][(I \[9;10])[(10;1)
3. f(g(A)) = (?1;0][f1g[[2;5][[6;8][[9;1)
4. g(f(g(A))) = (0;1)[(1;2)[(5;6)[(8;9)
5. (fg)2(A) = [0;2][[5;6][[8;9]
6. g(fg)2(A) = (?1;0)[(2;5)[(6;8)[(9;1)
7. (fg)3(A) = (?1;0][[2;5][[6;8][[9;1)
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8. g(fg)3(A) = (0;2)[(5;6)[(8;9)
Now we compute the images using compositions that start with f.
9. f(A) = [0;2][f3;4g[[5;6][f7g[[8;10]
10. g(f(A)) = (?1;0)[(2;3)[(3;4)[(4;5)[(6;7)[(7;8)[(10;1)
11. f(g(f(A))) = (?1;0][[2;5][[6;8][[10;1)
12. (gf)2(A) = (0;2)[(5;6)[(8;10)
13. f(gf)2(A) = [0;2][[5;6][[8;10]
14. (gf)3(A) = (?1;0)[(2;5)[(6;8)[(10;1)
Theorem 5.2: Suppose (S;T) is a topological space, and A is any subset of S. Then
the collection of all images of A using flnite compositions of the closure function f and the
complement function g, including the null composition, has at most fourteen elements.
Proof: From corollary 3.7, we know in the co-flnite case, there are only four types of
flnite compositions and all of them alternate f and g. We will start with g and compute
the seven images up to g(fg)3(A) (steps 2 to 8). These images may or may not be distinct,
depending on A. But when we go one more factor and compute (fg)4(A) we generate an
image we already have so the process repeats from that point on. Then we do the same
for the flrst six compositions that start with f (steps 9 to 14) and flnd that the seventh
composition f(gf)3(A) repeats. Thus, together with A itself, there are at most fourteen
difierent images of A.
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Those fourteen possibly difierent images of A are:
First, we flnd the compositions that start with g.
1. A
2. g(A) = S nA
3. f(g(A)) = S nA
4. g(f(g(A))) = S n(S nA) = S n[S nInt(A)] (by theorem 2.9 with A substituted for M)
This implies g(f(g(A))) = Int(A).
5. (fg)2(A) = Int(A)
6. g(fg)2(A) = S nInt(A)
7. (fg)3(A) = S nInt(A) = S nInt(Int(A)) (by theorem 2.9 with Int(A) substituted for
M)
8. g(fg)3(A) = S n(S nInt(Int(A))) = Int(Int(A))
The seven images above (from 2 to 8) are the most that can be generated by composition
functions starting with g since (fg)4(A) = Int(Int(A)) = Int(A) (by theorem 3.2).
This means that the solution of (fg)4(A) repeats the image in step 5.
Second, we flnd the compositions that start with f.
9. f(A) = A
10. g(f(A)) = S nA
11. f(g(f(A))) = S nA = S nInt(A) (by theorem 2.9 with A substituted for M)
12. (gf)2(A) = S n(S nInt(A)) = Int(A)
13. f(gf)2(A) = Int(A)
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14. (gf)3(A) = S nInt(A)
These six images (from 9 to 14) are the most that can be generated by composition func-
tions starting with f since f(gf)3(A) = S nInt(A) = S nInt(A) (by theorem 3.4).
This mean that the solution of f(gf)3(A) repeats the image in step 11.
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Bibliography
[1] James R. Munkres \Topology" Second edition, Pearson Education, 2000.
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