Component Intersection Graph of Finite Dimensional Vector Spaces

Proof. Suppose that $dim(\mathbb{V})=2$ and $\mathfrak{B}=\{u,v\}$ is a basis of $\mathbb{V}.$ Then $V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))$ can be partitioned into two sets ${\mathfrak{B}}_u=\{\alpha u|0\ne \alpha \in \mathbb{F}\}$ and ${\mathfrak{B}}_v=\{\beta v|0\ne \beta \in \mathbb{F}\}$ such that ${\mathfrak{B}}_u\cap {\mathfrak{B}}_v=\mathrm{\varnothing }$ and ${\mathfrak{B}}_u\cup {\mathfrak{B}}_v=V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})).$ If $u_1\in {\mathfrak{B}}_u$ and $u_2\in {\mathfrak{B}}_v,$ then $u_1\sim u_2$ in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ and any two vertices of ${\mathfrak{B}}_u$ are not adjacent in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}).$ Similarly, any two vertices of ${\mathfrak{B}}_v$ are not adjacent in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}).$ Hence ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ is complete bipartite.

Conversely, assume that ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ is complete bipartite and let $dim(\mathbb{V})=k.$ If $k>2,$ then there exists a basis $\mathfrak{B}=\{v_1,v_2,\cdots ,v_k\}$ of $\mathbb{V}$ such that $\mathfrak{S}(v_i)\cap \mathfrak{S}(v_j)=\mathrm{\varnothing }$ for $i\ne j$ and $1\le i,j\le k.$ Thus $v_1,v_2,v_3\in V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))$ such that $\mathfrak{S}(v_i)\cap \mathfrak{S}(v_j)=\mathrm{\varnothing }$ for $i\ne j$ and $1\le i,j\le 3$ and $v_1\sim v_2\sim v_3\sim v_1$ is a cycle of length three. Hence ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ is not complete bipartite. If $k=2,$ then by direct part of this theorem, ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ is complete bipartite. This complete the proof. 

Proof. Let $\mathfrak{B}$ be a basis of $\mathbb{V}$ and $u,v$ be two distinct vertices of ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}).$ We have the following cases;

1. (i) If ${\mathfrak{S}}_{\mathfrak{B}}(u)\cap {\mathfrak{S}}_{\mathfrak{B}}(v)=\mathrm{\varnothing },$ then $u\sim v$ is a path in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ and $d(u,v)=1.$

2. (ii) If ${\mathfrak{S}}_{\mathfrak{B}}(u)\cap {\mathfrak{S}}_{\mathfrak{B}}(v)\ne \mathrm{\varnothing },$ then we arrive at the following two subcases :

3. (a) If ${\mathfrak{S}}_{\mathfrak{B}}(u)\subseteq {\mathfrak{S}}_{\mathfrak{B}}(v)$ or ${\mathfrak{S}}_{\mathfrak{B}}(v)\subseteq {\mathfrak{S}}_{\mathfrak{B}}(u),$ then there exists $w\in \mathfrak{B}\setminus {\mathfrak{S}}_{\mathfrak{B}}(u)\cap {\mathfrak{S}}_{\mathfrak{B}}(v)$ such that ${\mathfrak{S}}_{\mathfrak{B}}(u)\cap {\mathfrak{S}}_{\mathfrak{B}}(w)=$ ${\mathfrak{S}}_{\mathfrak{B}}(v)\cap {\mathfrak{S}}_{\mathfrak{B}}(w)=\mathrm{\varnothing }.$ Thus $u\sim w\sim v$ is a path in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ and $d(u,w)=2.$

4. (b) If ${\mathfrak{S}}_{\mathfrak{B}}(u)⊈{\mathfrak{S}}_{\mathfrak{B}}(v)$ or ${\mathfrak{S}}_{\mathfrak{B}}(v)⊈{\mathfrak{S}}_{\mathfrak{B}}(u),$ then there exists $u_i\in {\mathfrak{S}}_{\mathfrak{B}}(u)\setminus {\mathfrak{S}}_{\mathfrak{B}}(v)$ $u_j\in {\mathfrak{S}}_{\mathfrak{B}}(v)\setminus {\mathfrak{S}}_{\mathfrak{B}}(u)$ such that $u_i\ne u_j$ and ${\mathfrak{S}}_{\mathfrak{B}}(u_i)\cap {\mathfrak{S}}_{\mathfrak{B}}(v)=$ ${\mathfrak{S}}_{\mathfrak{B}}(u_j)\cap {\mathfrak{S}}_{\mathfrak{B}}(u)=$ ${\mathfrak{S}}_{\mathfrak{B}}(u_i)\cap {\mathfrak{S}}_{\mathfrak{B}}(u_j)=\mathrm{\varnothing }.$ Thus $u\sim u_i\sim u_j\sim v$ is a path in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ and $d(u,v)=3.$ Therefore in all the cases, we find that ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ is connected and $diam({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\le 3.$ Hence proved.

Proof. Suppose that $\mathfrak{B}=\{v_1,v_2,\cdots ,v_n\}$ is a basis of $\mathbb{V}.$ We have the following cases;

1. (i) If $n=2$ and $\mathbb{F}={\mathbb{F}}_2,$ then by Theorem 2, ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})=K_2$ and $diam({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))=1.$

2. (ii) If $n=2$ and $\mathbb{F}\ne {\mathbb{F}}_2,$ then by Theorem 1, ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ is complete bipartite and $diam({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))=2.$

3. (iii) If $n\ge 3,$ then $w_1=v_1+v_2+v_3+\cdots +v_{n-1}$ and $w_2=v_2+v_3+\cdots +v_n\in$ $V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))$ such that $w_1\nsim w_2$. Note that ${\mathfrak{S}}_{\mathfrak{B}}(w_1)\cap {\mathfrak{S}}_{\mathfrak{B}}(w_2)\ne \mathrm{\varnothing }$ and $d(w_1,w_2)\ne 1$. If $d(w_1,w_2)=2,$ then there exists $w^{\ast }\in V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})))$ such that ${\mathfrak{S}}_{\mathfrak{B}}(w_1)\cap {\mathfrak{S}}_{\mathfrak{B}}(w^{\ast })$ $={\mathfrak{S}}_{\mathfrak{B}}(w^{\ast })\cap {\mathfrak{S}}_{\mathfrak{B}}(w_2)=\mathrm{\varnothing }$ and $w_1\sim w^{\ast }\sim w_2$ is a path in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}).$ This gives ${\mathfrak{S}}_{\mathfrak{B}}(w^{\ast })=\mathrm{\varnothing },$ a contradiction. Thus $d(w_1,w_2)\ne 2$ and $v_1$ and $v_n$ are two distinct vertices of ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ such that ${\mathfrak{S}}_{\mathfrak{B}}(v_1)\cap {\mathfrak{S}}_{\mathfrak{B}}(w_2)=$ ${\mathfrak{S}}_{\mathfrak{B}}(v_n)\cap {\mathfrak{S}}_{\mathfrak{B}}(w_1)=$ ${\mathfrak{S}}_{\mathfrak{B}}(v_n)\cap {\mathfrak{S}}_{\mathfrak{B}}(v_1)=\mathrm{\varnothing }.$ Therefore $w_1\sim u_n\sim u_1\sim w_2$ is a path in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ and $d(w_1,w_2)=3.$ By Lemma 1, we know that $diam({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\le 3$ and hence $diam({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))=3.$

Proof. Suppose that $\mathfrak{B}=\{v_1,v_2,\cdots ,v_n\}$ is a basis of a $n$-dimensional vector space $\mathbb{V}$ over a field $\mathbb{F}.$ Now we have the following three cases;

1. (i) If $n=2$ and $\mathbb{F}={\mathbb{F}}_2,$ then by Theorem 2, ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})=K_2$ and $gr({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))=\mathrm{\infty }.$

2. (ii) If $n=2$ and $\mathbb{F}\ne {\mathbb{F}}_2,$ then by Theorem 1, ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ is complete bipartite and ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ contain a cycle of length four and $gr({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))=4.$

3. (iii) If $n\ge 3,$ then $v_1,v_2,v_3\in V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))$ such that $v_1\sim v_2\sim v_3\sim v_1$ forms a triangle in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}).$ Hence $gr({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))=4.$

Proof. Let $\mathfrak{B}=\{v_1,v_2,\cdots ,v_n\}$ be a basis of $\mathbb{V}$ and $w\in V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})).$ We have the following two cases;

1. (i) If $|{\mathfrak{S}}_{\mathfrak{B}}(w)|=1,$ then there exist $u,v\in V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\setminus {\mathfrak{S}}_{\mathfrak{B}}(w)$ such that ${\mathfrak{S}}_{\mathfrak{B}}(v)\cap {\mathfrak{S}}_{\mathfrak{B}}(u)=$ ${\mathfrak{S}}_{\mathfrak{B}}(u)\cap {\mathfrak{S}}_{\mathfrak{B}}(w)=$ ${\mathfrak{S}}_{\mathfrak{B}}(v)\cap {\mathfrak{S}}_{\mathfrak{B}}(w)=\mathrm{\varnothing }.$

2. (ii) If $|{\mathfrak{S}}_{\mathfrak{B}}(w)|=2,$ then there exist $u_1,u_2\in V({\mathfrak{I}}_{\mathfrak{B}}(\mathbb{V}))\setminus {\mathfrak{S}}_{\mathfrak{B}}(w)$ such that ${\mathfrak{S}}_{\mathfrak{B}}(v_1)\cap {\mathfrak{S}}_{\mathfrak{B}}(u_1)=$ ${\mathfrak{S}}_{\mathfrak{B}}(v_1)\cap {\mathfrak{S}}_{\mathfrak{B}}(w)=$ ${\mathfrak{S}}_{\mathfrak{B}}(u_1)\cap {\mathfrak{S}}_{\mathfrak{B}}(w)=\mathrm{\varnothing }.$ Thus $u_1\sim w\sim v_1\sim u_1$ forms a triangle in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}).$ Thus in all the cases, $w$ is a vertex of some triangle and hence ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ is triangulated.

Proof. Suppose that $\mathfrak{B}=\{v_1,v_2,\cdots ,v_n\}$ is a basis of $\mathbb{V}.$ Note that $\mathfrak{B}\subseteq V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))$ is a clique in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}).$ If possible, suppose $\mathfrak{B}\cup \{w\}$ is a clique of ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}),$ where $w\in V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})).$ Then ${\mathfrak{S}}_{\mathfrak{B}}(v_i)\cap {\mathfrak{S}}_{\mathfrak{B}}(w)=\mathrm{\varnothing }$ for all $i=1,2,\cdots ,n$ and ${\mathfrak{S}}_{\mathfrak{B}}(w)=\mathrm{\varnothing },$ a contradiction. Finally, we prove that there does not exist any clique of size $n+1.$ If $\{w_1,w_2,\cdots ,w_{n+1}\}$ is a clique of size $n+1,$ then ${\mathfrak{S}}_{\mathfrak{B}}(w_i)\cap {\mathfrak{S}}_{\mathfrak{B}}(w_j)=\mathrm{\varnothing }$ for all $i\ne j$ and $1\le i,j\le n+1,$ $|\bigcup _{i=1}^{n+1}{\mathfrak{S}}_{\mathfrak{B}}(w_i)|>|\mathfrak{B}|,$ a contradiction. Therefore the clique number $\omega ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))$ of ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ is $n.$

Conversely, assume that $\omega ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))=n$ and dimension of $\mathbb{V}$ is $m.$ Then by direct part of this theorem, $\omega ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))=m.$ Therefore $m=n.$ 

Proof. It is obvious that if ${\mathbb{V}}_1$ and ${\mathbb{V}}_2$ are isomorphic (as vector spaces), then ${\mathfrak{I}}^{{\mathfrak{B}}_1}({\mathbb{V}}_1)$ and ${\mathfrak{I}}^{{\mathfrak{B}}_2}({\mathbb{V}}_2)$ are isomorphic (as graphs).

Conversely, assume that ${\mathfrak{I}}^{{\mathfrak{B}}_1}({\mathbb{V}}_1)$ and ${\mathfrak{I}}^{{\mathfrak{B}}_2}({\mathbb{V}}_2)$ are isomorphic (as graphs) and $dim({\mathbb{V}}_1),$ $dim({\mathbb{V}}_2)$ are $m,$ $n$ respectively. Then by Theorem 6, the clique number $\omega ({\mathfrak{I}}^{{\mathfrak{B}}_1}({\mathbb{V}}_1))$ and $\omega ({\mathfrak{I}}^{{\mathfrak{B}}_2}({\mathbb{V}}_2))$ are $n-1$ and $m-1$ respectively. However, as the two graphs are isomorphic, $m=n.$ Thus ${\mathbb{V}}_1$ and ${\mathbb{V}}_2$ are finite dimensional over the field $\mathbb{F}$ with $m=n$ and hence both are isomorphic (as vector spaces). 

Proof. Suppose that $\mathfrak{B}=\{u_1,u_2,\dots ,u_n\}$ is a basis of $\mathbb{V}.$

1. (i) Note that by Theorem 6, ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})[\mathfrak{B}]$ is a clique, i.e., a complete graph and every vertex $v$ of ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ such that ${\mathfrak{S}}_{\mathfrak{B}}(v)$ contains at least one ( possible more than one ) member of $\mathfrak{B}.$ Therefore there exists a map $\mathfrak{f}:V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\to \mathfrak{B}$ given by $\mathfrak{u}\mapsto {\mathfrak{u}}_1,$ where $\mathfrak{u}\in V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))$ and ${\mathfrak{u}}_1\in \mathfrak{B}$ such that ${\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{u}}_1)\subseteq {\mathfrak{S}}_{\mathfrak{B}}(\mathfrak{u}).$ By the axioms of choice the existence of such type of map is guaranteed. If $\mathfrak{w}\sim \mathfrak{u}$ in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}),$ i.e., ${\mathfrak{S}}_{\mathfrak{B}}(\mathfrak{v})\cap {\mathfrak{S}}_{\mathfrak{B}}(\mathfrak{w})=\mathrm{\varnothing },$ then ${\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{v}}_1)\cap {\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{w}}_1)=\mathrm{\varnothing }.$ Thus ${\mathfrak{v}}_1\ne {\mathfrak{w}}_1$ and since ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})[{\mathfrak{S}}^1]$ is a complete graph, ${\mathfrak{u}}_1\sim {\mathfrak{w}}_1.$ Therefore $\mathfrak{f}$ is an adjacency preserving map and a graph homomorphism.

2. (ii) Note that if $\mathfrak{u},\mathfrak{w}\in {\mathfrak{S}}^{k+1},$ then $|{\mathfrak{S}}_{\mathfrak{B}}(\mathfrak{u})\cap {\mathfrak{S}}_{\mathfrak{B}}(\mathfrak{w})|\ge 1.$ Therefore ${\mathfrak{S}}^{k+1}$ is an independent set. For maximality, if ${\mathfrak{u}}_1\in V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\setminus {\mathfrak{S}}^{k+1}$ with $|{\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{u}}_1)\le k,$ then there exists ${\mathfrak{w}}_1\in {\mathfrak{S}}^{k+1}$ such that ${\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{u}}_1)\cap {\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{w}}_1)=\mathrm{\varnothing }.$ Hence ${\mathfrak{S}}^{k+1}$ is a maximal independent set in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}).$

3. (iii) Note that if $\mathfrak{u},\mathfrak{w}\in {\mathfrak{S}}^{k+1}\cup$ ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})[{\mathfrak{S}}_{{\mathfrak{v}}_{\mathfrak{i}}}^k],$ then $|{\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{u}}_1)\cap$ ${\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{w}}_1)|\ge 1.$ Therefore, ${\mathfrak{S}}^{k+1}\cup [{\mathfrak{S}}_{v_i}^k]$ is an independent set. For maximality, let ${\mathfrak{u}}_1\in V({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\setminus ({\mathfrak{M}}_{k+1}\cup$ $[{\mathfrak{M}}_{v_i}^k]).$ If $|{\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{u}}_1)|\le k,$ then there exists ${\mathfrak{w}}_1\in [{\mathfrak{S}}_{v_i}^k]$ such that ${\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{u}}_1)\cap {\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{w}}_1)=\mathrm{\varnothing }.$ Hence ${\mathfrak{S}}^{k+1}\cup [{\mathfrak{S}}_{v_i}^k]$ is a maximal independent set in ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}).$

Proof. Let $\mathfrak{B}=\{v_1,v_2,\cdots ,v_n\}$ be a basis of $\mathbb{V}.$ Since the clique number $\omega ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))$ of ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})$ is $n$, $\lambda ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\ge n$. By Theorem 7 $(i)$, $\lambda ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\le |\mathfrak{B}|.$ Moreover, since $\mathfrak{B}$ is a maximal clique, $\omega ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\ge |\mathfrak{B}|.$ Since $\omega ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\le \lambda ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})),$ we have the following inequality. $$\omega ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\le \lambda ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}))\le |\mathfrak{B}|\le \omega ({\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V})),$$ and hence the theorem follows. 

Proof. Suppose that $\mathfrak{B}=\{{\mathfrak{v}}_1,{\mathfrak{v}}_2,\cdots ,{\mathfrak{v}}_n\}$ is a basis of a $n$-dimensional vector space $\mathbb{V}.$ Clearly, ${\mathfrak{u}}_i={\mathfrak{v}}_1+{\mathfrak{v}}_2+\cdots +{\mathfrak{v}}_{i-1}$ $+{\mathfrak{v}}_{i+1}+\cdots +\cdots ,{\mathfrak{v}}_n,$ for $1\le i\le n$ are distinct $n$ member of ${\mathfrak{S}}_{n-1}.$ Hence proved. 

Proof. Suppose that $\mathfrak{D}$ is a dominating set of ${\mathfrak{I}}^{\mathfrak{B}}(\mathbb{V}).$ We know that by Lemma 2, $|{\mathfrak{S}}_{n-1}|\ge n.$ If ${\mathfrak{S}}_{n-1}\subseteq \mathfrak{D},$ then $|\mathfrak{D}|\ge n$ and we are done. Otherwise, there exist ${\mathfrak{u}}_i^1\in {\mathfrak{S}}_{n-1}\setminus \mathfrak{D}$ and ${\mathfrak{w}}_1\in \mathfrak{D}\cap {\mathfrak{S}}_1$ such that ${\mathfrak{u}}_i^1\sim {\mathfrak{w}}_1.$ Clearly, $|{\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{w}}_1)|=1.$ Without loss of generality, we assume that ${\mathfrak{u}}_1^1={\mathfrak{v}}_2+{\mathfrak{v}}_3+\cdots +{\mathfrak{v}}_n.$ Therefore ${\mathfrak{u}}_j^2={\mathfrak{v}}_2+{\mathfrak{v}}_3+\cdots +{\mathfrak{v}}_{j-1}+{\mathfrak{w}}_1+{\mathfrak{v}}_{j+1}$ $+{\mathfrak{v}}_n,$ $2\le j\le n$ are distinct $(n-2)$ member of ${\mathfrak{S}}_{n-1}$ and ${\mathfrak{u}}_j^2\nsim {\mathfrak{w}}_1.$ If ${\mathfrak{u}}_j^2\in \mathfrak{D}$ for all $2\le j\le n,$ then $|\mathfrak{D}|\ge n$ and we are done. Otherwise, there exists ${\mathfrak{u}}_j^2\notin \mathfrak{D}.$ Without loss of generality we assume that ${\mathfrak{u}}_2^2\notin \mathfrak{D}.$ Since ${\mathfrak{u}}_2^2\notin \mathfrak{D}$ and ${\mathfrak{u}}_2^2\nsim {\mathfrak{w}}_1,$ there exists ${\mathfrak{w}}_2\in {\mathfrak{S}}_1\cap \mathfrak{D}$ such that ${\mathfrak{u}}_2^2\sim {\mathfrak{w}}_2.$ Clearly, $|{\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{w}}_2)|=1$ and ${\mathfrak{u}}_2^2={\mathfrak{w}}_1+{\mathfrak{v}}_3+\cdots +{\mathfrak{v}}_n.$ Therefore ${\mathfrak{u}}_k^3={\mathfrak{w}}_1+{\mathfrak{v}}_3+\cdots +{\mathfrak{v}}_{k-1}+{\mathfrak{w}}_2+{\mathfrak{v}}_{k+1}$ $+{\mathfrak{v}}_n,$ $3\le k\le n$ are distinct $(n-3)$ member of ${\mathfrak{S}}_{n-1}$ and ${\mathfrak{u}}_k^3\nsim {\mathfrak{w}}_1,{\mathfrak{u}}_k^3\nsim {\mathfrak{w}}_2.$ If ${\mathfrak{u}}_k^3\in \mathfrak{D}$ for all $3\le k\le n,$ then $|\mathfrak{D}|\ge n$ and we are done. Otherwise, there exists ${\mathfrak{u}}_k^3\notin \mathfrak{D}.$ Without loss of generality we assume that ${\mathfrak{u}}_3^3\notin \mathfrak{D}.$ Since ${\mathfrak{u}}_3^3\notin \mathfrak{D}$ and ${\mathfrak{u}}_3^3\nsim {\mathfrak{w}}_1,{\mathfrak{u}}_3^3\nsim {\mathfrak{w}}_2,$ there exists ${\mathfrak{w}}_3\in {\mathfrak{S}}_1\cap \mathfrak{D}$ such that ${\mathfrak{w}}_3\sim {\mathfrak{u}}_3^3.$ Clearly, $|{\mathfrak{S}}_{\mathfrak{B}}{\mathfrak{w}}_3)|=1$ and ${\mathfrak{u}}_3^3={\mathfrak{w}}_1+{\mathfrak{w}}_2+{\mathfrak{v}}_4+\cdots +{\mathfrak{v}}_n.$ Therefore ${\mathfrak{u}}_l^4={\mathfrak{w}}_1+{\mathfrak{w}}_2+{\mathfrak{v}}_4+\cdots +{\mathfrak{v}}_{l-1}$ $+{\mathfrak{w}}_3+{\mathfrak{v}}_{l+1}+{\mathfrak{v}}_n,$ $4\le l\le n$ are distinct $(n-4)$ member of ${\mathfrak{S}}_{n-1}$ and ${\mathfrak{u}}_l^4\nsim {\mathfrak{w}}_1,$ ${\mathfrak{u}}_l^4\nsim {\mathfrak{w}}_2.$ ${\mathfrak{u}}_l^4\nsim {\mathfrak{w}}_3.$ If ${\mathfrak{u}}_k^4\in \mathfrak{D}$ for all $4\le k\le n,$ then $|\mathfrak{D}|\ge n$ and we are done. Otherwise, there exists ${\mathfrak{u}}_k^4\notin \mathfrak{D}.$ Without loss of generality we assume that ${\mathfrak{u}}_4^4\notin \mathfrak{D}.$ Since ${\mathfrak{u}}_4^4\notin \mathfrak{D}$ and ${\mathfrak{u}}_4^4\nsim {\mathfrak{w}}_1,{\mathfrak{u}}_4^4\nsim {\mathfrak{w}}_2,{\mathfrak{u}}_4^4\nsim {\mathfrak{w}}_3,$ there exists ${\mathfrak{w}}_4\in {\mathfrak{S}}_1\cap \mathfrak{D}$ such that ${\mathfrak{w}}_4\sim {\mathfrak{u}}_4^4.$ Clearly, $|{\mathfrak{S}}_{\mathfrak{B}}({\mathfrak{w}}_4)|=1$ and ${\mathfrak{u}}_4^4={\mathfrak{w}}_1+{\mathfrak{w}}_2+{\mathfrak{w}}_3+{\mathfrak{v}}_5\cdots +{\mathfrak{v}}_n.$ Continuing in the similar way, there exist at least $n$ distinct members ${\mathfrak{w}}_1,{\mathfrak{w}}_2,\cdots ,{\mathfrak{w}}_n$ in $\mathfrak{D}$ and the lemma follows.