Merlin Thesis
Merlin Thesis
ABSTRACT 2
1 INTRODUCTION 3
1.1 Introduction to Topology . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Introduction to Semi and Semi-pre closed sets . . . . . . . . 4
1.3 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
APPLICATION OF TOPOLOGY 41
CONCLUSION 46
BIBLIOGRAPHY 47
PUBLICATION 50
1
ABSTRACT
In this dissertation a new class of generalized sets, namely β # g - closed sets and
β # g - open sets are introduced in Topological spaces and we also discuss some of
the properties of these sets. We also define a continuous mapping namely, β # g -
continuous, using the new generalized sets. From β # g - continuous function we
further extend it to contra - β # g - continuous function and analyze its properties.
Further we develop the concepts to define two new spaces, namely, β # g - connected
and β # g - compact.
2
Chapter 1
INTRODUCTION
Topology is one of the major areas of abstract mathematics. The word topology
is obtained from the Greek words “t̀opos” and “l̀ogos” meaning study of spaces.
The English form “topology” was used in 1883 in Listing’s obituary in the
journal Nature to distinguish “qualitative geometry from the ordinary geometry in
which quantitative relations chiefly are treated”.
3
Maurice Fréchet introduced the metric space in 1906. A metric space is now
considered a special case of a general topological space, with any given topological
space potentially giving rise to many distinct metric spaces. In 1914, Felix Haus-
dorff coined the term “topological space” and gave the definition for what is now
called a Hausdorff space. Currently, a topological space is a slight generalization of
Hausdorff spaces, given in 1922 by Kazimierz Kuratowski.
4
In 1987, Bhattacharyya and Lahiri defined and studied about semi- general-
ized closed sets and U.D. Tapi, et.al [20] have made a study on more concepts of
semi- generalized closed sets.
5
1.3 Preliminaries
1. φ and X belong to τ
Here the interior of A is denoted by int(A) and the closure of A is denoted by cl(A).
6
Definition 1.3.7.
Let (X,τ ) be a topological space. A subset A of the space X is said to be
Definition 1.3.8.
Let (X,τ ) be a topological space and A ⊆ X. The semi-closure of A, denoted by
scl(A) is defined to be the intersection of all semi-closed sets containing A
Definition 1.3.9.
Let (X,τ ) be a topological space. A subset A of the space X is said to be
7
Definition 1.3.10.
A map f : X → Y is said to be
1. continuous [6] if for every closed set F of Y, the inverse image f −1 (F ) is closed
in X.
4. pre continuous [21] if for every closed set F of Y, the inverse image f −1 (F ) is
pre closed in X.
5. semi-pre continuous [9] if for every closed set F of Y, the inverse image f −1 (F )
is semi-pre closed in X.
Definition 1.3.11.
A function f : X → Y is said to be
2. contra - semi - continuous [8] if f −1 (V ) is semi - closed in X for every open set
V in Y.
4. contra - pre - continuous [16] if f −1 (V ) is pre - closed in X for every open set
V in Y.
8
Definition 1.3.12. [18]
A topological space X is said to be connected, if X cannot be expressed as union of
any two disjoint non-empty open sets in X.
9
Chapter 2
β # GENERALIZED CLOSED AND OPEN SETS
In this chapter we introduce a set namely, β # g - closed set and we also introduce
and discuss the properties of β # g - open sets and β # g - continuous functions in a
Topological space.
Definition 2.1.1.
A subset A of a topological space (X,τ ) is called semi-pre# generalized closed (briefly
β # g - closed) if βcl(A) ⊆ U whenever A ⊆ U and U is semi*- open in (X,τ ).
Theorem 2.1.2.
Every closed set is β # g - closed.
Proof.
Let A be a closed set in (X,τ ).
Let A ⊆ U, and U is semi*- open in X.
Since A is closed, cl(A) = A.
But A ⊆ U.
This implies, cl(A) = A ⊆ U.
But βcl(A) ⊆ cl(A).
Thus, we have, βcl(A) ⊆ U whenever, A ⊆ U and U is semi*- open in X.
Therefore, A is β # g - closed.
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Remark 2.1.3.
The converse of the above theorem is not true, as seen from the following example.
Example 2.1.4.
Let X={a,b,c} with the topology τ ={φ, X, {a}}.
Let A={a}.
The sets which are β-closed are {{b},{c},{b,c}}.
βcl(A) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus A is β # g - closed, but A is not closed.
Theorem 2.1.5.
Every semi-closed set is β # g - closed.
Proof.
Let A be semi-closed in (X,τ ). Let A ⊆ U and U be semi*- open in X.
Since A is semi- closed, scl(A) = A.
But A ⊆ U.
This implies, scl(A) ⊆ U.
But, βcl(A) ⊆ scl(A) holds true always.
Thus we have, βcl(A) ⊆ U whenever, A ⊆ U and U is semi*- open in X.
Therefore, A is β # g - closed.
Remark 2.1.6.
The converse of the above theorem is not true, as seen from the following example.
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Example 2.1.7.
Let X={a,b,c} with the topology τ ={φ, X, {a}}.
Let A={a}.
The sets which are β-closed are {{b},{c},{b,c}}.
βcl(A) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus A is β # g - closed, but A is not semi- closed, since int(cl(A)) ⊆ A does not
hold true.
Theorem 2.1.8.
Every semi*-closed set is β # g - closed.
Proof.
Let A be semi*-closed in (X,τ ).
Let A ⊆ U and U be semi*- open in X.
Here A is semi*-closed, and
since every semi*-closed sets are semi-closed, we have A is semi-closed.
Thus, scl(A) = A.
But, βcl(A) ⊆ scl(A) holds true always.
Thus we have, βcl(A) ⊆ U whenever A ⊆ U and U is semi*-open in X.
Therefore, A is β # g - closed.
Remark 2.1.9.
The converse of the above theorem is not true, as seen from the following example.
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Example 2.1.10.
Let X={a,b,c} with the topology τ ={φ, X, {a}}.
Let A={a}.
The sets which are β-closed are {{b},{c},{b,c}}.
βcl(A) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus A is β # g - closed, but A is not semi*- closed, since int*(cl(A)) ⊆ A does not
hold true.
Theorem 2.1.11.
Every pre-closed set is β # g - closed.
Proof.
Let A be pre-closed in (X,τ ).
Let A ⊆ U and U be semi*-open in X.
Since A is pre-closed, pcl(A) = A.
But, βcl(A) ⊆ pcl(A) always holds true.
Thus, βcl(A) ⊆ U, whenever A ⊆ U and U is semi*-open in X.
Therefore, A is β # g - closed.
Remark 2.1.12.
The converse of the above theorem is not true, as seen from the following example.
Example 2.1.13.
Let X={a,b,c} with the topology τ ={φ, X, {a}}.
Let A={a}.
The sets which are β-closed are {{b},{c},{b,c}}.
βcl(A) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus A is β # g - closed, but A is not pre-closed, since cl(int(A)) ⊆ A does not hold
true.
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Theorem 2.1.14.
Every α - closed set is β # g - closed.
Proof.
Let A be α - closed in (X,τ ).
Let A ⊆ U and U be semi*-open in X.
Since every α - closed set is pre-closed,
A is pre-closed.
Thus by above theorem,
A is β # g - closed.
Remark 2.1.15.
The converse of the above theorem is not true, as seen from the following example.
Example 2.1.16.
Let X={a,b,c} with the topology τ ={φ, X, {a}}.
Let A={a}.
The sets which are β-closed are {{b},{c},{b,c}}.
βcl(A) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus A is β # g - closed, but A is not α - closed, since cl(int(cl(A))) ⊆ A does not
hold true.
Theorem 2.1.17.
Every semi-pre closed set is β # g - closed.
Proof.
Let A be semi-pre closed in (X,τ ).
Let A ⊆ U and U be semi*-open in X.
Since A is semi-pre closed, βcl(A) = A.
But, A ⊆ U.
Thus, βcl(A) ⊆ U whenever A ⊆ U and U is semi*-open in X.
Therefore, A is β # g - closed.
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Remark 2.1.18.
The converse of the above theorem is not true, as seen from the following example.
Example 2.1.19.
Let X={a,b,c} with the topology τ ={φ, X, {a}}.
Let A={a}.
The sets which are β-closed are {{b},{c},{b,c}}.
βcl(A) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus A is β # g - closed, but A is not semi- pre closed, since int(cl(int(A))) ⊆ A does
not hold true.
Remark 2.1.20.
Union of any two β # g - closed sets need not be β # g - closed.
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Example 2.1.21.
Let X={a,b,c} with the topology τ ={φ, X,{a},{b,c}}.
Let A={a} and B={b}.
First let us prove that A is β # g - closed.
The sets which are β-closed are {{a},{b},{c},{a,b},{b,c},{a,c}}.
Then, βcl(A) = ∩{{a},{a,b},{a,c}}.
=⇒ βcl(A) = {a}.
Let U = {a} ∈ τ .
The sets which are g - closed are {{a},{b},{c},{b,c}}.
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus A is β # g - closed.
Now let us prove, B is β # g - closed.
βcl(B) = ∩{{b},{a,b},{b,c}}.
=⇒ βcl(B) = {b}.
Let U = {b,c} ∈ τ .
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus B is β # g - closed.
Now, A∪B = {a}∪{b} = {a,b}.
Let C = {a,b}.
Then, βcl(C)= {a,b}.
Here there is no open set U in τ , other than X, which satisfy {a,b} ⊆ U.
Therefore, {a,b} is not β # g - closed.
Theorem 2.1.22.
If a subset of a topological space X is β # g - closed such that A ⊆ B ⊆ βcl(A), then
B is also β # g - closed.
Proof.
Let A be β # g - closed in X and let A ⊆ B ⊆ βcl(A).
Let U be semi*-open in X, and B ⊆ U.
Then A ⊆ U.
Since A is β # g - closed,
βcl(A) ⊆ U.
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By hypothesis,
βcl(B) ⊆ βcl(βcl(A)) = βcl(A) ⊆ U.
Thus βcl(B) ⊆ U where U is semi*-open and B ⊆ U.
Therefore, B is β # g - closed.
Remark 2.1.23.
The converse of the above theorem need not be true as seen from the following
example.
Example 2.1.24.
Let X={a,b,c} with the topology τ ={φ, X,{a},{b,c}}.
Let A = {b} and B = {b,c}.
First let us prove that A is β # g - closed.
The sets which are β-closed are {{a},{b},{c},{a,b},{b,c},{a,c}}.
βcl(A) = ∩{{b},{a,b},{b,c}}.
=⇒ βcl(A) = {b}.
Let U = {b,c} ∈ τ .
The sets which are g - closed are {{a},{b},{c},{b,c}}.
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus A is β # g - closed.
Now let us prove, B is β # g - closed.
βcl(B) = {b,c}.
Let U = {b,c} ∈ τ .
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus B is β # g - closed.
Therefore, A and B are β # g - closed.
But A ⊆ B 6⊆ βcl(A), since {b} ⊆ {b,c} 6⊆ {b}.
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Theorem 2.1.25.
If A is both open and g - closed in X, then it is β # g - closed in X.
Proof.
Let A be an open and g - closed set in X.
Let A ⊆ U and U be semi*-open in X.
Now, A ⊆ A.
By hypothesis,
Since A is g - closed and A is open, cl(A) ⊆ A.
But, βcl(A) ⊆ cl(A) holds true always.
Thus, βcl(A) ⊆ A.
But A ⊆ U.
Thus βcl(A) ⊆ U where A ⊆ U and U is semi*-open in X.
Therefore, A is β # g - closed.
Remark 2.1.26.
If A is both open and β # g - closed in X, then A need not be g - closed, which is
seen from the following example.
Example 2.1.27.
Let X = {a,b,c} and τ = {φ, X, {a}}.
Let A = {a}.
Then A is open.
The sets which are β-closed are {{b},{c},{b,c}}.
βcl(A) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open, since U ⊆ cl*(int(U)).
Thus A is β # g - closed.
But A is not g - closed since cl(A) ⊆ U does not hold true.
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2.2 β # generalized open sets
In this section we introduce the set namely, β # g - open in Topological spaces and
discuss some of its properties.
Definition 2.2.1.
A subset A of a topological space (X,τ ) is called semi-pre# generalized open (briefly
β # g - open) if its complement X\A is β # g - closed.
Theorem 2.2.2.
Every open set is β # g - open.
Proof.
The proof is obvious from the theorem 2.1.2.
Theorem 2.2.3.
Every semi-open set is β # g - open.
Proof.
The proof of this theorem is similar to the proof of the theorem 2.1.5.
Theorem 2.2.4.
Every semi*-open set is β # g - open.
Proof.
The proof of this theorem is obvious from theorem 2.1.8. since, the complement of
β # g - closed is β # g - open.
Theorem 2.2.5.
Every pre-open set is β # g - open.
Proof.
The proof of this theorem is similar to the proof of the theorem 2.1.11.
Theorem 2.2.6.
Every semi-pre open set is β # g - open.
Proof.
The proof of this theorem is obvious from theorem 2.1.17. since β # g - closed sets
are the complement of β # g - open
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Chapter 3
β # g - CONTINUOUS FUNCTIONS AND CONTRA - β # g -
CONTINUOUS FUNCTIONS
Definition 3.1.1.
A map f : X → Y is said to be β # g - continuous if for every closed(resp. open) set
F of Y, the inverse image f −1 (F ) is β # g - closed(resp. open) in X.
Definition 3.1.2.
A map f : X → Y is said to be β # g - irresolute if and only if the inverse image of
every β # g - closed(resp. open) set in Y, is β # g - closed(resp. open) in X.
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Theorem 3.1.3.
If a map f : X → Y is continuous, then it is β # g - continuous.
Proof.
Let f : X → Y be continuous and let V be closed in Y.
Since f is continuous,
f −1 (V ) is closed in X.
Since, every closed set is β # g - closed,
f −1 (V ) is β # g - closed in X.
Therefore, f is β # g - continuous.
Remark 3.1.4.
The converse of the above theorem is not true, as seen from the following example.
Example 3.1.5.
Let X = {a, b, c} = Y.
Let f : X → Y be defined as,
f(a) = a, f(b) = b, f(c) = c.
Let τ = {φ, X, {a}} and σ = {φ, Y, {b, c}} be the topologies of X and Y respectively.
The sets which are closed in X are {φ, X, {b, c}}.
The sets which are closed in Y are {φ, Y, {a}}.
Let V = {a} which is closed in Y.
Then, f −1 (V ) = {a} is in X.
The sets which are β - closed in X are {{b}, {c}, {b, c}}.
βcl(f −1 (V )) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open in X, since U ⊆ cl*(int(U)).
Thus, f −1 (V ) = {a} is β # g - closed in X but it is not closed in X.
Then, f is β # g - continuous but not continuous.
21
Theorem 3.1.6.
If a map f : X → Y is semi- continuous, then it is β # g - continuous.
Proof.
Let f : X → Y be semi- continuous and let V be closed in Y.
Since f is semi- continuous,
f −1 (V ) is semi- closed in X.
Also, since every semi- closed set is β # g - closed,
f −1 (V ) is β # g - closed in X.
Therefore, f is β # g - continuous.
Remark 3.1.7.
The converse of the above theorem is not true, as seen from the following example.
Example 3.1.8.
Let X = {a, b, c} = Y.
Let f : X → Y be defined as,
f(a) = a, f(b) = b, f(c) = c.
Let τ = {φ, X, {a}} and σ = {φ, Y, {b, c}} be the topologies of X and Y respectively.
The sets which are closed in X are {φ, X, {b, c}}.
The sets which are closed in Y are {φ, Y, {a}}.
Let V = {a} which is closed in Y.
Then, f −1 (V ) = {a} is in X.
The sets which are β - closed in X are {{b}, {c}, {b, c}}.
βcl(f −1 (V )) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open in X, since U ⊆ cl*(int(U)).
Thus, f −1 (V ) = {a} is β # g - closed in X.
Then, f is β # g - continuous.
But int(cl(f −1 (V ))) ⊆ f −1 (V ) does not hold in X.
Thus f −1 (V ) = {a} is not semi- closed in X.
Then, f is not semi- continuous.
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Theorem 3.1.9.
If a map f : X → Y is α - continuous, then it is β # g - continuous.
Proof.
Let f : X → Y be α - continuous and let V be closed in Y.
Since f is α - continuous,
f −1 (V ) is α - closed in X.
Also, since every α - closed set is β # g - closed,
f −1 (V ) is β # g - closed in X.
Therefore, f is β # g - continuous.
Remark 3.1.10.
The converse of the above theorem is not true, as seen from the following example.
Example 3.1.11.
Let X = {a, b, c} = Y.
Let f : X → Y be defined as,
f(a) = a, f(b) = b, f(c) = c.
Let τ = {φ, X, {a}} and σ = {φ, Y, {b, c}} be the topologies of X and Y respectively.
The sets which are closed in X are {φ, X, {b, c}}.
The sets which are closed in Y are {φ, Y, {a}}.
Let V = {a} which is closed in Y.
Then, f −1 (V ) = {a} is in X.
The sets which are β - closed in X are {{b}, {c}, {b, c}}.
βcl(f −1 (V )) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open in X, since U ⊆ cl*(int(U)).
Thus, f −1 (V ) = {a} is β # g - closed in X.
Then, f is β # g - continuous.
But cl(int(cl(f −1 (V )))) ⊆ f −1 (V ) does not hold in X.
Thus f −1 (V ) = {a} is not α - closed in X.
Then, f is α - continuous.
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Theorem 3.1.12.
If a map f : X → Y is pre- continuous, then it is β # g - continuous.
Proof.
Let f : X → Y be pre- continuous and let V be closed in Y.
Since f is pre- continuous,
f −1 (V ) is pre- closed in X.
Also, since every pre- closed set is β # g - closed,
f −1 (V ) is β # g - closed in X.
Therefore, f is β # g - continuous.
Remark 3.1.13.
The converse of the above theorem is not true, as seen from the following example.
Example 3.1.14.
Let X = {a, b, c} = Y.
Let f : X → Y be defined as,
f(a) = a, f(b) = b, f(c) = c.
Let τ = {φ, X, {a}} and σ = {φ, Y, {b, c}} be the topologies of X and Y respectively.
The sets which are closed in X are {φ, X, {b, c}}.
The sets which are closed in Y are {φ, Y, {a}}.
Let V = {a} which is closed in Y.
Then, f −1 (V ) = {a} is in X.
The sets which are β - closed in X are {{b}, {c}, {b, c}}.
βcl(f −1 (V )) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open in X, since U ⊆ cl*(int(U)).
Thus, f −1 (V ) = {a} is β # g - closed in X.
Then, f is β # g - continuous.
But cl(int(f −1 (V ))) ⊆ f −1 (V ) does not hold in X.
Thus f −1 (V ) = {a} is not pre- closed in X.
Then, f is not pre- continuous.
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Theorem 3.1.15.
If a map f : X → Y is semi-pre continuous, then it is β # g - continuous.
Proof.
Let f : X → Y be semi-pre continuous and let V be closed in Y.
Since f is semi-pre continuous,
f −1 (V ) is semi-pre closed in X.
Also, since every semi-pre closed set is β # g - closed,
f −1 (V ) is β # g - closed in X.
Therefore, f is β # g - continuous.
Remark 3.1.16.
The converse of the above theorem is not true, as seen from the following example.
Example 3.1.17.
Let X = {a, b, c} = Y.
Let f : X → Y be defined as,
f(a) = a, f(b) = b, f(c) = c.
Let τ = {φ, X, {a}} and σ = {φ, Y, {b, c}} be the topologies of X and Y respectively.
The sets which are closed in X are {φ, X, {b, c}}.
The sets which are closed in Y are {φ, Y, {a}}.
Let V = {a} which is closed in Y.
Then, f −1 (V ) = {a} is in X.
The sets which are β - closed in X are {{b}, {c}, {b, c}}.
βcl(f −1 (V )) = φ.
Let U = {a}, where φ ⊆ {a}.
Here U is semi*-open in X, since U ⊆ cl*(int(U)).
Thus, f −1 (V ) = {a} is β # g - closed in X.
Then, f is β # g - continuous.
But int(cl(int(f −1 (V )))) ⊆ f −1 (V ) does not hold in X.
Thus f −1 (V ) = {a} is not semi-pre closed in X.
Then, f is not semi-pre continuous.
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Theorem 3.1.18.
If a map f : X → Y is β # g - irresolute, then it is β # g - continuous.
Proof.
Let f : X → Y be β # g - irresolute.
Let V be closed in Y.
Since every closed set is β # g - closed,
V is β # g - closed in Y.
Since f is β # g - irresolute,
f −1 (V ) is β # g - closed in X.
Thus, for a set V which is closed in Y, f −1 (V ) is β # g - closed in X.
Therefore, f is β # g - continuous.
Theorem 3.1.19.
If f : X → Y and g : Y → Z are both β # g - irresolute, then g◦ f : X → Z is β # g -
irresolute.
Proof.
Let f : X → Y and g : Y → Z be β # g - irresolute.
Let A ⊂ Z and let A be β # g - closed in Z.
Since g is β # g - irresolute,
g −1 (A) is β # g - closed in Y.
Since f : X → Y is β # g - irresolute,
f −1 (g −1 (A)) is β # g - closed in X.
Thus (g◦ f )−1 (A) = f −1 (g −1 (A)) is β # g - closed in X.
Therefore, g◦ f : X → Z is β # g - irresolute.
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Theorem 3.1.20.
Let X,Y and Z be any topological spaces. For any β # g - irresolute map f : X → Y
and any β # g - continuous map g : Y → Z, the composition g◦ f : X → Z is β # g -
continuous.
Proof.
Let f : X → Y be β # g - irresolute and let g : Y → Z be β # g - continuous.
Let A ⊂ Z and let A be closed in Z.
Since g is β # g - continuous,
g −1 (A) is β # g - closed in Y.
Since f : X → Y is β # g - irresolute,
f −1 (g −1 (A)) is β # g - closed in X.
Thus (g◦ f )−1 (A) = f −1 (g −1 (A)) is β # g - closed in X.
Therefore, g◦ f : X → Z is β # g - continuous.
Theorem 3.1.21.
Let X,Y and Z be any topological spaces. For any β # g - continuous map f : X → Y
and any continuous map g : Y → Z, the composition g◦ f : X → Z is β # g -
continuous.
Proof.
Let f : X → Y be β # g - continuous and let g : Y → Z be continuous.
Let A ⊂ Z and let A be closed in Z.
Since g is continuous,
g −1 (A) is closed in Y.
Since f : X → Y is β # g - continuous,
f −1 (g −1 (A)) is β # g - closed in X.
Thus (g◦ f )−1 (A) = f −1 (g −1 (A)) is β # g - closed in X.
Therefore, g◦ f : X → Z is β # g - continuous.
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Thus the above theorems can be represented in a tabular form as
f g g◦ f
β # g - continuous continuous β # g - continuous
β # g - irresolute β # g - continuous β # g - continuous
β # g - irresolute β # g - irresolute β # g - irresolute
28
3.2 Contra - β # g - continuous functions
Definition 3.2.1.
A function f : X → Y is said to be contra - β # g - continuous if f −1 (V ) is β # g -
closed in X for each open set V of Y.
Definition 3.2.2.
A space (X,τ ) is said to be β # g - locally indiscrete if every β # g - open set in it is
closed.
Theorem 3.2.3.
Every contra - continuous function is contra - β # g - continuous.
Proof.
Let f : X → Y be a contra - continuous function.
Let V be open in Y.
Since f is contra - continuous,
f −1 (V ) is closed in X.
Since every closed set is β # g - closed,
f −1 (V ) is β # g - closed in X.
Thus for a set V open in Y, f −1 (V ) is β # g - closed in X.
Therefore, f is contra - β # g - continuous.
Remark 3.2.4.
If a function is contra - β # g - continuous then it need not be a contra - continuous
function.
29
Theorem 3.2.5.
Every contra - semi - continuous function is contra - β # g - continuous.
Proof.
Let f : X → Y be a contra - semi - continuous function.
Let V be open in Y.
Since f is contra - semi - continuous,
f −1 (V ) is semi - closed in X.
Since every semi - closed set is β # g - closed,
f −1 (V ) is β # g - closed in X.
Thus for a set V open in Y, f −1 (V ) is β # g - closed in X.
Therefore, f is contra - β # g - continuous.
Remark 3.2.6.
If a function is contra - β # g - continuous then it need not be a contra - semi -
continuous function.
Theorem 3.2.7.
Every contra - pre - continuous function is contra - β # g - continuous.
Proof.
Let f : X → Y be a contra - pre - continuous function.
Let V be open in Y.
Since f is contra - pre - continuous,
f −1 (V ) is pre-closed in X.
Since every pre-closed set is β # g - closed,
f −1 (V ) is β # g - closed in X.
Thus for a set V open in Y, f −1 (V ) is β # g - closed in X.
Therefore, f is contra - β # g - continuous.
Remark 3.2.8.
If a function is contra - β # g - continuous then it need not be a contra - pre -
continuous function.
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Theorem 3.2.9.
Every contra - α - continuous function is contra - β # g - continuous.
Proof.
Let f : X → Y be a contra - α - continuous function.
Let V be open in Y.
Since f is contra - α - continuous,
f −1 (V ) is α - closed in X.
Since every α - closed set is β # g - closed,
f −1 (V ) is β # g - closed in X.
Thus for a set V open in Y, f −1 (V ) is β # g - closed in X.
Therefore, f is contra - β # g - continuous.
Remark 3.2.10.
If a function is contra - β # g - continuous then it need not be a contra - α - continuous
function.
Theorem 3.2.11.
Every contra - semi-pre - continuous function is contra - β # g - continuous.
Proof.
Let f : X → Y be a contra - semi-pre - continuous function.
Let V be open in Y.
Since f is contra - semi-pre - continuous,
f −1 (V ) is semi-pre - closed in X.
Since every semi-pre - closed set is β # g - closed,
f −1 (V ) is β # g - closed in X.
Thus for a set V open in Y, f −1 (V ) is β # g - closed in X.
Therefore, f is contra - β # g - continuous.
Remark 3.2.12.
If a function is contra - β # g - continuous then it need not be a contra - semi-pre -
continuous function.
31
Theorem 3.2.13.
If a function f : X → Y is β # g - continuous and X is β # g - locally indiscrete, then
f is contra - continuous.
Proof.
Let f : X → Y be β # g - continuous and let X be β # g - locally indiscrete.
Let V be open in Y.
Since f is β # g - continuous,
f −1 (V ) is β # g - open in X.
Since X is β # g - locally indiscrete,
f −1 (V ) is closed in X.
Thus if V is open in Y, then f −1 (V ) is closed in X.
Therefore, f is contra - continuous.
32
Chapter 4
β # g - CONNECTED AND β # g - COMPACT
Definition 4.1.1.
A topological space X is said to be β # g - connected if X cannot be expressed as
union of any two disjoint non-empty β # g - open sets in X.
Theorem 4.1.2.
If a space X is β # g - connected, then it is connected.
Proof.
Let X be β # g - connected.
Suppose X is not connected.
Then there exists two disjoint non-empty open sets A and B, such that A ∩ B = φ
and X = A ∪ B.
Since every open set is β # g - open,
A and B are β # g - open.
Then A and B forms a separation of X.
This is a contradiction to the fact that X is β # g - connected.
Therefore, X is connected.
33
Theorem 4.1.3.
For a topological space X, the following are equivalent
1. X is β # g - connected.
2. X and φ are the only subsets of X which are both β # g - open and β # g - closed.
Proof.
First, we prove,
(1) =⇒ (2)
Let X be β # g - connected.
Suppose S is a proper subset of X which is both β # g -open and β # g - closed.
Then its complement X - S is also both β # g - open and β # g - closed.
Thus X = S ∪ (X - S), is the union of two disjoint non-empty β # g - open sets.
This is a contradiction, since X is β # g - connected.
Therefore, S = φ or X are the only subsets of X which are both β # g - open and β # g
- closed.
Now, we prove,
(2) =⇒ (1)
Suppose X = A ∪ B, where A and B are disjoint non-empty β # g - open subsets of
X.
Let A be both β # g - open and β # g - closed.
By (2), A = φ or X.
If A = φ then B = X, or if A = X then B = φ.
Therefore, X is β # g - connected.
34
Theorem 4.1.4.
If f : X → Y is β # g - continuous onto and X is β # g - connected, then Y is connected.
Proof.
Let X be β # g - connected and let f : X → Y be β # g - continuous.
Suppose Y is not connected.
Then Y = A ∪ B,
where A and B are disjoint non-empty open sets in Y.
Since f is β # g - continuous and onto,
f −1 (Y ) = f −1 (A) ∪ f −1 (B)
That is, X = f −1 (A) ∪ f −1 (B),
where f −1 (A) and f −1 (B) are β # g - open in X.
This is a contradiction, since X is β # g - connected.
Therefore, Y is connected.
Theorem 4.1.5.
If f : X → Y is β # g - irresolute onto and X is β # g - connected, then Y is β # g -
connected.
Proof.
Let f : X → Y be β # g - irresolute and let X be β # g - connected.
Suppose Y is not β # g - connected.
Then Y = A ∪ B, where A and B are disjoint non-empty β # g - open sets in Y.
Since f is β # g - irresolute and onto,
f −1 (Y ) = f −1 (A) ∪ f −1 (B)
That is, X = f −1 (A) ∪ f −1 (B),
where f −1 (A) and f −1 (B) are disjoint non-empty β # g - open sets in Y.
This is a contradiction, since X is β # g - connected.
Therefore, Y is β # g - connected.
35
Theorem 4.1.6.
If the sets C and D are β # g - open and forms a separation of X and if Y is a β # g -
connected subspace of X, then Y lies entirely in C or in D.
Proof.
Let Y be a β # g - connected subspace of X.
Let C and D form a separation of X, where both C and D are β # g - open.
Then C ∩ Y and D ∩ Y are β # g - open in Y.
These two sets are disjoint.
Also, Y = (C ∩ Y)∪(D ∩ Y).
If both C ∩ Y and D ∩ Y are non-empty, then it forms a separation of Y.
This is a contradiction, since Y is β # g - connected.
Thus either the set C ∩ Y or D ∩ Y must be empty.
Therefore, Y lies entirely in C or in D.
36
4.2 β # g - Compact
Definition 4.2.1.
A collection A of β # g - open sets in X is called a β # g - open cover of B, which is a
subset of X, if B ⊆ ∪ {Uα : Uα ∈A} holds.
Definition 4.2.2.
A space X is said to be β # g - compact if every β # g - open cover of X has a finite
subcover.
Definition 4.2.3.
A subset B of X is said to be β # g - compact relative to X if for every β # g - open
cover A of B, there is a finite subcollection of A that covers B.
Definition 4.2.4.
A space X is said to be β # g - Lindelof if every cover of X by β # g - open sets contains
a countable subcover.
Remark 4.2.5.
Theorem 4.2.6.
Let f : X → Y be a β # g - open function and let Y be β # g - compact. Then X is
compact.
Proof.
Let Y be β # g - compact.
Let {Vα } be an open cover of X.
Then {f (Vα )} is a cover of Y consisting of β # g - open sets.
Since Y is β # g - compact,
{f (Vα )} contains a finite subcover.
(i.e.,) {f (Vα1 ), f (Vα2 ), . . . , f (Vαn )} forms a finite subcover of f (Vα ).
37
=⇒ {Vα1 , Vα2 , . . . , Vαn } is a finite subcover of {Vα }.
Therefore, X is compact.
Theorem 4.2.7.
Let f : X → Y be a β # g - continuous surjection and let X be β # g - compact. Then
Y is compact.
Proof.
Let f : X → Y be a β # g - continuous surjection and let X be β # g - compact.
Let {Vα } be an open cover of Y.
Then {f −1 (Vα )} is a cover of X, consisting of β # g - open sets.
Since X is β # g - compact,
{f −1 (Vα )} contains a finite subcover.
(i.e.,) {f −1 (Vα1 ), f −1 (Vα2 ), . . . , f −1 (Vαn )} is a finite subcover of {f −1 (Vα )}.
=⇒ {Vα1 , Vα2 , . . . , Vαn } is a finite subcover of {Vα }.
Therefore, Y is compact.
Theorem 4.2.8.
If f : X → Y is β # g - irresolute and onto and if X is β # g - compact, then Y is β # g
- compact.
Proof.
Let f : X → Y be β # g - irresolute and onto and let X be β # g - compact.
Let {Vα } be a β # g - open cover of Y.
Then, {f −1 (Vα )} is a cover of X, consisting β # g - open sets.
Since X is β # g - compact,
{f −1 (Vα )} contains a finite subcover.
(i.e.,) {f −1 (Vα1 ), f −1 (Vα2 ), . . . , f −1 (Vαn )} is a finite subcover of {f −1 (Vα )}.
=⇒ {Vα1 , Vα2 , . . . , Vαn } is a finite subcover of {Vα }.
Therefore, Y is β # g - compact.
38
Theorem 4.2.9.
Let f : X → Y be β # g - continuous onto and let X be β # g - Lindelof. Then, Y is
Lindelof.
Proof.
Let f : X → Y be β # g - continuous onto and let X be β # g - Lindelof.
Let {Vα } be an open cover of Y.
Then {f −1 (Vα )} is a cover of X, consisting of β # g - open sets.
Since X is β # g - Lindelof,
{f −1 (Vα )} contains a countable subcover.
(i.e.,) {f −1 (Vαn )} is a countable subcover of {f −1 (Vα )}.
=⇒ {Vαn } is a countable subcover of {Vα }.
Therefore, Y is Lindelof.
Theorem 4.2.10.
Let f : X → Y be β # g - irresolute onto and let X be β # g - Lindelof. Then, Y is β #
- Lindelof.
Proof.
Let f : X → Y be β # g - irresolute onto map and let X be β # g - Lindelof.
Let {Vα } be an β # g - open cover of Y.
Then {f −1 (Vα )} is a cover of X, consisting of β # g - open sets.
Since X is β # g - Lindelof,
{f −1 (Vα )} contains a countable subcover.
(i.e.,) {f −1 (Vαn )} is a countable subcover of {f −1 (Vα )}.
=⇒ {Vαn } is a countable subcover of {Vα }.
Therefore, Y is β # g - Lindelof.
39
Theorem 4.2.11.
Let f : X → Y be a β # g - open mapping and let Y be a β # g - Lindelof space, then
X is Lindelof.
Proof.
Let Y be β # g - Lindelof.
Let {Vα } be an open cover of X.
Then {f (Vα )} is a cover of Y which consists of β # g - open sets.
Since Y is β # g - Lindelof,
{f (Vα )} contains a countable subcover.
(i.e.,) {f (Vαn )} is a countable subcover of f (Vα ).
=⇒ {Vαn } is a countable subcover of {Vα }.
Therefore, X is Lindelof.
40
Application of Topology
Introduction
Topology in general is deliberated to be one of the headstones of modern abstract
mathematics.
Initially, the results in topology were inspired by real-life problems, and after
its official emergence, the prominence curved to its abstract growth. Nonetheless,
from past few eras there has been a momentous improvement of the applied topology
to other fields also. Nowadays, mathematicians and scientists use topology to model
and comprehend the real world occurrences.
Topology basically has emerged out of geometry, where the concept of distances
and angles are excluded. Here objects are taken and treated as if they are made of
rubber which one can stretch, crumble, twist and even deform but without cutting
and pasting.
In topology, the objects we take are called topological spaces.
Topologists cannot distinguish between a circle and a square, sphere and a
cube, because they share topological properties throughout the deformation math-
ematically called Homeomorphism (topological isomorphism). The properties that
remain unchanged during homeomorphism are what we call topological properties
or topological invariants.
Here we describe some applications of topology in other fields of Science and
Technology. Here we discuss topological applications to Biology, and Geographic
Information System (GIS)
41
Applications
1. Biology
Topology has influenced the world of Science and Technology with much great effects.
Topology besides being a branch of mathematics, it plays a very good role in Biology
also. Here we will discuss of how topology has influenced the study of DNA.
Deoxyribonucleic acid (DNA) is a molecule composed of two polynucleotide
chains that coil around each other to form a double helix carrying genetic instructions
for the development, functioning, growth and reproduction of all known organisms
and many viruses.
The two DNA strands are known as polynucleotides as they are composed of
simpler monomeric units called nucleotides. Each nucleotide is composed of one of
four nitrogen-containing nucleobases
i. Cytosine [C]
ii. Guanine [G]
iii. Adenine [A]
iv. Thymine [T],
a sugar called deoxyribose, and a phosphate group.
The nitrogenous bases of the two separate polynucleotide strands are bound
together, according to base pairing rules (A with T and C with G), with hydrogen
bonds to make double-stranded DNA. The complementary nitrogenous bases are
divided into two groups, pyrimidines and purines. In DNA, the pyrimidines are
thymine and cytosine; the purines are adenine and guanine.
Since genotypes-phenotypes are of primary importance in biology, we see how
topology is even useful in sequencing the right nucleotides in DNA strand.
The four nucleotides are arranged in a manner that they resemble a sequence.
The sequence of nucleotides on every single chain of DNA decides the sequence of
the other chain. The problem found in DNA research is in the comparison of distinct
DNA sequences.
In topology, we define something called Metric, which is basically a distance
function used to measure the distance between the elements of a set. Here the sets
42
Figure 4.1:
DL (x,y) = min{ iP + dP + rP },
where the minimum taken is from all sequences P that turn x into y.
Thus by measuring the distance between DNA strands using Topology, com-
parison of distinct DNA sequences are being done and hence Topology plays an
43
important role in DNA modelling by metric spaces.
• First, by looking at a map and working out exactly where the medium is
located and then entering that information manually to the image.
44
Figure 4.2:
45
Conclusion
Thus we have introduced a new closed set β # g - closed in a Topological space
and we have extended this concept to β # g - open sets. Using these sets we have
dicussed about β # g - continuous and contra - β # g - continuous functions. Also we
have defined two new spaces β # g -connected and β # g - compact.
46
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