EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 6117 ISSN 1307-5543 – ejpam.com Published by New York Business Global Weakly Connected Independence Number of a Graph Reignver Merontos1,∗, Imelda S. Aniversario1,2, Michael B. Frondoza1,2, Sergio R. Canoy, Jr.1,2 1 Department of Mathematics and Statistics, College of Science and Mathematics, MSU-Iligan Institute of Technology, Iligan City, Philippines 2 Center of Mathematical and Theoretical Physical Sciences - PRISM, MSU-Iligan Institute of Technology, Iligan City, Philippines Abstract. Let G be a simple undirected connected graph with vertex and edge sets V (G) and E(G), respectively. The subgraph ⟨S⟩w of S ⊆ V (G) is the graph whose vertex set is N [S] and whose edge set Ew consists of edges in E(G) incident to some vertex in S. A subset S of V (G) is a weakly connected set of G if ⟨S⟩w is connected. S is called a weakly connected independent set (WCIS) of G if it is both weakly connected and independent. In this paper, we characterize the weakly connected independent sets in the join, corona, and the lexicographic product of two graphs. From these characterizations the weakly connected independence numbers of the corresponding graphs are easily determined. Also, characterization of graphs G with weakly connected independence numbers αw(G) equal to 1, n−1 and n are given. It is also shown that for any non-negative integers k, m, and n with k > m+1 and n ≥ k+m+2, there exists a connected graph G such that |V (G)| = n, αw(G) = k and α(G) = k +m. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Independent, weakly connected set, weakly connected independence number, join, corona, lexicographic 1. Introduction The concept of weakly connected domination was introduced by Grossman [1] and was studied by Dunbar et.al [2] where upper and lower bounds for γw(G) were obtained. This parameter extends domination by ensuring weak connectivity within the dominating set. Graph theory plays a vital role in modeling networks, where balancing independence and connectivity is crucial. Weakly connected domination has been widely studied, with works by Alzoubi et al. [3] on minimal sets and Bendali et al. [4] on computational ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6117 Email addresses: reignver.merontos@g.msuiit.edu.ph (R. Merontos) imelda.aniversario@g.msuiit.edu.ph (I. Aniversario) michael.frondoza@g.msuiit.edu.ph (M. Frondoza) sergio.canoy@g.msuiit.edu.ph (S. Canoy Jr.) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) R. Merontos et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6117 2 of 9 complexity. Recent studies, such as those by Hamja et al. [5] and Militante and Eballe [6], further explored variations in weakly connected parameters. Motivated by the growing interest in weak connectivity and its role in network the- ory, this paper introduces and investigates the weakly connected independence number. The study of this new parameter involves addressing the inherent difficulty of combin- ing independence and connectivity, two properties that often counteract each other in graph structures. Similar to weakly connected domination, we believe that understanding the weakly connected independence number will yield significant contributions to inde- pendence theory and stimulate further research in graph operations, and combinatorial optimization. 2. Terminologies and Notations Let G = (V (G), E(G)) be a simple undirected graph. The distance between two vertices v, w ∈ V (G), denoted dG(v, w), is the length of a shortest v-w path connecting v and w. Any v-w path of length dG(v, w) is called a v-w geodesic. The open neighborhood of a vertex v of G is the set NG(v) = {u ∈ V (G) : uv ∈ E(G)}, while its closed neighborhood is the set NG[v] = NG(v) ∪ {v}. The open neighborhood of a set S ⊆ V (G) is the set NG(S) = ∪v∈SNG(v) and its closed neighborhood is the set NG[S] = S ∪ NG(S). Any v ∈ V (G) with |NG(v)| = 0 is called an isolated vertex. Vertex v is a leaf or an endvertex if |NG(v)| = 1. A vertex w of G is a support vertex if wv ∈ E(G) for some leaf v in G. The sets I(G), L(G), and S(G) will, respectively, denote the sets containing all the isolated vertices, leaves, and support vertices in G. A subset A of V (G) is an independent set if for every pair of distinct vertices in G do not form an edge. The maximum cardinality of an independent set in G, denoted by α(G), is called the independence number of G. Any independent set with cardinality equal to α(G) is called an α-set in G. A set S ⊆ V (G) is a dominating set in G if NG[S] = V (G). It is a super dominating set if for every v ∈ V (G) \S there exists w ∈ S such that NG(w)∩ [V (G) \S] = {v}. The domination number (super domination number) of G, denoted γ(G) (resp. γsp(G)) is the minimum cardinality of a dominating (resp. super dominating) set in G. Any dominating set (super dominating set) with cardinality γ(G) (resp. γsp(G)) is called a γ-set (resp. γsp-set). The subgraph ⟨S⟩w of S ⊆ V (G) is the graph whose vertex set is NG[S] and whose edge set Ew consists of edges in E(G) incident to some vertex in S. A subset S of V (G) is a weakly connected set of G if ⟨S⟩w is connected. S is called a weakly connected independent set (WCIS) ofG if it is both weakly connected and independent. The maximum cardinality of a WCIS of G is called the weakly connected independence number of G and is denoted by αw(G). A WCIS of G having cardinality αw(G) is called a maximum WCIS of G. A set S that is both a WCIS and a dominating set of G is called a weakly connected independent dominating set (WCIDS) of G. Note that a minimum WCIDS of G always exists (see Dunbar et al. [2]). Denote by ιc(G) the cardinality of a minimum WCIDS (or ιG-set) of G. Let G and H be any two graphs. The join G + H is the graph with vertex set R. Merontos et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6117 3 of 9 V (G+H) = V (G)∪ V (H) and edge set E(G+H) = E(G)∪E(H)∪ {uv : u ∈ V (G), v ∈ V (H)}. The corona G◦H of two graphs G and H is the graph obtained by taking one copy of G and |V (G)| copies of H, and then joining the ith vertex of G to every vertex of the ith copy of H. Denote by Hv the copy of H in G ◦H, every vertex of which is adjacent to a unique vertex v ∈ G. The lexicographic product G[H] of two graphs G and H is the graph with V (G[H]) = V (G)×V (H), and (u, u′)(v, v′) ∈ E(G[H]) if and only if either uv ∈ E(G) or u = v and u′v′ ∈ E(H). Observe that any non-empty subset C of V (G) × V (H) (in fact, any set of ordered pairs) can be written as C = ∪x∈S({x} × Tx) ⊆ V (G[H]), where S ⊆ V (G) and Tx ⊆ V (H) for all x ∈ S. Henceforth, we shall use this form to denote any subset C of V (G)× V (H). Readers are referred to [7] for other basic definitions that are not given here. 3. Results It is worth noting that if G is a graph and S ⊆ V (G), then ⟨S⟩w is simply obtained from the ⟨N [S]⟩ by deleting all the edges e = xy in E(G) with x, y ∈ N(S) \ S. The first result is easy and almost follows from the definitions. Theorem 1. For any graph G of order n, 1 ≤ αw(G) ≤ α(G). Moreover, (i) αw(G) = 1 if and only if every component H of G is complete; and (ii) αw(G) = α(G) if and only if ⟨S⟩w is connected for some α-set Sin G. Proof. Clearly, 1 ≤ αw(G). Since every weakly connected independent set is indepen- dent, it follows that αw(G) ≤ α(G). (i) Suppose that αw(G) = 1 and assume on the contrary that G has a component H which is not complete. Then there exist distinct vertices v and w of H such that dH(v, w) = dG(v, w) = 2. Let S = {v, w} and let u ∈ NG(v) ∩ NG(w). Then S is independent and ⟨S⟩w is connected. This implies that αw(G) ≥ |S| = 2, contrary to our assumption that αw(G) = 1. Thus, every component H of G is complete. For the converse, suppose that every component of G is complete. Let S be an αw-set in G. Since ⟨S⟩w is connected, S ⊆ V (H) for a unique complete component H of G. Since S is an independent set in H, it follows that |S| = 1. Thus, αw(G) = |S| = 1. (ii) Suppose that αw(G) = α(G), say S is an αw-set in G. Then S is an independent set and and ⟨S⟩w is connected. Since |S| = α(G), it follows that S is an α-set in G. Conversely, suppose ⟨S⟩w is connected for some α-set S in G. Then S is a weakly connected independent set. Therefore, αw(G) = |S| = α(G). The next result follows from Theorem 1. Corollary 1. Let G be a connected graph of order n. Then αw(G) = 1 if and only if G = Kn. We now characterize all connected graphs G of order n ≥ 2 such that αw(G) = n− 1. R. Merontos et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6117 4 of 9 Theorem 2. Let G be a connected graph of order n ≥ 2. Then αw(G) = n − 1 if and only if G = K1,n−1. Proof. Assume that G is a connected graph of order n ≥ 2 and αw(G) = n − 1. If n = 2, then αw(G) = 1. By Corollary 1, G = K2 = K1,1. Suppose n ≥ 3. Let S be an αw-set of G, say S = V (G) ∖ {w}. Since S is independent, it follows that uv /∈ E(G) for every pair of vertices u, v ∈ S. Now, since S is weakly connected in G, it follows that w ∈ NG(v) for each v ∈ S. Consequently, G = K1,n−1. For the converse, suppose that G = K1,n−1. Let w be the central vertex of G. Then, clearly, S = V (G) ∖ {w} is a weakly connected independent set in G. Since S is also an α-set in G, it follows from Theorem 1(ii) that αw(G) = n− 1. Theorem 3. Let k, m, and n be non-negative integers with k > m+1 and n ≥ k+m+2. Then there exists a connected graph G such that |V (G)| = n, αw(G) = k and α(G) = k +m. Proof. Let us consider the following cases: Case 1. Suppose that n = k +m+ 2. Suppose first that m = 0. Let H1 = K1,k, where u is the central vertex of H1 and v1, v2, . . . , vk are the remaining vertices (see Figure 1). Let G be the graph obtained from H1 by adding the vertex x and the edges xu and xvk. Clearly, the set S1 = {v1, v2, . . . , vk} is both an α-set and an αw-set of G. Thus, |V (G)| = k + 2 = n, αw(G) = α(G) = k. .................................... .................................... .................................... .................................... .................................... .................................... ............................................... ........ ........ ........ ........ ........ ........ ... ............ ........... ........... ..... ............ ........... ........... ..... ....................................... u v3 v2 v1 vkv4 . . . H1 : .................................... .................................... .................................... .................................... .................................... .................................... .................................... ............................................... ........ ........ ........ ........ ........ ........ ... ............ ........... ........... ..... ............ ........... ........... ..... ....................................... ........................................................... ............ ........... ........... ..... xu v3 v2 v1 vkv4 . . . G : Figure 1: Graph G with αw(G) = α(G) = k Next, suppose that m > 0. Let H2 be the union of K1,k−1 and K1,m+1, where u is the central vertex of K1,k−1 and x1, x2, . . . , xk−1 are its remaining vertices, v is the central vertex of K1,m+1 and y1, y2, . . . , ym+1 are its remaining vertices. Let G be the graph obtained from H2 by adding the edge uv (see Figure 2). The set S2 = {x1, x2, . . . , xk−1}∪ {y1, y2, . . . , ym+1} is the unique α-set of G and S3 = {x1, x2, . . . , xk−1, v} is an αw-set of G. Hence, |V (G)| = k +m+ 2 = n, αw(G) = |S3| = k and α(G) = |S2| = k +m. .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ............................................................................................................................... ............................................... ........ ........ ........ ........ ........ ........ ... ............ ........... ........... ..... ............ ........... ........... ..... ....................................... ............................................... ........ ........ ........ ........ ........ ........ ... ............ ........... ........... ..... ............ ........... ........... ..... ....................................... uv y3 y2 y1 ym+1y4 x3 x2 x1 xk−1x4 . . . . . . G : Figure 2: Graph G with αw(G) = k and α(G) = k +m R. Merontos et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6117 5 of 9 Case 2. Suppose that n > k +m+ 2. Consider the graphs G1 and G2 in Figure 3. Suppose that m = 0. .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ............................................... ........ ........ ........ ........ ........ ........ ... ............ ........... ........... ..... ............ ........... ........... ..... ....................................... .......... .......... .......... .......... .......... .......... .......... .......... .......... ........ ................ ............... ............... ............... ............... ............... ....... ................................................................ .................. ............................................................... ....................................... ........... .......... .......... .......... .......... ...... ............................................................................. ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ........ ......... ........ ........ ........ ... .......................... ... u x3 x2 x1 xkx4 . . . G1 : z1 z2 z3 zt .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... .................................... ........... .......... .......... .......... .......... .......... .......... .......... .......... ....... ................ ............... ............... ............... ............... ............... ....... ............................................................... ....................................... ......... ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ........ ...... .......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ........ ........... .......... .......... .......... .......... ...... ......... ........ ........ ........ ... .................................................................................................................... ............................................... ........ ........ ........ ........ ........ ........ ... ............ ........... ........... ..... ............ ........... ........... ..... ....................................... ........................................... ........ ........ ........ ........ ........ ........ ... ............ ........... ........... ..... ............ ........... ........... . ....................................... ................................................................ .................. ............................................................................. .......................... z1 z2 z3 zr uv y3 y2 y1 ym+1y4 x3 x2 x1 xk−1x4 ... . . . . . . G2 : Figure 3: Graph G with αw(G) = k and α(G) = k +m Let t = n − k − 1 and take G = G1. Then, clearly, S4 = {x1, x2, . . . , xk} is a both an α-set and an αw-set of G. Thus, |V (G)| = k + t + 1 = n, αw(G) = α(G) = k. If m > 0, then set r = n − k − m − 2 and take G = G2. It can easily be verified that |V (G)| = (k − 1) + (m+ 1) + r + 2 = n, αw(G) = k and α(G) = k +m. The next result is a consequence of Theorem 3. Corollary 2. The difference α(G)− αw(G) can be made arbitrarily large. Proof. Let m, n, and k be positive integers such that k > m+1 and n = k+m+2. By Theorem 3, there exists a connected graph with |V (G)| = n, αw(G) = k and α(G) = k+m. Therefore, α(G)− αw(G) = m. Next, we give the weakly connected independence number of paths and cycles. Theorem 4. (i) αw(Pn) = ⌈n 2 ⌉ = α(Pn) for all n ≥ 1. (ii) αw(Cn) = ⌊n 2 ⌋ = α(Cn) for all n ≥ 3. Proof. (i) Let Pn = [v1, v2, . . . , vn]. If n is even, then SE = {vi ∈ V (Pn) : i is even} and SO = {vj ∈ V (Pn) : j is odd} are α-sets in Pn. Since ⟨NG[SE ]⟩ = ⟨NG[SO]⟩ = Pn, it follows that SE and SO are weakly connected independent sets in Pn. By Theorem 1(ii), we have αw(Pn) = α(Pn) = |SE | = n 2 . If n is odd, then S = {vj ∈ V (Pn) : j is odd} is the unique α-set in Pn. Again, since ⟨NG[S]⟩ = Pn, it follows that S is a weakly connected independent set in Pn. By Theorem 1(ii), we have αw(Pn) = |S| = n+1 2 . (ii) Let Cn = [v1, v2, . . . , vn, v1]. If n is even, the S1 = {v1, v3, · · · , vn−1} is an α- set in Cn. Since ⟨NG[S1]⟩ = Cn, it follows that S1 is a weakly connected independent set in Cn. By Theorem 1(ii), we have αw(Cn) = α(Cn) = |S1| = n 2 . If n is odd, then S2 = {v1, v3, · · · , vn−2} is an α-set in Cn. Since ⟨NG[S]⟩ = Pn, it follows that S2 is a weakly connected independent set in Cn. By Theorem 1(ii), we have αw(Cn) = α(Cn) = |S| = n−1 2 . In what follows, we characterize the WCIS in G +H and determine the weakly con- nected independent number of G+H. R. Merontos et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6117 6 of 9 Theorem 5. Let G and H be any two graphs. Then S is a WCIS in G+H if and only if either S is an independent set in G or S is an independent set in H. Proof. Assume that S is a WCIS in G+H. Since S is an independent set in G+H, it follows that either S ⊆ V (G) or S ⊆ V (H). Thus, S is an independent set in G or an independent set in H. Conversely, let S be an independent set in G. Clearly, S is an independent set of G+H. Since V (H) ⊆ NG+H(x) for every x ∈ NG[S], it follows that ⟨NG+H [S]⟩ is connected. This implies that ⟨S⟩w is weakly connected in G +H. Similarly, ⟨S⟩w is weakly connected in G+H if S be an independent set in G. Corollary 3. Let G and H be graphs. Then αw(G+H) = max{α(G), α(H)} = α(G+H). Example 1. Let G be any graph and m be a positive integer. Then (i) αw(Km +G) = max{m,α(G)}, (ii) αw(Km +G) = α(G), (iii) αw(Km,n) = αw(Km +Kn) = max{m,n}. The next result characterizes the WCIS of G ◦H. Theorem 6. Let G be a connected graph and H be any graph. A subset S of V (G ◦H) is a WCIS in G ◦H if and only if one of the following holds: (i) S is an independent set in Hv for some v ∈ V (G). (ii) S = C ∪ (∪v∈NG(C)Sv), where (a) C is a WCIS in G, and (b) Sv is an independent set (may be empty) in Hv for each v ∈ NG(C). Proof. Suppose S is a WCIS in G ◦ H. Then S is an independent set in G ◦ H and ⟨S⟩w is connected. Consider the following cases: Case 1: V (G) ∩ S = ∅. Then S ⊆ ∪u∈V (G)V (Hu). Since ⟨S⟩w is connected, it follows that S is in exactly one of the components of ⟨∪u∈V (G)V (Hu)⟩, that is, S ⊆ V (Hv) for a unique vertex v ∈ V (G). Consequently, S is an independent subset of V (Hv). This shows that (i) holds. Case 2: V (G) ∩ S ̸= ∅. Let C = V (G) ∩ S and Sv = S ∩ V (Hv) for each v ∈ V (G). Since S is a WCIS in G ◦H, C is a WCIS in G. Let v ∈ V (G) such that Sv ̸= ∅. Then Sv is an independent set in Hv R. Merontos et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6117 7 of 9 and v /∈ C because S is an independent set in G ◦H. Moreover, since ⟨S⟩w is connected, v ∈ NG(C). Thus, S = C ∪ (∪v∈Sv) and (a) and (b) hold. For the converse, suppose first that (i) holds, that is, suppose that S is an independent set of Hv for a unique v ∈ V (G). Then, clearly, S is a WCIS of G◦H. Next, suppose that (ii) holds, i.e., S = C ∪ (∪v∈X⊆NG(C)Sv) and satisfies (a) and (b). Since C ∩NG(C) = ∅ and C and Sv and are independent sets, S is an independent set in G◦H. Moreover, since ⟨C⟩w is connected in G and v ∈ NG(C) for each non-empty set Sv, it follows that ⟨S⟩w is connected in G ◦H. Thus, S is a WCIS in G ◦H. αw(G ◦H) ≥ |S′| =≤ α(H)|V (G)|+ (1− α(H))ιc(G). Corollary 4. Let G be a connected graph and H be any graph. Then αw(G◦H) = α(H) if G = K1. Otherwise, αw(G ◦H) = α(H)(|V (G)|+ (1− α(H))ιc(G)). Proof. Clearly, αw(K1 ◦H) = αw(K1+H) = α(H) (see Example 1(ii)). . So suppose G ̸= K1 and let S be an αw-set in G◦H. Then S = C∪ (∪v∈NG(C)Sv), where C is a WCIS in G and Sv is an independent set of Hv for each v ∈ NG(C) by Theorem 6. Suppose C is not a dominating set of G. Then V (G) \ NG[C] ̸= ∅. Choose w ∈ V (G) \ NG[C] such that wx ∈ E(G) for some x ∈ NG(C). Then C∗ = C ∪ {w} is a WCIS in G and NG(C ∗) = NG(C) ∪ NG(w). Let Lv be an α-set in Hv for each v ∈ NG(C ∗). Then, by Theorem 6, S∗ = C∗ ∪ (∪v∈NG(C∗)Lv) is a WCIS of G ◦H. This implies that αw(G ◦H) = |S| < |S∗| which is not possible. Therefore, C is a WCIDS of G. From this and the fact that 1− α(H) ≤ 0, we have αw(G ◦H) = |S| ≤ |C|+ α(H)(|V (G)| − |C|) = α(H)|V (G)|+ (1− α(H))|C| ≤ α(H)|V (G)|+ (1− α(H))ιc(G). Next, let C0 be a minimum WCIDS of G and let Sv be an α-set in Hv for each v ∈ NG(C0). Then, by Theorem 6, S′ = C0 ∪ (∪v∈NG(C0)Sv) is a WCIS of G ◦H. Hence, αw(G ◦H) ≥ |C| =≤ α(H)|V (G)|+ (1− α(H))ιc(G). This proves the desired equality. 4. Lexicographic product of graphs Sandueta and Canoy characterized the weakly connected sets in the lexicographic product of graphs. R. Merontos et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6117 8 of 9 Theorem 7. [8] Let G and H be connected non-trivial graphs and let C = ∪x∈S({x} × Tx) ⊆ V (G[H]), where S ⊆ V (G) and Tx ⊆ V (H) for all x ∈ S. Then C is weakly connected in G[H] if and only if S is a weakly connected in G The next result characterizes WCIS in G[H]. Theorem 8. Let G and H be connected non-trivial graphs and let C = ∪x∈S({x}×Tx) ⊆ V (G[H]), where S ⊆ V (G) and Tx ⊆ V (H) for all x ∈ S. Then C is WCIS in G[H] if and only if S is a WCIS of G and Tx is an independent set of H for each x ∈ S. Proof. Suppose C = ∪x∈S({x} × Tx) ⊆ V (G[H]), where S ⊆ V (G) and Tx ⊆ V (H) for all x ∈ S and that C is a WCIS in G[H]. By Theorem 7, S is weakly connected in G. Next, let x, y ∈ S such that x ̸= y. Pick a ∈ Tx and b ∈ Ty. Then (x, a), (y, b) ∈ C and (x, a) ̸= (y, b). Since C is independent, (x, a)(y, b) /∈ E(G[H]). This implies that xy /∈ E(G). Thus, S is independent in G. Hence, S a WCIS in G. Now, let x ∈ S and let c, d ∈ Tx such that c ̸= d. Then (x, c), (x, d) ∈ C and (x, c) ̸= (x, d). Since C is an independent set of G[H], we have (x, c)(x, d) /∈ E(G[H]). This implies that cd /∈ E(H). Thus, Tx is an independent set of H. Conversely, assume that S is WCIS in G and Tx is independent in H for all x ∈ S. By Theorem 7, C is a weakly connected set in G[H]. Let (x, a), (y, b) ∈ C such that (x, a) ̸= (y, b). Suppose x ̸= y. Since S is an independent set in G, xy /∈ E(G). Thus, (x, a)(y, b) /∈ E(G[H]). Now, assume x = y. Then, a, b ∈ Tx and a ̸= b. Since Tx is an independent set of H, ab /∈ E(H). Hence, (x, a)(y, b) /∈ E(G[H]). Thus, C is an independent set of G[H]. Therefore, C is a weakly connected independent set in G[H]. Corollary 5. Let G and H be connected non-trivial graphs. Then αw(G[H]) = αw(G)α(H). Proof. Let C = ∪x∈S({x} × Tx) ⊆ V (G[H]), where S ⊆ V (G) and Tx ⊆ V (H) for all x ∈ S, be an αw-set in G[H]. By Theorem 8, S is a WCIS in G and Tx is an independent set in H for every x ∈ S. Hence, αw(G[H]) = |C| = |Σx∈S({x} × Tx)| ≤ αw(G)α(H). Next, let S be an αw-set in G and A be an α-set in H. For each x ∈ S, let Tx = A. By Theorem 8, C = ∪x∈S({x} × Tx) is a WCIS in G[H]. Consequently, αw(G[H]) ≥ |C| = |Σx∈S({x} × Tx)| = αw(G)α(H). Therefore, αw(G[H]) = αw(G)α(H). 5. Conclusion The concept of weakly connected independent set as well as the parameter weakly connected independence number were introduced and initially investigated in this study. R. Merontos et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 6117 9 of 9 Graphs for which the weakly connected independence number and independence num- ber are equal were characterized. It was shown that the difference between these two parameters can be made arbitrarily large. 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