Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 3s (2025) 

 

 

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Minimal and Maximal 𝒈𝜼-Continuous Functions in Topological Spaces 

 

D. Subbulakshmi 

Associate Professor, Department of Mathematics, RVS College of Arts and Science, Tamil Nadu, India, Email: 

subbulakshmi169@gmail.com. 

 

Article History: 

Received: 24-09-2024 

Revised: 05-11-2024 

Accepted: 20-11-2024 

Abstract:  

The concept of maximal and minimal gη-continuous functions and some new results 

are given. 

Keywords: minimal gη-continuous, maximal gη-continuous functions. 

 

1. INTRODUCTION 

Levine [2] proposed some properties in 1963, s-open sets were introduced into topological spaces. In 

1984, Andrijevic [1] described some of the topological properties of alpha sets. The concept of 

generalized closed sets in topological spaces was presented by Norman Levine [3]. [7, 8, 9] 

introduced the concept of 𝑔𝜂-closed sets and 𝑔𝜂-continuous functions and their various 

characterizations.  

Nakaoka and Oda developed two subclasses of open sets: maximal and minimal open sets [4,5,6]. 

Later, numerous authors concentrated on this subject, developing the concept of minimal and 

maximal open sets. Following these improvements, we investigate minimal and maximal gη-

continuous functions. 

2. MINIMAL 𝒈𝜼-CONTINUOUS FUNCTIONS 

This section introduces and establishes various properties of minimal 𝑔𝜂-continuous topological 

spaces. 

Definition 2.1[10]:  A minimal gη-open is a proper, nonempty gη-open subset U' of]= (𝑋′, 𝜏′) if any 

of its 𝑔𝜂-opens are 𝑈′ is 𝜑′ or 𝑈′.  

Definition 2.2: If �̇�−1(𝑀′) is a 𝑔𝜂-open in (𝑋′, 𝜏′) for any minimal open 𝑀′ in (𝑌′, 𝜎′) then a 

function �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) is minimal 𝑔𝜂-continuous. 

Example 2.3: Let𝑋′ = 𝑌′ = {𝑒′, 𝑓′, 𝑔′}, 𝜏′ = {𝑋′, 𝜑′, {𝑒′}, {𝑔′}, {𝑒′, 𝑔′}}, 𝜎′ =
{𝑌′, 𝜑′, {𝑒′}, {𝑓′}, {𝑒′, 𝑓′}}. Define  �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) by �̇�(𝑒′) = 𝑓′, �̇�(𝑓′) = 𝑔′, �̇�(𝑔′) = 𝑒′. Here 
{𝑒′}, {𝑓′} are minimal open in(𝑌′, 𝜎′). Therefore  �̇�  is minimal 𝑔𝜂-continuous. 

Theorem 2.4: All 𝑔𝜂-continuonus is minimal 𝑔𝜂-continuous. 

Proof: Let 𝑀′ be a minimal open in (𝑌′, 𝜎′) and �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) be a 𝑔𝜂-contenuous. 𝑀′ Is an 

open in (𝑌′, 𝜎′) as all minimal opens are open. Consequently, �̇� is 𝑔𝜂-continuous, �̇� is minimal 𝑔𝜂-

continuous as a result.  

However, the converse of this theorem is not necessarily true, as shown by the following example. 



Communications on Applied Nonlinear Analysis 

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Example 2.5: Let 𝑋′ = 𝑌′ = {𝑒′, 𝑓′, 𝑔′}, 𝜏′ = {𝑋′, 𝜑′, {𝑒′}, {𝑔′}, {𝑒′, 𝑔′}}, 𝜎′ =

{𝑌′, 𝜑′, {𝑒′}, {𝑓′}, {𝑒′, 𝑓′}}. Define �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) as �̇�(𝑒′) = 𝑓′, �̇�(𝑓′) = 𝑔′, �̇�(𝑔′) = 𝑒′. Then 

�̇�−1(𝑒′) = 𝑔′, �̇�−1(𝑓′) = 𝑒′. Therefore �̇� is minimal 𝑔𝜂-continuous. Hence �̇� is not                   𝑔𝜂-

continuous. 

Theorem 2.6: If any maximal closed in's inverse image 𝑌′, 𝜎′) is 𝑔𝜂-closed in (𝑋′, 𝜏′) then  let 

�̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) be minimal 𝑔𝜂-continuous 

Proof: �̇� be a minimal 𝑔𝜂-continuous and let 𝑁′ be a maximal closed in (𝑌′, 𝜎′). (𝑌′, 𝜎′) − 𝑁′ is a 

minimal open in (𝑌′, 𝜎′). When �̇� is minimal 𝑔𝜂-continuous, �̇�−1((𝑌′, 𝜎′) − 𝑁′) is 𝑔𝜂-open in 

(𝑋′, 𝜏′). So �̇�−1((𝑌′, 𝜎′) − 𝑁′) = (𝑋′, 𝜏′) − �̇�−1(𝑁′) is 𝑔𝜂-open in (𝑋′, 𝜏′). �̇�−1(𝑁′) is 𝑔𝜂-closed in 

(𝑋′, 𝜏′).  

On the contrary, �̇�−1(𝑁′) is 𝑔𝜂-closed in (𝑋′, 𝜏′) for all maximal closed 𝑁′ in (𝑌′, 𝜎′). Let 𝑀′ be a 

minimal open in (𝑌′, 𝜎′). So �̇�−1((𝑌′, 𝜎′) − 𝑀′) = (𝑋′, 𝜏′) − 𝕒−1(𝑀′) is 𝑔𝜂-closed in (𝑋′, 𝜏′). 

Therefore �̇� is minimal 𝑔𝜂-continuous. 

Theorem 2.7: Let �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) is minimal 𝑔𝜂-continuous iff, 𝑞′ ∈ (𝑋′, 𝜏′)  and minimal open 

𝑀′ in (𝑌′, 𝜎′) holding �̇�(𝑞′), there is 𝑔𝜂-open 𝑁′ in (𝑋′, 𝜏′) like that 𝑞′ ∈ 𝑁′, �̇�(𝑁′) ⊂ 𝑀′. 

Proof: 𝑀′ be minimal open in (𝑌′, 𝜎′) holding �̇�(𝑞′), 𝑞′ ∈ 𝑁′ where 𝑁′ is 𝑔𝜂-open in (𝑋′, 𝜏′). Since 

�̇� is minimal 𝑔𝜂-continuous, �̇�−1(𝑁′) is 𝑔𝜂-open in(𝑋′, 𝜏′). And 𝑁′ = �̇�−1(𝑀′). Therefore �̇�(𝑁′) ⊂

𝑀′. 

On the contrary, 𝑀′ be minimal open in (𝑌′, 𝜎′). Then there is 𝑔𝜂-open 𝑁′ in (𝑋′, 𝜏′), such that 𝑞′ ∈

𝑁′,  �̇�(𝑞′) ∈ �̇�(𝑁′) ⊂ 𝑀′, 𝑞 ∈ �̇�−1(�̇�(𝑁′)) ⊂ �̇�−1(𝑀′). Therefore �̇�−1(𝑀′) is 𝑔𝜂-open in (𝑋′, 𝜏′). 

Hence �̇� is minimal 𝑔𝜂-continuous. 

Theorem 2.8: Let 𝐵′ be a subset of (𝑋′, 𝜏′). If �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) is minimal 𝑔𝜂-continuous then 

the restriction function �̇� | 𝐵′: 𝐵′ → (𝑌′, 𝜎′) is minimal 𝑔𝜂-continuous. where 𝐵′ has the relative 

topology. 

Proof: Assume 𝐴′ is subset of (𝑋′, 𝜏′) and 𝑀′ be minimal open in (𝑌′, 𝜎′). When �̇� is minimal              

𝑔𝜂-continuous, �̇�−1(𝑀′) is 𝑔𝜂-open in (𝑋′, 𝜏′). So (�̇�|𝐵′)−1(𝑀′) = 𝐵′ ∩ �̇�−1(𝑀′). Hence                         

𝐵′ ∩ �̇�−1(𝑀′) is 𝑔𝜂-open in 𝐵′. Thus �̇�|B′ is minimal 𝑔𝜂-continuous. 

Remark 2.9: Minimal 𝑔𝜂-continuous do not always have to be minimal 𝑔𝜂-continuous in 

composition.  

Theorem 2.10: If �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′)  and �̇�: (𝑌′, 𝜎′) → (𝑍′, μ′) be 𝑔𝜂-continuous and minimal                

𝑔𝜂-continuous so �̇�o�̇�: (𝑋′, 𝜏′) → (𝑍′, μ′) is minimal 𝑔𝜂-continuous. 

Proof: 𝐴′ be minimal open in(𝑍′, μ′), when �̇� is minimal 𝑔𝜂-continuous, �̇�−1(𝐴) is 𝑔𝜂-open in 

(𝑌′, 𝜎′). So �̇� is 𝑔𝜂-continuous, �̇�−1(�̇�−1(𝐴)) = (�̇�o�̇�)−1(𝐴) is 𝑔𝜂-open in (𝑋′, 𝜏′). Hence �̇�o�̇� be 

𝑔𝜂-continuous. 

 

 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

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3. MAXIMAL 𝒈𝜼-CONTINUOUS FUNCTIONS  

  Introducing maximal 𝑔𝜂-continuous is the goal of this section. Examples are used to obtain 

some attributes of such functions. 

Definition 3.1: A Maximal 𝑔𝜂-open is a proper nonempty 𝑔𝜂-open subset 𝑈′ of (𝑋′, 𝜏′) if any of its 

𝑔𝜂-opens are 𝑈′ is either (𝑋′, 𝜏′) or 𝑈′.  

Definition 3.2: If �̇�−1(𝑀′) is a 𝑔𝜂-open in (𝑋′, 𝜏′) for any maximal open 𝑀′ in (𝑌′, 𝜎′) then a 

function �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) is maximal 𝑔𝜂-continuous. 

Example 3.3: Let 𝑋′ = 𝑌′ = {𝑒′, 𝑓′, 𝑔′}, 𝜏′ = {𝑋′, 𝜑′, {𝑒′}, {𝑓′, 𝑔′}},  𝜎′ = {𝑌′, 𝜑′, {𝑒′}, {𝑔′}, {𝑒′, 𝑔′}}. 

Define  �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) by �̇�(𝑒′) = 𝑒′, �̇�(𝑓′) = 𝑔′, �̇�(𝑔′) = 𝑓′. Here {𝑒′, 𝑔′} is maximal open in 

(𝑌′, 𝜎′). Therefore �̇� is maximal 𝑔𝜂-continuous. 

Theorem 3.4: All 𝑔𝜂-continuous is maximal 𝑔𝜂-continuous. 

Proof: Let 𝑀′ be a maximal open in (𝑌′, 𝜎′) and �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) be a 𝑔𝜂-continuous, 𝑀′ be a 

maximal open in (𝑌′, 𝜎′). All maximal open is an open when 𝑀′ is an open in (𝑌′, 𝜎′). So �̇� is      𝑔𝜂-

continuous. Therefore �̇� is maximal 𝑔𝜂-continuous. 

The converse of the preceding theorem does not necessarily have to be true, as demonstrated by the 

example that follows. 

Example 3.5: Take 𝑋′ = 𝑌′ = {𝑒′, 𝑓′, 𝑔′}, 𝜎′ = {𝑌′, 𝜑′, {𝑒′}, {𝑔′}, {𝑒′, 𝑔′}}, 𝜏′ =

{𝑋′, 𝜑′, {𝑒′}, {𝑓′}, {𝑒′, 𝑓′}}.  Assign  �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) by  �̇�(𝑒′) = 𝑓′, �̇�(𝑓′) = 𝑒′, �̇�(𝑔′) = 𝑔′. 

Then �̇�−1({𝑒′, 𝑔′}) =  {𝑓′, 𝑔′} is 𝑔𝜂-open in (𝑋′, 𝜏′). Consequently �̇� is maximal                𝑔𝜂-

continuous. So �̇� is not 𝑔𝜂-continuous.  

Remark 3.6: There is independence between maximal-continuous and minimal-continuous. 

Example 3.7: Take 𝑋′ = 𝑌′ = {𝑒′, 𝑓′, 𝑔′}, 𝜏′ = {𝑋′, 𝜑′, {𝑒′}, {𝑔′}, {𝑒′, 𝑔′}}, 𝜎′ = {𝑌′, 𝜑′, {𝑒′}, {𝑓′, 𝑔′}}. 

Assign  �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) by �̇�(𝑒) = 𝑓, �̇�(𝑓) = 𝑒, �̇�(𝑔) = 𝑔. Now {𝑓′, 𝑔′} is maximal open in 

(𝑌′, 𝜎′)). For that reason �̇� is maximal 𝑔𝜂-continuous. Consequently, �̇� is not minimal  𝑔𝜂-

continuous.  

Example 3.8: Take 𝑋′ = 𝑌′ = {𝑒′, 𝑓′, 𝑔′}, 𝜏′ = {𝑋′, 𝜑′, {𝑒′}}, 𝜎′ = {𝑌′, 𝜑′, {𝑒′}, {𝑔′}, {𝑒′, 𝑔′}}. Assign  

�̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) by �̇�(𝑒) = 𝑓′,  �̇�(𝑓′) = 𝑒′, �̇�(𝑔′) = 𝑔′. Now {𝑒′}, {𝑔′} is minimal open in 

(𝑌′, 𝜎′). Consequently, �̇� is minimal 𝑔𝜂-continuous and �̇� is not maximal 𝑔𝜂-continuous.  

Theorem 3.9: Assign �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) is maximal 𝑔𝜂-continuous iff the inverse image of each 

minimal closed in (𝑌′, 𝜎′)is a 𝑔𝜂-closed in (𝑋′, 𝜏′). 

Proof: Take �̇� be a maximal 𝑔𝜂-continuous, 𝑁′ be a minimal closed in (𝑌′, 𝜎′). So (𝑌′, 𝜎′) − 𝑁′ is a 

maximal open in (𝑌′, 𝜎′). As �̇� is maximal 𝑔𝜂-continuous, �̇�−1((𝑌′, 𝜎′) − 𝑁′) is an 𝑔𝜂-open in 

(𝑋′, 𝜏′). But �̇�−1((𝑌′, 𝜎′) − 𝑁′) = (𝑋′, 𝜏′) − �̇�−1(𝑁′) is 𝑔𝜂-open in (𝑋′, 𝜏′). Consequently, �̇�−1(𝑁′) 

is 𝑔𝜂-closed in (𝑋′, 𝜏′).  

On the contrary, �̇�−1(𝑁′) is 𝑔𝜂-closed in (𝑋′, 𝜏′) for all minimal closed 𝑁′ in (𝑌′, 𝜎′). Let 𝑀′ be 

maximal open in (𝑌′, 𝜎′). So (𝑌′, 𝜎′) − 𝑀′  is minimal closed in (𝑌′, 𝜎′). But                          



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�̇�−1((𝑌′, 𝜎′) − 𝑀′) = (𝑋′, 𝜏′) − �̇�−1(𝑀′) is 𝑔𝜂-closed in (𝑋′, 𝜏′). Hence �̇�−1(𝑀′) is 𝑔𝜂-open in 

(𝑋′, 𝜏′). Thus �̇� is maximal 𝑔𝜂-continuous. 

Theorem 3.10: Assign �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) is maximal 𝑔𝜂-continuous iff 𝑞′ ∈ (𝑋′, 𝜏′) and maximal 

open 𝑀′ in (𝑌′, 𝜎′) holding  �̇�(𝑞′), there is 𝑔𝜂-open 𝑁′ in (𝑋′, 𝜏′) in order that 𝑞′ ∈ 𝑁′, �̇�(𝑁′) ⊂ 𝑀′. 

Proof: Take 𝑀′ be maximal open in (𝑌′, 𝜎′) holding �̇�(𝑞′), 𝑞 ′ ∈ 𝑁′ where 𝑁′ is an 𝑔𝜂-open in 

(𝑋′, 𝜏′), �̇� is maximal 𝑔𝜂-continuous and �̇�−1(𝑁′) is 𝑔𝜂-open in (𝑋′, 𝜏′). Then 𝑁′ = �̇�−1(𝑀′). So  

�̇�(𝑁′) ⊂ 𝑀′. 

On the contrary, 𝑀′ be a maximal open in (𝑌′, 𝜎′). Then there is 𝑔𝜂-open 𝑁′ in (𝑋′, 𝜏′), 𝑞′ ∈ 𝑁′,                           

�̇�(𝑞′) ∈  �̇�(𝑁′) ⊂ 𝑀′, 𝑞′ ∈ �̇�−1(�̇�(𝑁′)) ⊂ �̇�−1(𝑀′). So �̇�−1(𝑀′) is 𝑔𝜂-open in (𝑋′, 𝜏′). 

Consequently,  �̇� is maximal 𝑔𝜂-continuous. 

Theorem 3.11: Take 𝐵′ be a non-empty subset of (𝑋′, 𝜏′) and �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) is maximal                       

𝑔𝜂-continuous then the restriction �̇� | 𝐵′: 𝐵′ → 𝑌′ is maximal 𝑔𝜂-continuous. When 𝐵′ has the 

relative topology. 

Proof: Conclude 𝐴′ is a non-empty subset of a (𝑋′, 𝜏′)and 𝑀′ be any maximal open in (𝑌′, 𝜎′). So �̇� 

is maximal 𝑔𝜂-continuous, �̇�−1(𝑀′) is 𝑔𝜂-open in (𝑋′, 𝜏′). Using relative topology, 

(�̇� | 𝐵′)−1(𝑀′) = 𝐵′ ∩ �̇�−1(𝑀′). So 𝐵′ ∩ �̇�−1(𝑀′) is 𝑔𝜂-open in 𝐵′. Thus �̇� | 𝐵′ is maximal           

𝑔𝜂-continuous. 

Remark 3.12: Maximal 𝑔𝜂-continuous functions do not always have to be maximal                        

𝑔𝜂-continuous in composition. 

Theorem 3.13: The maximal gη-continuousness is  �̇�o�̇� ̇: (𝑋′, 𝜏′)→(𝑍′, μ′). If �̇�: (𝑋′, 𝜏′) → (𝑌′, 𝜎′) is 

gη-continuous and �̇�: (𝑌′, 𝜎′)→(𝑍′, μ′) is maximal gη-continuous. 

Proof:  Since �̇� is a maximal 𝑔𝜂-continuous, �̇�−1(𝐴′) is 𝑔𝜂-open in (𝑌′, 𝜎′) , assuming that 𝐴′ be a 

maximal open in (𝑍′, μ′). Every open is 𝑔𝜂-open. But �̇� is 𝑔𝜂-continuous, �̇�−1(�̇�−1(𝐴′)) =

(�̇�o�̇�)−1(𝐴′) is 𝑔𝜂-open in (𝑋′, 𝜏′). Hence �̇�o�̇�  is 𝑔𝜂-continuous. 

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[11] D Little Femilin Jana, R Jaya, M Arokia Ranjithkukar, S Krishnakumar, R Senthil Kumar. RESOLVING SETS 

AND DIMENSION IN SPECIAL GRAPHS. Advances and Application of Mathematical Sciences Volume 21, 

Issue 7(2022); 3709-3717. 

[12] R Senthil Kumar, RV Shanmathi, G Mageswaran, J Manikandan. Power Flow Analysis of Transient Stability in 

Microgrids used in Power Stations. 2022 Sixth International Conference on I-SMAC (IoT in Social, Mobile, 

Analytics and Cloud)(I-SMAC),(2022); 936-942. 

[13] Y Rosemathy, K Alli, T Thanigasalam, E Rajesh, R Senthil Kumar. On Soft SIgδs-closed sets. E3S Web of 

Conferences 376, 01112 (2023). 

[14] R Senthil Kumar, BVS Acharyulu, PK Dhal, Richa Adlakha, Sonu Kumar, C Saravanan, Krishna Bikram Shah. 

Optimization Technique for Renewable Energy Storage Systems for Power Quality Analysis with Connected 

Grid. International Transactions on Electrical Energy Systems Volume 2023, Article ID 4675421. 

[15] Senthil Kumar R4 and Tharmar S4 Rajeev Gandhi S1, Prabhavathi K2*, VeeraSivaji R3. Efficient Domination 

In Fuzzy Graphs and Intuitionistic Fuzzy Graphs in Strong and weak forms. E3S Web of Conferences 399, 

04026 (2023)