S}}\leq\theta\eta_{k}\big{\\}}$ $\displaystyle~{}~{}~{}~{}+\max\limits_{1\leq
j\leq
p}\sum_{i=1}^{p}\big{\|}\widehat{\Sigma}_{yy,ij}^{(k)}-\Sigma_{yy,ij}^{(k)}\big{\|}_{{\cal
S}}I\big{\\{}\|\widehat{\Sigma}_{yy,ij}^{(k)}\|_{{\cal
S}}\geq\eta_{k},\theta\eta_{k}<\|\Sigma_{yy,ij}^{(k)}\|_{{\cal
S}}<\eta_{k}\big{\\}}$ $\displaystyle\leq Z\cdot\max\limits_{1\leq j\leq
p}\sum_{i=1}^{p}I\big{\\{}\|\hat{\Sigma}_{yy,ij}^{(k)}-\Sigma_{yy,ij}^{(k)}\|_{{\cal
S}}\geq(1-\theta)\eta_{k}\big{\\}}+Z\cdot\max\limits_{1\leq j\leq
p}\sum_{i=1}^{p}I\big{\\{}\|\Sigma_{yy,ij}^{(k)}\|_{{\cal
S}}>\theta\eta_{k}\big{\\}}$ $\displaystyle\lesssim Z\cdot
N\big{\\{}(1-\theta)\eta_{k}\big{\\}}+Z(\theta\eta_{k})^{-\tau}c_{1}(p).$

The above bounds imply that

$\displaystyle I_{2}\lesssim Z\cdot
N\big{\\{}(1-\theta)\eta_{k}\big{\\}}+\eta_{k}^{1-\tau}c_{1}(p)+Z(\theta\eta_{k})^{-\tau}c_{1}(p).$

We finally bound $I_{3}.$ By the triangle inequality and Lemma 1,

$\displaystyle I_{3}$ $\displaystyle\leq\max\limits_{1\leq j\leq
p}\sum_{i=1}^{p}\big{\|}\widehat{\Sigma}_{yy,ij}^{(k)}-\Sigma_{yy,ij}^{(k)}\big{\|}_{{\cal
S}}I\big{\\{}\|\widehat{\Sigma}_{yy,ij}^{(k)}\|_{{\cal
S}}<\eta_{k},\|\Sigma_{yy,ij}^{(k)}\|_{{\cal S}}\geq\eta_{k}\big{\\}}$
$\displaystyle~{}~{}~{}~{}+\max\limits_{1\leq j\leq
p}\sum_{i=1}^{p}\big{\|}\widehat{\Sigma}_{yy,ij}^{(k)}\big{\|}_{{\cal
S}}I\big{\\{}\|\widehat{\Sigma}_{yy,ij}^{(k)}\big{\|}_{{\cal
S}}<\eta_{k},\|\Sigma_{yy,ij}^{(k)}\|_{{\cal S}}\geq\eta_{k}\\}$
$\displaystyle\leq Z\cdot\max\limits_{1\leq j\leq
p}\sum_{i=1}^{p}I\big{\\{}\|\widehat{\Sigma}_{yy,ij}^{(k)}\|_{{\cal
S}}\geq\eta_{k}\big{\\}}+\eta_{k}\max\limits_{1\leq j\leq
p}\sum_{i=1}^{p}I\big{\\{}\|\Sigma_{yy,ij}^{(k)}\|_{{\cal
S}}\geq\eta_{k}\big{\\}}$ $\displaystyle\lesssim
Z\eta_{k}^{-\tau}c_{1}(p)+\eta_{k}^{1-\tau}c_{1}(p).$

Combining the bounds for $I_{1},I_{2}$ and $I_{3}$ yields that

$\big{\|}\mathcal{T}_{\eta_{n}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}-\mathcal{T}_{\eta_{k}}\\{\boldsymbol{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}\big{\|}_{{\cal
S},1}\lesssim
Z\eta_{k}^{-\tau}c_{1}(p)+\eta_{k}^{1-\tau}c_{1}(p)+Z(\theta\eta_{k})^{-\tau}c_{1}(p)+Z\cdot
N\big{\\{}(1-\theta)\eta_{k}\big{\\}}.$ (S.12)

By Lemma B3 and the choice of $\eta_{k}=c_{3}{\cal M}_{y}(\log
p/n)^{1/2}=o(1)$ for sufficiently large $c_{3},$

$\displaystyle\mathbb{P}\Big{(}N\big{\\{}(1-\theta)\eta_{k}\big{\\}}>0\Big{)}$
$\displaystyle=\mathbb{P}\Big{(}\max\limits_{1\leq i,j\leq
p}\|\widehat{\Sigma}_{yy,ij}^{(k)}-\Sigma_{yy,ij}^{(k)}\|_{{\cal
S}}\geq(1-\theta)\eta_{k}\Big{)}$ $\displaystyle\leq
c_{1}p^{2}\exp\\{-c_{2}(1-\theta)^{2}c_{3}^{2}\log p\\}\to 0.$

Hence $N\big{\\{}(1-\theta)\eta_{k}\big{\\}}=o_{P}(1).$ Applying Lemma B3
again, we obtain that $Z=O_{P}(\eta_{k}),$ which together with (S.11) and
(S.12) imply that

$\big{\|}\mathcal{T}_{\eta_{n}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}-\boldsymbol{\Sigma}^{(k)}_{yy}\big{\|}_{{\cal
S},1}=O_{P}\left\\{c_{1}(p)\mathcal{M}_{y}^{1-\tau}\Big{(}\frac{\log
p}{n}\Big{)}^{\frac{1-\tau}{2}}\right\\}.$

By the similar procedure, the second result of (S.10) can be derived and the
proof is omitted.

Step 2. We will show that

$\big{\|}\widetilde{\bf M}-{\mathchoice{{\ooalign{\hbox{\raise
7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\displaystyle\widehat{\vrule
width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\displaystyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\textstyle\widehat{\vrule width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\textstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 6.40926pt\hbox{\scalebox{1.0}[-1.0]{\lower
6.40926pt\hbox{$\scriptstyle\widehat{\vrule width=0.0pt,height=4.78333pt\vrule
height=0.0pt,width=6.41667pt}$}}}}\cr\hbox{$\scriptstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 5.9537pt\hbox{\scalebox{1.0}[-1.0]{\lower
5.9537pt\hbox{$\scriptscriptstyle\widehat{\vrule
width=0.0pt,height=3.41666pt\vrule
height=0.0pt,width=4.58334pt}$}}}}\cr\hbox{$\scriptscriptstyle{\bf
M}$}}}}}\big{\|}=O_{P}\left[c_{1}(p)\big{\\{}c_{1}(p)+c_{2}(p)\big{\\}}\mathcal{M}_{y}^{1-\tau}\Big{(}\frac{\log
p}{n}\Big{)}^{\frac{1-\tau}{2}}\right].$ (S.13)

Note that

$\displaystyle\big{\|}\widetilde{\bf M}-{\mathchoice{{\ooalign{\hbox{\raise
7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\displaystyle\widehat{\vrule
width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\displaystyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\textstyle\widehat{\vrule width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\textstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 6.40926pt\hbox{\scalebox{1.0}[-1.0]{\lower
6.40926pt\hbox{$\scriptstyle\widehat{\vrule width=0.0pt,height=4.78333pt\vrule
height=0.0pt,width=6.41667pt}$}}}}\cr\hbox{$\scriptstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 5.9537pt\hbox{\scalebox{1.0}[-1.0]{\lower
5.9537pt\hbox{$\scriptscriptstyle\widehat{\vrule
width=0.0pt,height=3.41666pt\vrule
height=0.0pt,width=4.58334pt}$}}}}\cr\hbox{$\scriptscriptstyle{\bf
M}$}}}}}\big{\|}$ $\displaystyle=$ $\displaystyle
O_{P}\Bigg{(}\Big{\|}\iint\big{[}\mathcal{T}_{\eta_{k}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}(u,v)-\boldsymbol{\Sigma}^{(k)}_{yy}(u,v)\Big{]}\widehat{\bf
W}(u,v)\big{[}\mathcal{T}_{\eta_{k}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}(u,v)-\boldsymbol{\Sigma}^{(k)}_{yy}(u,v)\Big{]}^{{\mathrm{\scriptscriptstyle
T}}}\,{\rm d}u{\rm d}v\Big{\|}$
$\displaystyle~{}~{}~{}~{}+\Big{\|}\iint\big{[}\mathcal{T}_{\eta_{k}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}(u,v)-\boldsymbol{\Sigma}^{(k)}_{yy}(u,v)\big{]}\widehat{\bf
W}(u,v)\boldsymbol{\Sigma}^{(k)}_{yy}(u,v)^{{\mathrm{\scriptscriptstyle
T}}}\,{\rm d}u{\rm d}v\Big{\|}\Bigg{)}.$

By Lemma B4 and (S.5), (S.10), we obtain that

$\displaystyle\Big{\|}\iint\big{[}\mathcal{T}_{\eta_{k}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}(u,v)-\boldsymbol{\Sigma}^{(k)}_{yy}(u,v)\big{]}\widehat{\bf
W}(u,v)\big{[}\mathcal{T}_{\eta_{k}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}(u,v)-\boldsymbol{\Sigma}^{(k)}_{yy}(u,v)\big{]}^{{\mathrm{\scriptscriptstyle
T}}}\,{\rm d}u{\rm d}v\Big{\|}$ $\displaystyle\lesssim$
$\displaystyle\big{\|}\mathcal{T}_{\eta_{k}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}-\boldsymbol{\Sigma}^{(k)}_{yy}\big{\|}_{{\cal
S},1}\big{\|}\mathcal{T}_{\eta_{k}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}-\boldsymbol{\Sigma}^{(k)}_{yy}\big{\|}_{{\cal
S},\infty}\big{\\{}1+o_{P}(1)\big{\\}}$ $\displaystyle=$ $\displaystyle
O_{P}\left[c_{1}(p)\big{\\{}c_{1}(p)+c_{2}(p)\big{\\}}\mathcal{M}_{y}^{2(1-\tau)}\Big{(}\frac{\log
p}{n}\Big{)}^{1-\tau}\right]$

and

$\displaystyle\Big{\|}\iint\big{[}\mathcal{T}_{\eta_{k}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}(u,v)-\boldsymbol{\Sigma}^{(k)}_{yy}(u,v)\big{]}\widehat{\bf
W}(u,v)\boldsymbol{\Sigma}^{(k)}_{yy}(u,v)^{{\mathrm{\scriptscriptstyle
T}}}\,{\rm d}u{\rm d}v\Big{\|}$ $\displaystyle\lesssim$
$\displaystyle\Big{[}\|\mathcal{T}_{\eta_{n}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}-\boldsymbol{\Sigma}^{(k)}_{yy}\|_{{\cal
S},1}\|\mathcal{T}_{\eta_{n}}\\{\widehat{\boldsymbol{\Sigma}}^{(k)}_{yy}\\}-\boldsymbol{\Sigma}^{(k)}_{yy}\|_{{\cal
S},\infty}\Big{]}^{1/2}\Big{[}\|\boldsymbol{\Sigma}^{(k)}_{yy}\|_{{\cal
S},1}\|\boldsymbol{\Sigma}^{(k)}_{yy}\|_{{\cal
S},\infty}\Big{]}^{1/2}\big{\\{}1+o_{P}(1)\big{\\}}.$

For the second term above, it follows that

$\displaystyle\|\boldsymbol{\Sigma}^{(k)}_{yy}\|_{{\cal
S},1}=\max\limits_{1\leq j\leq p}\sum_{i=1}^{p}\|\Sigma_{yy,ij}^{(k)}\|_{{\cal
S}}\leq\max\limits_{1\leq j\leq
p}\sum_{i=1}^{p}\|\Sigma_{yy,ij}^{(k)}\|^{\tau}_{{\cal
S}}\|\Sigma_{yy,ij}^{(k)}\|^{1-\tau}_{{\cal S}}\lesssim c_{1}(p),$

where $\tau\in(0,1)$ and $\|\Sigma_{yy,ij}^{(k)}\|_{{\cal S}}$ is uniformly
bounded under Condition 7. Following the similar procedure, we can show that
$\|\boldsymbol{\Sigma}^{(k)}_{yy}\|_{{\cal S},\infty}\lesssim
c_{1}(p)+c_{2}(p)$. For the truly sparse case with $\tau=0$, it follows from
the derivation in Section B.3 under Conditions 1, 2, 6 with fixed $r$ and $L$
that

$\displaystyle\max\limits_{1\leq j\leq
p}\sum_{i=1}^{p}\|\Sigma_{yy,ij}^{(k)}\|_{{\cal S}}$
$\displaystyle\leq\max\limits_{1\leq j\leq p}\sum_{1\leq m,m^{\prime}\leq
r}\sum_{i=1}^{p}|A_{im}A_{jm^{\prime}}|\|\Sigma_{xx,mm^{\prime}}^{(k)}\|_{{\cal
S}}+\max\limits_{1\leq j\leq
p}\sum_{m=1}^{r}\sum_{i=1}^{p}|A_{im}|\|\Sigma^{(k)}_{x\epsilon,mj}\|_{{\cal
S}}$ $\displaystyle\lesssim r^{2}Lc_{1}(p)+rc_{1}(p)\asymp c_{1}(p).$

Similarly, under Condition 4′, we can show that

$\displaystyle\max\limits_{1\leq i\leq
p}\sum_{j=1}^{p}\|\Sigma_{yy,ij}^{(k)}\|_{{\cal S}}\lesssim
r^{2}Lc_{1}(p)+rLc_{2}(p)\asymp c_{1}(p)+c_{2}(p).$

The above results together with (S.10) from Step 1 implies (S.13).

Step 3. Denote $\hat{\epsilon}=\mathcal{D}\big{(}\mathcal{C}({\bf
K}),\mathcal{C}(\widehat{\bf K})\big{)}.$ By the fact that
$\hat{\epsilon}^{2}=(2r)^{-1}\|{\bf K}{\bf K}^{\mathrm{\scriptscriptstyle
T}}-\widehat{\bf K}\widehat{\bf K}^{\mathrm{\scriptscriptstyle
T}}\|_{\text{F}}^{2}$ and Corollary 4.1 in Vu and Lei (2013), we obtain that

$\displaystyle 2r\nu_{r}\hat{\epsilon}^{2}$
$\displaystyle\leq|\langle\widetilde{\bf
M}-{\mathchoice{{\ooalign{\hbox{\raise
7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\displaystyle\widehat{\vrule
width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\displaystyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\textstyle\widehat{\vrule width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\textstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 6.40926pt\hbox{\scalebox{1.0}[-1.0]{\lower
6.40926pt\hbox{$\scriptstyle\widehat{\vrule width=0.0pt,height=4.78333pt\vrule
height=0.0pt,width=6.41667pt}$}}}}\cr\hbox{$\scriptstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 5.9537pt\hbox{\scalebox{1.0}[-1.0]{\lower
5.9537pt\hbox{$\scriptscriptstyle\widehat{\vrule
width=0.0pt,height=3.41666pt\vrule
height=0.0pt,width=4.58334pt}$}}}}\cr\hbox{$\scriptscriptstyle{\bf
M}$}}}}},\widehat{\bf K}\widehat{\bf K}^{\mathrm{\scriptscriptstyle T}}-{\bf
K}{\bf K}^{\mathrm{\scriptscriptstyle T}}\rangle|$
$\displaystyle\leq\left|\Big{\langle}\widetilde{\bf
M}-{\mathchoice{{\ooalign{\hbox{\raise
7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\displaystyle\widehat{\vrule
width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\displaystyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\textstyle\widehat{\vrule width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\textstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 6.40926pt\hbox{\scalebox{1.0}[-1.0]{\lower
6.40926pt\hbox{$\scriptstyle\widehat{\vrule width=0.0pt,height=4.78333pt\vrule
height=0.0pt,width=6.41667pt}$}}}}\cr\hbox{$\scriptstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 5.9537pt\hbox{\scalebox{1.0}[-1.0]{\lower
5.9537pt\hbox{$\scriptscriptstyle\widehat{\vrule
width=0.0pt,height=3.41666pt\vrule
height=0.0pt,width=4.58334pt}$}}}}\cr\hbox{$\scriptscriptstyle{\bf
M}$}}}}},\frac{\widehat{\bf K}\widehat{\bf K}^{\mathrm{\scriptscriptstyle
T}}-{\bf K}{\bf K}^{\mathrm{\scriptscriptstyle T}}}{\|\widehat{\bf
K}\widehat{\bf K}^{\mathrm{\scriptscriptstyle T}}-{\bf K}{\bf
K}^{\mathrm{\scriptscriptstyle
T}}\|_{\text{F}}}\Big{\rangle}\right|\sqrt{2r}\hat{\epsilon},$

which further implies that

$\nu_{r}\hat{\epsilon}\leq(2r)^{-1/2}\Big{|}\big{\langle}\widetilde{\bf
M}-{\mathchoice{{\ooalign{\hbox{\raise
7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\displaystyle\widehat{\vrule
width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\displaystyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\textstyle\widehat{\vrule width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\textstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 6.40926pt\hbox{\scalebox{1.0}[-1.0]{\lower
6.40926pt\hbox{$\scriptstyle\widehat{\vrule width=0.0pt,height=4.78333pt\vrule
height=0.0pt,width=6.41667pt}$}}}}\cr\hbox{$\scriptstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 5.9537pt\hbox{\scalebox{1.0}[-1.0]{\lower
5.9537pt\hbox{$\scriptscriptstyle\widehat{\vrule
width=0.0pt,height=3.41666pt\vrule
height=0.0pt,width=4.58334pt}$}}}}\cr\hbox{$\scriptscriptstyle{\bf
M}$}}}}},\boldsymbol{\widehat{\Delta}}\big{\rangle}\Big{|},$

where $\boldsymbol{\widehat{\Delta}}=\frac{\widehat{\bf K}\widehat{\bf
K}^{\mathrm{\scriptscriptstyle T}}-{\bf K}{\bf K}^{\mathrm{\scriptscriptstyle
T}}}{\|\widehat{\bf K}\widehat{\bf K}^{\mathrm{\scriptscriptstyle T}}-{\bf
K}{\bf K}^{\mathrm{\scriptscriptstyle T}}\|_{\text{F}}}$ and
$\|\boldsymbol{\widehat{\Delta}}\|_{\text{F}}=1.$

Note that $\boldsymbol{\widehat{\Delta}}$ is a matrix with rank at most $2r$
and the associated singular values
$\hat{\sigma}_{1}\geq\hat{\sigma}_{2}\geq\cdots\geq\hat{\sigma}_{2r}\geq 0.$
Then by the von-Neumann’s trace inequality, we obtain that

$\displaystyle\nu_{r}\hat{\epsilon}$
$\displaystyle\leq(2r)^{-1/2}\big{\|}\widetilde{\bf
M}-{\mathchoice{{\ooalign{\hbox{\raise
7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\displaystyle\widehat{\vrule
width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\displaystyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\textstyle\widehat{\vrule width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\textstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 6.40926pt\hbox{\scalebox{1.0}[-1.0]{\lower
6.40926pt\hbox{$\scriptstyle\widehat{\vrule width=0.0pt,height=4.78333pt\vrule
height=0.0pt,width=6.41667pt}$}}}}\cr\hbox{$\scriptstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 5.9537pt\hbox{\scalebox{1.0}[-1.0]{\lower
5.9537pt\hbox{$\scriptscriptstyle\widehat{\vrule
width=0.0pt,height=3.41666pt\vrule
height=0.0pt,width=4.58334pt}$}}}}\cr\hbox{$\scriptscriptstyle{\bf
M}$}}}}}\big{\|}\sum_{i=1}^{2r}\hat{\sigma}_{i}$
$\displaystyle\leq(2r)^{-1/2}\big{\|}\widetilde{\bf
M}-{\mathchoice{{\ooalign{\hbox{\raise
7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\displaystyle\widehat{\vrule
width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\displaystyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\textstyle\widehat{\vrule width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\textstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 6.40926pt\hbox{\scalebox{1.0}[-1.0]{\lower
6.40926pt\hbox{$\scriptstyle\widehat{\vrule width=0.0pt,height=4.78333pt\vrule
height=0.0pt,width=6.41667pt}$}}}}\cr\hbox{$\scriptstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 5.9537pt\hbox{\scalebox{1.0}[-1.0]{\lower
5.9537pt\hbox{$\scriptscriptstyle\widehat{\vrule
width=0.0pt,height=3.41666pt\vrule
height=0.0pt,width=4.58334pt}$}}}}\cr\hbox{$\scriptscriptstyle{\bf
M}$}}}}}\big{\|}\sqrt{2r\sum_{i=1}^{2r}\hat{\sigma}_{i}^{2}}$
$\displaystyle=\big{\|}\widetilde{\bf M}-{\mathchoice{{\ooalign{\hbox{\raise
7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\displaystyle\widehat{\vrule
width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\displaystyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\textstyle\widehat{\vrule width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\textstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 6.40926pt\hbox{\scalebox{1.0}[-1.0]{\lower
6.40926pt\hbox{$\scriptstyle\widehat{\vrule width=0.0pt,height=4.78333pt\vrule
height=0.0pt,width=6.41667pt}$}}}}\cr\hbox{$\scriptstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 5.9537pt\hbox{\scalebox{1.0}[-1.0]{\lower
5.9537pt\hbox{$\scriptscriptstyle\widehat{\vrule
width=0.0pt,height=3.41666pt\vrule
height=0.0pt,width=4.58334pt}$}}}}\cr\hbox{$\scriptscriptstyle{\bf
M}$}}}}}\big{\|}\|\boldsymbol{\widehat{\Delta}}\|_{\text{F}}=\big{\|}\widetilde{\bf
M}-{\mathchoice{{\ooalign{\hbox{\raise
7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\displaystyle\widehat{\vrule
width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\displaystyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 7.09259pt\hbox{\scalebox{1.0}[-1.0]{\lower
7.09259pt\hbox{$\textstyle\widehat{\vrule width=0.0pt,height=6.83331pt\vrule
height=0.0pt,width=9.16669pt}$}}}}\cr\hbox{$\textstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 6.40926pt\hbox{\scalebox{1.0}[-1.0]{\lower
6.40926pt\hbox{$\scriptstyle\widehat{\vrule width=0.0pt,height=4.78333pt\vrule
height=0.0pt,width=6.41667pt}$}}}}\cr\hbox{$\scriptstyle{\bf
M}$}}}}{{\ooalign{\hbox{\raise 5.9537pt\hbox{\scalebox{1.0}[-1.0]{\lower
5.9537pt\hbox{$\scriptscriptstyle\widehat{\vrule
width=0.0pt,height=3.41666pt\vrule
height=0.0pt,width=4.58334pt}$}}}}\cr\hbox{$\scriptscriptstyle{\bf
M}$}}}}}\big{\|}.$

By (S.13) from Step 2, we complete the proof of Theorem 3. $\square$

###  B.5 Auxiliary lemmas

In this section, we present some auxiliary lemmas that are used in the proofs
of theoretical results in Sections 3 and 4. The first lemma is a direct
corollary of Hahn-Banach theorem.

######  Lemma B1.

For a function $f:\mathcal{D}\rightarrow\mathcal{B}$, where
$\mathcal{D}\in\mathbb{R}^{p}$ is compact set and $\mathcal{B}$ is Banach
space with norm $\|\cdot\|$, then
$\|\textstyle\int_{\mathcal{D}}f(x)\mathrm{d}x\|\leq\textstyle\int_{\mathcal{D}}\|f(x)\|\mathrm{d}x$.

The second lemma specifies the order of $\boldsymbol{\Sigma}^{(k)}_{xx}(u,v)$.

######  Lemma B2.

Suppose that Condition 1 holds, then for $k=1,\dots,k_{0},$
$\iint\|\boldsymbol{\Sigma}^{(k)}_{xx}(u,v)\|^{2}\,{\rm d}u{\rm d}v=O(1)$ and
$\|\iint\boldsymbol{\Sigma}^{(k)}_{xx}(u,v)\,{\rm d}u{\rm d}v\|=O(1).$

###### Proof of Lemma B2.

By Cauchy-Schwartz inequality, Condition 1 and Fubini Theorem,

$\displaystyle\iint\|\boldsymbol{\Sigma}^{(k)}_{xx}(u,v)\|^{2}\,{\rm d}u{\rm
d}v$ $\displaystyle\leq\iint\|E\\{{\bf X}_{t+k}(u){\bf
X}_{t}(v)^{{\mathrm{\scriptscriptstyle T}}}\\}\|_{\text{F}}^{2}\,{\rm d}u{\rm
d}v=\sum_{i=1}^{r}\sum_{j=1}^{r}\iint\big{[}{\mathbb{E}}\\{X_{(t+k)i}(u)X_{tj}(v)\\}\big{]}^{2}\,{\rm
d}u{\rm d}v$
$\displaystyle\leq\Big{\\{}\sum_{i=1}^{r}{\mathbb{E}}\big{(}\|X_{(t+k)i}\|^{2}\big{)}\Big{\\}}\Big{\\{}\sum_{j=1}^{r}{\mathbb{E}}\big{(}\|X_{tj}\|^{2}\big{)}\Big{\\}}=O(1).$

On the other hand, by Lemma B1, we have that

$\left\|\iint\boldsymbol{\Sigma}^{(k)}_{xx}(u,v)\,{\rm d}u{\rm
d}v\right\|^{2}\leq\Big{(}\iint\big{\|}\boldsymbol{\Sigma}^{(k)}_{xx}(u,v)\big{\|}\,{\rm
d}u{\rm
d}v\Big{)}^{2}\leq\text{Vol}(\mathcal{U}^{2})\iint\big{\|}\boldsymbol{\Sigma}^{(k)}_{xx}(u,v)\big{\|}^{2}\,{\rm
d}u{\rm d}v=O(1),$

where $\text{Vol}(\cdot)$ represents the volume. ∎

The third lemma provides the non-asymptotic error bound on
$\widehat{\Sigma}_{yy,ij}^{(k)}(u,v).$

######  Lemma B3.

Suppose that Conditions 7 and 8 hold. Then there exists some positive
constants $c_{1}$ and $c_{2}$ such that for $k=1,\dots,k_{0}$ and $t>0,$

$\mathbb{P}\left(\max\limits_{1\leq i,j\leq
p}\big{\|}\widehat{\Sigma}_{yy,ij}^{(k)}-\Sigma_{yy,ij}^{(k)}\big{\|}_{{\cal
S}}>\mathcal{M}_{y}t\right)\leq
c_{1}p^{2}\exp\big{\\{}-c_{2}n\min(t,t^{2})\big{\\}},$

If $n\geq\rho^{2}\log p$ with $\rho>\sqrt{2}c_{2}^{-1/2},$ then

$\max\limits_{1\leq i,j\leq
p}\big{\|}\widehat{\Sigma}_{yy,ij}^{(k)}-\Sigma_{yy,ij}^{(k)}\big{\|}_{{\cal
S}}\leq\rho\mathcal{M}_{y}\sqrt{\frac{\log p}{n}}$

holds with probability greater than $1-c_{1}p^{2-c_{2}\rho^{2}}$.

###### Proof of Lemma B3.

This lemma follows directly from Guo and Qiao (2021); Fang et al. (2021) and
the choice of $t=\rho(\log p/n)^{1/2}\leq 1,$ and hence the proof is omitted
here. ∎

The following lemma is used in the proof of Theorem 3.

######  Lemma B4.

For ${\bf S}(u,v)=\\{S_{ij}(u,v)\\}_{p\times p}$ and ${\bf
T}(u,v)=\\{T_{ij}(u,v)\\}_{p\times p}$ with each $S_{ij}$ and
$T_{ij}\in{\mathbb{H}}\otimes{\mathbb{H}},$ it holds that

$\Big{\|}\iint{\bf S}(u,v){\bf T}^{{\mathrm{\scriptscriptstyle T}}}(u,v)\,{\rm
d}u{\rm d}v\Big{\|}\leq\Big{(}\|{\bf S}\|_{{\cal S},\infty}\|{\bf S}\|_{{\cal
S},1}\Big{)}^{1/2}\Big{(}\|{\bf T}\|_{{\cal S},\infty}\|{\bf T}\|_{{\cal
S},1}\Big{)}^{1/2}.$

###### Proof of Lemma B4.

Notice that

$\displaystyle\Big{\|}\iint{\bf S}(u,v){\bf
T}(u,v)^{{\mathrm{\scriptscriptstyle T}}}\,{\rm d}u{\rm d}v\Big{\|}_{1}$
$\displaystyle=\max\limits_{1\leq j\leq
p}\sum_{i=1}^{p}\left|\iint\sum_{k=1}^{p}S_{ik}(u,v)T_{jk}(u,v)\,{\rm d}u{\rm
d}v\right|$ (S.14) $\displaystyle\leq\max\limits_{1\leq j\leq
p}\sum_{i=1}^{p}\sum_{k=1}^{p}\|S_{ik}\|_{{\cal S}}\|T_{jk}\|_{{\cal S}}$
$\displaystyle\leq\Big{\\{}\max\limits_{1\leq j\leq
p}\sum_{k=1}^{p}\|T_{jk}\|_{{\cal S}}\Big{\\}}\Big{\\{}\max\limits_{1\leq
k\leq p}\sum_{i=1}^{p}\|S_{ik}\|_{{\cal S}}\Big{\\}}=\|{\bf S}\|_{{\cal
S},1}\|{\bf T}\|_{{\cal S},\infty}.$

By the similar arguments, we obtain that

$\Big{\|}\iint{\bf S}(u,v){\bf T}(u,v)^{{\mathrm{\scriptscriptstyle T}}}\,{\rm
d}u{\rm d}v\Big{\|}_{\infty}\leq\|{\bf S}\|_{{\cal S},\infty}\|{\bf
T}\|_{{\cal S},1}.$ (S.15)

Lemma B4 follows immediately from (S.14), (S.15), the matrix norm inequality
$\|{\bf E}\|^{2}\leq\|{\bf E}\|_{\infty}\|{\bf E}\|_{1}$ for any $p\times p$
matrix ${\bf E}$ and the choice of ${\bf E}=\iint{\bf S}(u,v){\bf
T}(u,v)^{{\mathrm{\scriptscriptstyle T}}}\,{\rm d}u{\rm d}v.$ ∎

##  Appendix C Additional empirical results

Figure 7 provides spatial heatmaps of varimax-rotated loading matrix and
sparse loading matrices with different sparsity levels for Japanese males.

(a) 1st factor (rotated)

(b) 2nd factor (rotated)

(c) 1st factor (27 zeros)

(d) 2nd factor (27 zeros)

(e) 1st factor (32 zeros)

(f) 2nd factor (32 zeros)

Figure 7: Spatial heatmaps based on varimax-rotated loadings (left column) and
sparse loadings with 27 zeros (middle column) and 32 zeros (right column) of
47 prefectures on two factors for Japanese males.

## References

Fang, Q., Guo, S. and Qiao, X. (2021). Finite sample theory for high-
dimensional functional/scalar time series with applications, Electronic
Journal of Statistics, in press.

Guo, S. and Qiao, X. (2021). On consistency and sparsity for high-dimensional
functional time series with application to autoregressions,
arXiv:2003.11462v2.

Lam, C., Yao, Q. and Bathia, N. (2011). Estimation of latent factors for high-
dimensional time series, Biometrika 98: 901–918.

Vu, V. Q. and Lei, J. (2013). Minimax sparse principal subspace estimation in
high dimensions, The Annals of Statistics 41: 2905–2947.

Wang, B.-Y. and Xi, B.-Y. (1997). Some inequalities for singular values of
matrix products, Linear Algebra and Its Applications 264: 109–115.


# On Fuzzification of $n$-Lie Algebras

Shadi Shaqaqha  Department of Mathematics, Yarmouk University, Irbid, Jordan
<EMAIL_ADDRESS>
###### Abstract.

The aim of this paper is to introduce the notion of intuitionistic fuzzy Lie
subalgebras and intutionistic fuzzy Lie ideals of $n$-Lie algebras. It is a
generalization of intuitionistic fuzzy Lie algebras. Then, we investigate some
of characteristics of intutionistic fuzzy Lie ideals (resp. subalgebras) of
$n$-Lie algeras. Finally, we define the image and the inverse image of
intuitionistic fuzzy Lie subalgebra under $n$-Lie algebra homomorphism. The
properties of intuitionistic fuzzy $n$-Lie subalgebras and intuitionistic
fuzzy Lie ideals under homomorphisms of $n$-Lie algebras are studied. Finally,
we define the intuitionistic fuzzy quotient $n$-Lie algebra by an
intuitionistic fuzzy ideal of n-Lie algebra and prove that it is a $n$-Lie
algebra.

###### Key words and phrases:

$n$-Lie algebras; $n$-Lie homomorphism; intutionistic fuzzy set;
intuitionistic fuzzy $n$-Lie subalgebra; intuitionistic fuzzy $n$-Lie ideal.

###### 2000 Mathematics Subject Classification:

08A72, 03E72, 20N25

##  1\. Introduction

The concept of $n$-Lie algebras was introduced by Filippov [15] in 1985. If
$n=2$, then we get Lie algebra structure which was introduced by Sophus Lie
(1842-1899) while he was attempting to classify certain smooth subgroups of
general linear groups that are now called Lie groups. The case where $n=3$ was
initially appeared in Nanmbu’s work [26] while he was attempting to describe
simultaneous classical dynamics of three particles. Takhtajan [29]
investigated the geometrical and algebraic aspects of the generalized Nambu
mechanics, and established the connection between the Nambu mechanics and
Filippov’s theory of $n$-Lie algebras. More applictions to the $n$-Lie
algebras can be found in [7, 16, 17, 23, 24, 25, 28, 34]. So, our results are
expected to be useful in various applications.
The notion of fuzzy sets was firstly introduced by Zadeh [37]. The fuzzy set
theory states that there are propositions with an infinite number of truth
values, assuming two extreme values, $1$ (totally true), $0$ (totally false)
and a continuum in between, that justify the term fuzzy. Applications of this
theory can be found, for example, in artificial intelligence, computer
science, control engineering, decision theory, logic and management science.
After introducing of fuzzy sets by Zadeh, many researches were conducted on
the generalizations of this fundamental concept. Among these generalizations
was the concept of intuitionistic fuzzy sets, which was introduced by
Attanasov [8] in 1986, is the most important and interesting one. The elements
of the intuitionistic fuzzy sets are distinguished by an additional degree
called the degree of uncertainty. There are numerous applications to this new
concept include computer science, mathematics, medicine, chemistry, economics,
astronomy etc.. Many mathematicians have involved in extending the concept of
intuitionistic fuzzy sets to border of abstract algebra. Biswas applied the
concepts of intuitionistic fuzzy sets to the theory of groups and studied
intuitionistic fuzzy subgroups in [10]. Also, Akram studied Lie algebra in
intuitionistic fuzzy sets and obtained some results in [3].
The fuzzy Lie subalgebras and fuzzy Lie ideals are considered in [21] by Kim
and Lee, and in [35, 36] by Yehia. They established the analogues of most of
the fundamental ground results involving Lie algebras in the fuzzy setting.
The study of fuzzy subalgebras (resp. ideals) of $n$-Lie algebras was
initiated by B. Davvaz and WA. Dudek [13]. Recently, the complex
(intuitionistic) fuzzy Lie algebras is studied in [32, 33] as a generalization
of (intutionistic) fuzzy Lie algebras.
In this paper we describe intuitionistic fuzzy n-Lie algebras. We will
introduce n-Lie algebras into intuitionistic fuzzy set. Our work will
generalize the theory of (intu- itionstic) fuzzy Lie algebras ([3, 4, 5, 6,
33, 35, 36]).

##  2\. Preliminaries

$n$-Lie algebras were originally introduced by Filippov [15] in 1985. They
generalize Lie algebras. In this article the ground field $F$ is arbitrary.

######  Definition 2.1.

Let $n\in\mathbb{N}$, $n\geq 2$. An $n$-Lie algebra is a pair $(L,[~{}])$
where $L$ is a vector space and

$[~{}]:L^{n}\rightarrow L;~{}(x_{1},\ldots,x_{n})\mapsto[x_{1},\ldots,x_{n}]$

is an $n$-linear map, called $n$-Lie bracket, that satisfies the following
identities for all $\sigma$ in the symmetric group $S_{n}$ and
$x_{1},\ldots,x_{n},y_{2},\ldots,y_{n}\in L$:

  * (i)

Skew symmetry:

$[x_{\sigma(1)},\ldots,x_{\sigma(n)}]=\mathrm{sign}(\sigma)[x_{1},\ldots,x_{n}].$

  * (ii)

The generalized Jacobi identity (called also the Filippov identity):

$[[x_{1},\ldots,x_{n}],y_{2},\ldots,y_{n}]=\sum_{i=1}^{n}[x_{1},\ldots,x_{i-1},[x_{i},y_{2},\ldots,y_{n}],x_{i},\ldots,x_{n}].$

Subalgebras of $n$-Lie algebras and homomorphisms (or isomorphisms) between
$n$-Lie algebras are defined as usual ([15]).
Throughout this paper, $L$ is a $n$-Lie algebra over field $F$.
The concept of intuitionistic fuzzy set was introduced by Atanassov [8], where
he added a new component (which determines the degree of non-membership) in
the definition of fuzzy set (FS) that was given by Zadeh. Let $X$ be a non-
empty set, and let
$A=(\mu_{A},\lambda_{A})=\\{(x,\mu_{A}(x),\lambda_{A}(x))~{}:~{}x\in X\\}$
where $\mu_{A}:X\rightarrow[0,1]$ and $\lambda_{A}:X\rightarrow[0,1]$ be
mappings such that $\mu_{A}(x)+\lambda_{A}(x)\leq 1$. Then $A$ is called an
intuitionistic fuzzy set. In this case the mappings $\mu_{A}$ and
$\lambda_{A}$ denote the degree of membership and the degree of non-membership
to $A$ respectively, for each element $x\in X$. The value
$\pi_{A}(x)=1-\mu_{A}(x)-\lambda_{A}(x)$ is called uncertainty or
intuitionistic index of the element $x\in X$ to the intuitionistic fuzzy set
$A$. It is obvious that each fuzzy set $A=\mu_{A}=\\{(x,\mu_{A}(x))~{}:~{}x\in
X\\}$ can be represented as an intuitionistic fuzzy set where
$A=\\{(x,\mu_{A}(x),1-\mu_{A}(x))~{}:~{}x\in X\\}$.

######  Definition 2.2.

Let $A=(\mu_{A},\lambda_{A})$ and $B=(\mu_{B},\lambda_{B})$ be two
intuitionistic fuzzy sets of a set $X$. Then

  * (i)

The complement of $A$ is $\bar{A}=(\lambda_{A},\mu_{A})$,

  * (ii)

$A\subseteq B$ if and only if $\mu_{A}(x)\leq\mu_{B}(x)$ and
$\lambda_{A}(x)\geq\lambda_{B}(x)$ for all $x\in X$,

  * (iii)

the intersection of $A$ and $B$ is $A\cap
B=\\{(x,\mathrm{min}\\{\mu_{A}(x),\mu_{B}(x)\\},\mathrm{max}\\{\lambda_{A}(x),\lambda_{B}(x)\\}~{}|~{}x\in
X$,

  * (iv)

the union of $A$ and $B$ is $A\cup
B=\\{(x,\mathrm{max}\\{\mu_{A}(x),\mu_{B}(x)\\},\mathrm{min}\\{\lambda_{A}(x),\lambda_{B}(x)\\}~{}|~{}x\in
X$,

  * (v)

$\Box A=\\{(x,~{}\mu_{A}(x),~{}\mu_{A}^{\complement}(x)):~{}x\in X\\}$ where
$\mu_{A}^{\complement}(x)=(1-\mu_{A}(x))$ for all $x\in X$,

  * (vi)

$\diamondsuit
A=\\{(x,~{}\lambda_{A}^{\complement}(x),~{}\lambda_{A}(x)):~{}x\in X\\}$ where
$\lambda_{A}^{\complement}(x)=(1-\lambda_{A}(x))$ for all $x\in X$.

Let $L$ be a $n$-Lie algebra over $F$. A fuzzy set
$A=\mu_{A}=\\{(x,~{}\mu_{A}(x)):x\in L\\}$ on $L$ is a fuzzy Lie subalgebra if
the following conditions are satisfied:

  * (i)

$\mu_{A}(x+y)\geq\mathrm{min}\\{\mu_{A}(x),\mu_{A}(y)\\}$ for all $x,y\in L$,

  * (ii)

$\mu_{A}(\alpha x)\geq\mu_{A}(x)$ for all $x\in L$ and $\alpha\in F$,

  * (iii)

$\mu_{A}([x_{1},~{}x_{2},\ldots,x_{n}])\geq\mathrm{min}\\{\mu_{A}(x_{1}),\mu_{A}(x_{2}),\cdots,\mu_{A}(x_{n})\\}$
for all $x_{1},~{}x_{2},\ldots,~{}x_{n}\in L$.

It is called a fuzzy Lie ideal if the condition

$\mu_{A}([x_{1},~{}x_{2},\ldots,x_{n}])\geq\mathrm{min}\\{\mu_{A}(x_{1}),\mu_{A}(x_{2}),\cdots,\mu_{A}(x_{n})\\}$

is replaced by

$\mu_{A}([x_{1},~{}x_{2},\ldots,x_{n}])\geq\mathrm{max}\\{\mu_{A}(x_{1}),\mu_{A}(x_{2}),\cdots,\mu_{A}(x_{n})\\}$

([13]).

##  3\. Intuitionistic Fuzzy $n$-Lie Algebras

For the sake of simplicity, we shall use the symbols $a\wedge
b=\mathrm{min}\left\\{a,b\right\\}$ and $a\vee
b=\mathrm{max}\left\\{a,b\right\\}$.

######  Definition 3.1.

Let $A=(\mu_{A},\lambda_{A})=\\{(x,~{}\mu_{A}(x),~{}\lambda_{A}(x)):x\in L\\}$
be an intuitionistic fuzzy set of $L$. Then $A$ is called an intuitionistic
fuzzy Lie subalgebra of $L$ if the following conditions are satisfied:

  * (i)

$\mu_{A}(x+y)\geq\mu_{A}(x)\wedge\mu_{A}(y)$ and
$\lambda_{A}(x+y)\leq\lambda_{A}(x)\vee\lambda_{A}(y)$ for all $x,y\in L$,

  * (ii)

$\mu_{A}(\alpha x)\geq\mu_{A}(x)$ and $\lambda_{A}(\alpha
x)\leq\lambda_{A}(x)$ for all $x\in L$ and $\alpha\in F$,

  * (iii)

$\mu_{A}([x_{1},~{}x_{2},\ldots,~{}x_{n}])\geq\mu_{A}(x_{1})\wedge\mu_{A}(x_{2})\wedge\cdots\wedge\mu_{A}(x_{n})$
and
$\lambda_{A}([x_{1},~{}x_{2},\ldots,~{}x_{n}])\leq\lambda_{A}(x_{1})\vee\lambda_{A}(x_{2})\vee\cdots\vee\lambda_{A}(x_{n})$
for all $x_{1},x_{2},\ldots,x_{n}\in L$.

An intuitionistic fuzzy set $A$ on $L$ is called an intuitionistic fuzzy Lie
ideal if the conditions $(i)$ and $(ii)$ are satisfied together with the
following addition condition:
$(iii)^{\prime}\mu_{A}([x_{1},~{}x_{2},\ldots,~{}x_{n}])\geq\mu_{A}(x_{1})\vee\mu_{A}(x_{2})\vee\cdots\vee\mu_{A}(x_{n})$
and
$\lambda_{A}([x_{1},~{}x_{2},\ldots,~{}x_{n}])\leq\lambda_{A}(x_{1})\wedge\lambda_{A}(x_{2})\wedge\cdots\wedge\lambda_{A}(x_{n})$
for all $x_{1},x_{2},\ldots,x_{n}\in L$.

In the special case that $n=2$, we obtain intuitionistic fuzzy Lie algebras
([3]). When first two conditions hold, we say that $A$ is an intuitionistic
fuzzy vector subspace of $L$. The second condition implies
$\mu_{A}(0)\geq\mu_{A}(x)$, $\mu_{A}(-x)\geq\mu_{A}(x)$,
$\lambda_{A}(0)\leq\lambda_{A}(x)$, and $\lambda_{A}(-x)\leq\lambda_{A}(x)$
for all $x\in L$. Also, every intuitionistic fuzzy Lie ideal of an $n$-Lie
algebra is an intuitionistic fuzzy Lie subalgebra, but the converse is not
necessary true ( see [32, Example 3.1]).

Let $\\{A_{i}=(\mu_{A_{i}},\lambda_{A_{i}})~{}|~{}i\in I\\}$ be a collection
of intuitionistic fuzzy sets on a nonempty set $X$. Then

$\bigcap_{i\in I}A_{i}=\\{(x,~{}\mu_{\bigcap_{i\in
I}A_{i}}(x),~{}\lambda_{\bigcap_{i\in I}A_{i}}(x)):~{}x\in X\\},$

where

$\mu_{\bigcap_{i\in I}A_{i}}(x)=\inf_{i\in I}\\{\mu_{A_{i}}(x)\\}$

and

$\lambda_{\bigcap_{i\in I}A_{i}}(x)=\sup_{i\in I}\\{\lambda_{A_{i}}(x)\\},$

is an intuitionistic fuzzy set too (see [33]). We shall give the proof of the
following theorem, established in [32] to the case of intuitionistic fuzzy Lie
algebras, which proves that the arbitrary intersection of intuitionistic fuzzy
Lie subalgebras (resp. ideals) of an $n$-Lie algebra $L$ is an intuitionistic
fuzzy Lie subalgebras (resp. ideal) of $L$ too. However it was proved in the
case that $L$ is a Lie (super)algebras and where the family is finite (see
[11]).

######  Theorem 3.1.

Let $\\{A_{i}\\}_{i\in I}$ $(A_{i}=(\mu_{A_{i}},\lambda_{A_{i}}),i\in I)$ be a
collection of intuitionistic fuzzy Lie subalgebras (resp. ideals) on $L$. Then
$\bigcap_{i\in I}A_{i}$ is an intuitionistic fuzzy Lie subalgebra (resp.
ideal) of $L$.

Proof. Here we will prove the case of intuitionistic fuzzy Lie subalgera. For
$x,~{}y\in L$ and $\alpha\in F$, we have

$\displaystyle\mu_{\bigcap_{i\in I}A_{i}}(x+y)=$ $\displaystyle\inf_{i\in
I}\\{\mu_{A_{i}}(x+y)\\}$ $\displaystyle\geq$ $\displaystyle\inf_{i\in
I}\\{\mu_{A_{i}}(x)\wedge\mu_{A_{i}}(y)\\}$ $\displaystyle=$
$\displaystyle\inf_{i\in I}\\{\mu_{A_{i}}(x)\\}\wedge\inf_{i\in
I}\\{\mu_{A_{i}}(y)\\}$ $\displaystyle=$ $\displaystyle\mu_{\bigcap_{i\in
I}A_{i}}(x)\wedge\mu_{\bigcap_{i\in I}A_{i}}(y).$

Also,

$\displaystyle\mu_{\bigcap_{i\in I}A_{i}}(\alpha x)=$ $\displaystyle\inf_{i\in
I}\\{\mu_{A_{i}}(\alpha x)\\}$ $\displaystyle\geq$ $\displaystyle\inf_{i\in
I}\\{\mu_{A_{i}}(x)\\}$ $\displaystyle=$ $\displaystyle\mu_{\bigcap_{i\in
I}A_{i}}(x).$

Similarly, we can prove that $\lambda_{\bigcap_{i\in
I}A_{i}}(x+y)\leq\lambda_{\bigcap_{i\in I}A_{i}}(x)\vee\lambda_{\bigcap_{i\in
I}A_{i}}(y)$ and $\lambda_{\bigcap_{i\in I}A_{i}}(\alpha
x)\leq\lambda_{\bigcap_{i\in I}A_{i}}(x)$. Next, if $x_{1},\ldots,x_{n}\in L$,
then

$\displaystyle\mu_{\bigcap_{i\in I}A_{i}}([x_{1},\ldots,x_{n}])=$
$\displaystyle\inf_{i\in I}\\{\mu_{A_{i}}([x_{1},\ldots,x_{n}])\\}$
$\displaystyle\geq$ $\displaystyle\inf_{i\in
I}\\{\mu_{A_{i}}(x_{1})\wedge\cdots\wedge\mu_{A_{i}}(x_{n})\\}$
$\displaystyle=$ $\displaystyle\inf_{i\in
I}\\{\mu_{A_{i}}(x)\\}\wedge\cdots\wedge\inf_{i\in I}\\{\mu_{A_{i}}(x_{n})\\}$
$\displaystyle=$ $\displaystyle\mu_{\bigcap_{i\in
I}A_{i}}(x_{1})\wedge\cdots\wedge\mu_{\bigcap_{i\in I}A_{i}}(x_{n}).$

In a similar way, one can show $\lambda_{\bigcap_{i\in
I}A_{i}}([x_{1},\ldots,x_{n}])\leq\lambda_{\bigcap_{i\in
I}A_{i}}(x)\vee\cdots\vee\lambda_{\bigcap_{i\in I}A_{i}}(x_{n})$. Therefore,
$\bigcap_{i\in I}A_{i}$ is an intuitionistic fuzzy Lie subalgebra. The proof
of the case of intuitionistic fuzzy Lie ideal is same, so we omit it. $\Box$

Let $A=\\{(x,~{}\mu_{A}(x),~{}\lambda_{A}(x)):~{}x\in X\\}$ be an
intuitionistic fuzzy set. For $s,t\in[0,~{}1]$ the set $A^{t}_{s}=\\{x\in
X:\mu_{A}(x)\geq s,~{}\lambda_{A}(x)\leq t\\}$ is called the upper level
subset of the intuitionistic fuzzy subset $A$. In particular if $t=1$, then we
get the upper $s$-level cut $A_{s}^{1}=U(\mu_{A};s)=\\{x\in
X~{}:~{}\mu_{A}(x)\geq s\\}$. Also, if $s=0$, then we get the lower $t$-level
cut $A_{0}^{t}=L(\lambda_{A};t)=\\{x\in X~{}:~{}\lambda_{A}(x)\leq t\\}$.
The following two theorems will show relations between intuitionistic fuzzy
Lie subalgebras of $L$ and Lie subalgebras of $L$. They are very similar to
the case that suggested by Kondo and Dudek in [22].

######  Theorem 3.2.

Let $A=(\mu_{A},~{}\lambda_{A})$ be an intuitionistic fuzzy set of an $n$-Lie
algebra $L$. Then A is an intuitionistic fuzzy Lie subalgebra of $L$ if and
only if the non-empty set $A^{t}_{s}$ is Lie subalgebra for all
$s,t\in[0,~{}1]$.

Proof. Let $A=\\{(x,~{}\mu_{A}(x),~{}\lambda_{A}(x)):~{}x\in L\\}$ be an
intuitionistic fuzzy Lie subalgebra. For $x,~{}y\in A^{s}_{t}$ and $\gamma\in
F$

  * (i)

$\mu_{A}(x+y)\geq\mu_{A}(x)\wedge\mu_{A}(y)\geq s$ and
$\lambda_{A}(x+y)\leq\lambda_{A}(x)\vee\lambda_{A}(y)\leq t$,

  * (ii)

$\mu_{A}(\gamma x)\geq\mu_{A}(x)\geq s$ and $\lambda_{A}(\gamma
x)\leq\lambda_{A}(x)\leq t$.

Thus $x+y,~{}\gamma x\in A_{s}^{t}$. Also for $x_{1},\ldots,x_{n}\in L$, we
have
$\mu_{A}([x_{1},,\ldots,x_{n}])\geq\mu_{A}(x_{1})\wedge\cdots\wedge\mu_{A}(x_{n}\geq
s$ and
$\lambda_{A}([x_{1},\ldots,x_{n}])\leq\lambda_{A}(x_{1})\vee\cdots\vee\lambda_{A}(x_{n})\leq
t$. That is $[x_{1},\ldots,x_{n}]\in A^{t}_{s}$. Therefore $A^{t}_{s}$ is Lie
subalgebra of $L$. Conversely, suppose that $A^{s}_{t}\neq\emptyset$ is a Lie
subalgebra of $L$ for every $s,t\in[0,~{}1]$. Let $x,y\in L$ and $\alpha\in
F$. Fix $s_{1}=\mu_{A}(x)\wedge\mu_{A}(x)$ and
$t_{1}=\lambda_{A}(x)\vee\lambda_{A}(y)$, so that $x,y\in A_{s_{1}}^{t_{1}}$.
Since $A_{s_{1}}^{t_{1}}$ is a subspace of $L$, we have $x+y$ and $\alpha x$
are in $A_{s_{1}}^{t_{1}}$, and so $\mu_{A}(x+y)\geq
s_{1}=\mu_{A}(x)\wedge\mu_{A}(y)$, $\lambda_{A}(x+y)\leq
t_{1}=\lambda_{A}(x)\vee\lambda_{A}(y)$. Also for $x\in L$, set
$s_{1}=\mu_{A}(x)$ and $t_{1}=\lambda_{A}(x)$. Then $x\in A_{s_{1}}^{t_{1}}$,
and so $\gamma x\in A^{t_{1}}_{s_{1}}$. Hence $\mu_{A}(\gamma
x)\geq\mu_{A}(x)$ and $\lambda_{A}(\gamma x)\leq\lambda_{A}(x)$. Finally let
$x_{1},x_{2},\ldots,x_{n}\in L$. Fix
$t=\mu_{A}(x_{1})\wedge\cdots\wedge\mu_{A}(x_{n})$ and
$s=\lambda_{A}(x_{1})\vee\cdots\vee\lambda_{A}(x_{n})$. Thus $x_{i}\in
A^{t}_{s}$ for all $i=1,\ldots,n$. Since $A^{t}_{s}$ is a subalgebra of $L$,
we have $[x_{1},\ldots,x_{n}]\in A^{t}_{s}$, so that
$\mu_{A}([x_{1},\ldots,x_{n}])\geq
s=\mu_{A}(x_{1})\wedge\cdots\wedge\mu_{A}(x_{2})$ and
$\lambda_{A}([x_{1},\ldots,x_{n}])\leq
t=\lambda_{A}(x_{1})\vee\cdots\vee\lambda_{A}(x_{n})$. The case of
intuitionistic fuzzy ideals is almost same. $\Box$

Let $A=\\{(x,~{}\mu_{A}(x),~{}\lambda_{A}(x)):~{}x\in X\\}$ be an
intuitionistic fuzzy set. For $s,t\in[0,~{}1]$ the set
$A^{t^{<}}_{s^{>}}=\\{x\in X:\mu_{A}(x)>s,~{}\lambda_{A}(x)<t\\}$ is called
the strong upper level subset of the intuitionistic fuzzy subset $A$.

######  Theorem 3.3.

Let $A=(\mu_{A},~{}\lambda_{A})$ be an intuitionistic fuzzy subset of $L$.
Then $A$ is an intuitionistic fuzzy Lie subalgebra of $L$ if and only if the
non empty set $A^{t^{<}}_{s^{>}}$ is Lie subalgebra, for all $s,t\in[0,~{}1]$.

Proof. The proof of the forward direction is almost identical to the proof in
Theorem 3.2. Conversely, suppose that the non-empty set $A^{t^{<}}_{s^{>}}$ is
a Lie subalgebra for all $s,t\in[0,~{}1]$. We need to show that the conditions
of Definition 3.1 are satisfied. Let $x,y\in L$. If $\mu_{A}(x)=0$ or
$\mu_{A}(y)=0$, then it is clear that
$\mu_{A}(x+y)\geq\mu_{A}(x)\wedge\mu_{A}(y)$, so we may assume that
$\mu_{A}(x)\neq 0$ and $\mu_{A}(y)\neq 0$. Let $s_{0}$ be the largest number
on the interval $[0,1]$ such that $s_{0}<\mu_{A}(x)\wedge\mu_{A}(y)$ and there
is no $a\in L$ satisfying $s_{0}<\mu_{A}(a)<\mu_{A}(x)\wedge\mu_{A}(y)$.
Having $x,y\in A_{s_{0}^{>}}^{t^{<}}$ where
$1>t>\lambda_{A}(x)\vee\lambda_{A}(y)$ (such $t$ exists because $\mu_{A}(x)$
and $\mu_{A}(y)$ are greater than $0$ in addition to
$\mu_{A}(x)+\lambda_{A}(x)$ and $\mu_{A}(y)+\lambda_{A}(y)$ are less than or
equals to $1$) implies that $x+y\in A^{t}_{s_{0}}$, and hence
$\mu_{A}(x+y)>s_{0}$. Since there exist no $a\in L$ with
$s_{0}<\mu_{A}(a)<\mu_{A}(x)\wedge\mu_{A}(y)$, it follows that
$\mu_{A}(x+y)\geq\mu_{A}(x)\wedge\mu_{A}(y)$. If $\lambda_{A}(x)=1$ or
$\lambda_{A}(y)=1$, then it is obvious that
$\lambda_{A}(x+y)\leq\lambda_{A}(x)\vee\lambda_{A}(y)$. Assume now that
$\lambda_{A}(x)\neq 1\neq\lambda_{A}(y)$. Let $t_{0}$ be the smallest number
on the interval $[0,1]$ such that $t_{0}>\lambda_{A}(x)\vee\lambda_{A}(y)$ and
there is no $a\in L$ with
$t_{0}>\lambda_{A}(a)>\lambda_{A}(x)\vee\lambda_{A}(y)$. Hence $x,y\in
A^{t_{0}^{<}}_{s^{>}}$ where $s<\mu_{A}(x)\wedge\mu_{A}(y)$, and so $x+y\in
A^{t_{0}^{<}}_{s^{>}}$. Therefore $\lambda_{A}(x+y)<t_{0}$. Since there is no
$a\in L$ such that $t_{0}>\lambda_{A}(a)>\lambda_{A}(x)\vee\lambda_{A}(y)$, it
follows that $\lambda_{A}(x+y)\leq\lambda_{A}(x)\vee\lambda_{A}(y)$ as
desired. In a similar way we can show that $\mu_{A}(\gamma x)\geq\mu_{A}(x)$
and $\lambda_{A}(\gamma x)\leq\lambda_{A}(x)$ for all $x\in L$ and $\gamma\in
F$. Next let $x_{1},\ldots,x_{n}\in L$. Again we may assume that neither of
$\mu_{A}(x_{1}),\ldots,\mu_{A}(x_{n})$ is $0$. Let $s_{1}$ be the greatest
number on the interval $[0,1]$ such that
$s_{1}<\mu_{A}(x)\wedge\cdots\wedge\mu_{A}(x_{n})$ and there is no $a\in L$
such that $s_{1}<\mu_{A}(a)<\mu_{A}(x_{1})\wedge\cdots\wedge\mu_{A}(x_{n})$.
Then $x_{i}\in A^{t^{<}}_{s_{1}^{>}}$, where
$1>t>\lambda_{A}(x_{1})\vee\cdots\vee\mu_{A}(x_{n})$, for each $i=1,\ldots,n$.
Since $A^{t^{<}}_{s_{1}^{>}}$ is subalgebra of $L$, we have
$[x_{1},\ldots,x_{n}]\in A^{t^{<}}_{s_{1}^{>}}$. Hence
$\mu_{A}(x_{1},\ldots,x_{n})\geq\mu_{A}(x_{1})\wedge\cdots\wedge\mu_{A}(x_{n})$.
In a similar fashion, we can prove that
$\lambda_{A}([x_{1},\ldots,x_{n}])\leq\lambda_{A}(x_{1})\vee\cdots\vee\lambda_{A}(x_{n})$
for all $x_{1},\ldots,x_{n}\in L$. $\Box$

From the proofs of Theorem 3.2 and Theorem 3.3, we can immediately obtain the
following result.

######  Corollary 3.1.

Let $A=(\mu_{A},\lambda_{A})$ be an intuitionistic fuzzy subset of an $n$-Lie
algebra $L$. The following statements are equivalent for every $s,t\in[0,1]$:

  * (i)

$A$ is an intuitionistic fuzzy Lie subalgebra (resp. ideal) of $L$,

  * (ii)

The nonempty subsets $A^{t}_{s}$ are $n$-Lie subalgebras (resp. ideals) of
$L$,

  * (iii)

The nonempty subsets $A^{t^{<}}_{s}=\\{x\in L~{}|~{}\mu_{A}(x)\geq
s~{}\mathrm{and}~{}\lambda_{A}(x)<t\\}$ are $n$-Lie subalgebras (resp. ideals)
of $L$,

  * (iv)

The nonempty subsets $A^{t}_{s^{>}}=\\{x\in
L~{}|~{}\mu_{A}(x)>s~{}\mathrm{and}~{}\lambda_{A}(x)\leq t\\}$ are $n$-Lie
subalgebras (resp. ideals) of $L$,

  * (v)

The The nonempty subsets $A^{t^{<}}_{s^{>}}$ are $n$-Lie subalgebras (resp.
ideals) of $L$.

##  4\. Operations On Intuitionistic Fuzzy $n$-Lie ideals

We omit the proofs for the following two results because they are
straightforward.

######  Theorem 4.1.

Let $A=(\mu_{A},~{}\lambda_{A})$ be an intuitionistic fuzzy set in an $n$-Lie
algebra $L$. Then $A$ is an intuitionistic fuzzy Lie subalgebra (resp. ideal)
if and only if $\Box A$ and $\diamondsuit A$ are intuitionistic fuzzy Lie
subalgebras (resp. ideals).

Let $A=(\mu_{A},~{}\lambda_{A})$ and $B=(\mu_{B},~{}\lambda_{B})$ be two
intuitionistic fuzzy subsets on a $n$-Lie algebra $L$. Let us introduce the
following sum of $A$ and $B$, which was defined by Chen and Zhang [11] in the
case of intuitionistic fuzzy Lie superalgebras:

$A+B=(\mu_{A+B},~{}\lambda_{A+B}),$

where

$\mu_{A+B}(x)=\sup_{x=a+b}\\{\mu_{A}(a)\wedge\mu_{B}(b)\\},$

and

$\lambda_{A+B}(x)=\inf_{x=a+b}\\{\lambda_{A}(a)\vee\lambda_{B}(b)\\}.$

If $A=(\mu_{A},~{}\lambda_{A})$ and $B=(\mu_{B},~{}\lambda_{B})$ are
intuitionistic fuzzy sets on $L$, then $A+B$ is an intuitionistic fuzzy set of
$L$. Indeed if $x\in L$ with $\mu_{A+B}(x)+\lambda_{A+B}(x)>1$, then
$\sup_{x=a+b}\\{\mu_{A}(a)\wedge\mu_{B}(b)\\}+\inf_{x=a+b}\\{\lambda_{A}(a)\vee\lambda_{B}(b)\\}>1$,
and so there exist $a_{1},~{}b_{1}\in L$ such that $x=a_{1}+b_{1}$ with
$\mu_{A}(a_{1})\wedge\mu_{B}(b_{1})+\inf_{x=a+b}\\{\lambda_{A}(a)\vee\lambda_{B}(b)\\}>1$.
It follows that
$\mu_{A}(a_{1})\wedge\mu_{B}(b_{1})+\lambda_{A}(a_{1})\vee\lambda_{B}(b_{1})>1$.
Without loss of generality we may assume $\mu_{A}(a_{1})\leq\mu_{B}(b_{1})$.
If $\lambda_{A}(a_{1})\geq\lambda_{B}(b_{1})$, then it is a clear
contradiction. If $\lambda_{A}(a_{1})\leq\lambda_{B}(b_{1})$, then
$\mu_{A}(a_{1})+\lambda_{B}(b_{1})>1$. Hence
$\mu_{A}(a_{1})+1-\mu_{B}(b_{1})>1$, and so $\mu_{A}(a_{1})>\mu_{B}(b_{1})$.
Contradiction. Therefore, $A+B$ is an intuitionistic fuzzy set of $L$.

It is well known that if $A$ and $B$ are ideals of $L$, then $A+B$ is an ideal
of $L$ too. We are going to obtain an analogous result in the case of
intuitionistic fuzzy Lie ideals. Similar result was obtained for complex fuzzy
Lie subalgebra in [32], and for intuitionistic fuzzy Lie sub-superalgebras in
[11].

######  Theorem 4.2.

Let $A=(\mu_{A},~{}\lambda_{A})$ and $B=(\mu_{B},~{}\lambda_{B})$ be two
intuitionistic fuzzy $n$-Lie ideals on $L$. Then $A+B$ is an intuitionistic
fuzzy $n$-Lie ideal of $L$ too.

Proof. We proceed as in the proof of corresponding result on complex fuzzy Lie
algebras [33]. The only difference appears in the proof is to show that
$\mu_{A}([x_{1},\ldots,x_{n}])\geq\mu_{A}(x_{1})\vee\cdots\vee\mu_{A}(x_{n})$
and
$\lambda_{A}([x_{1},\ldots,x_{n}])\geq\lambda_{A}(x_{1})\wedge\mu_{A}(x_{n})$
for each $x_{1},\ldots,x_{n}\in L$. Let $x_{1},\ldots,x_{n}\in L$. Suppose
that
$\mu_{A+B}([x_{1},\ldots,x_{n}])<\mu_{A+B}(x_{1})\vee\cdots\vee\mu_{A+B}(x_{n})$.
Then there exists $t$ such that
$\mu_{A+B}([x_{1},\ldots,x_{n}])<t<\mu_{A+B}(x_{1})\vee\ldots\vee\mu_{A+B}(x_{n})$.
Without loss of generality we may assume
$\mu_{A+B}(x_{1})=\mu_{A+B}(x_{1})\vee\cdots\vee\mu_{A+B}(x_{n})$. Thus
$\mu_{A+B}([x_{1},\ldots,x_{n}])<t<\sup_{x_{1}=a+b}\\{\mu_{A}(a)\wedge\mu_{B}(b)\\}$,
and so there exist $a_{1},~{}b_{1}\in L$ such that
$\mu_{A+B}([x,~{}y])<t<\mu_{A}(a_{1})\wedge\mu_{B}(b_{1})$. Therefore
$\mu_{A}(a_{1})>t$ and $\mu_{B}(b_{1})>t$. Hence

$\displaystyle\mu_{A+B}([x_{1},\ldots,x_{n}])$ $\displaystyle=$
$\displaystyle\mu_{A+B}([a_{1}+b_{1},\ldots,x_{n}])$ $\displaystyle\geq$
$\displaystyle\sup_{[x_{1},\ldots,x_{n}]=[a,\ldots,x_{n}]+[b,\ldots,x_{n}]}\\{\mu_{A}([a,\ldots,x_{n}])\wedge\mu_{B}([b,\ldots,x_{n}])\\}$
$\displaystyle\geq$
$\displaystyle\mu_{A}([a_{1},\ldots,x_{n}])\wedge\mu_{B}([b_{1},\ldots,x_{n}])$
$\displaystyle\geq$
$\displaystyle(\mu_{A}(a_{1})\vee\cdots\vee\mu_{A}(x_{n}))\wedge(\mu_{B}(b_{1})\vee\cdots\vee\mu_{B}(x_{n}))$
$\displaystyle>$ $\displaystyle t$ $\displaystyle>$
$\displaystyle\mu_{A+B}([x_{1},\ldots,x_{n}]).$

Contradiction. Thus
$\mu_{A+B}([x_{1},\ldots,x_{n}]\geq\mu_{A+B}(x_{1})\vee\cdots\vee\mu_{A+B}(x_{n})$.
Finally, suppose
$\lambda_{A+B}([x_{1},\ldots,x_{n}])>\lambda_{A+B}(x_{1})\wedge\cdots\wedge\lambda_{A+B}(x_{n})$.
Then there exists $s$ such that
$\lambda_{A+B}([x_{1},\ldots,x_{n}])>s>\lambda_{A+B}(x_{1})\wedge\cdots\wedge\lambda_{A+B}(x_{n})$.
Without loss of generality we may assume
$\lambda_{A+B}(x_{1})=\lambda_{A+B}(x_{1})\wedge\cdots\wedge\mu_{A+B}(x_{n})$.
Thus
$\lambda_{A+B}([x_{1},\ldots,x_{n}])>s>\inf_{x_{1}=a+b}\\{\lambda_{A}(a)\vee\lambda_{B}(b)\\}$,
and so there exist $a_{1},~{}b_{1}\in L$ such that
$\lambda_{A+B}([x_{1},\ldots,x_{n}])>s>\lambda_{A}(a_{1})\vee\lambda_{B}(b_{1})$.
Therefore $\lambda_{A}(a_{1})<s$ and $\lambda_{B}(b_{1})<s$. Now

$\displaystyle\lambda_{A+B}([x_{1},\ldots,x_{n}])$ $\displaystyle=$
$\displaystyle\lambda_{A+B}([a_{1}+b_{1},\ldots,x_{n}])$ $\displaystyle\leq$
$\displaystyle\inf_{[x,~{}y]=[a,\ldots,x_{n}]+[b,\ldots,x_{n}]}\\{\lambda_{A}([a,\ldots,x_{n}])\vee\lambda_{B}([b,\ldots,x_{n}])\\}$
$\displaystyle\leq$
$\displaystyle\lambda_{A}([a_{1},\ldots,x_{n}])\vee\lambda_{B}([b_{1},\ldots,x_{n}])$
$\displaystyle\leq$
$\displaystyle(\lambda_{A}(a_{1})\wedge\cdots\wedge\lambda_{A}(x_{n}))\vee(\lambda_{B}(b_{1})\wedge\cdots\wedge\lambda_{B}(x_{n}))$
$\displaystyle<$ $\displaystyle s$ $\displaystyle<$
$\displaystyle\lambda_{A+B}([x_{1},\ldots,x_{n}]).$

Contradiction. Thus $A+B$ is a complex intuitionistic fuzzy $n$-Lie ideal of
$L$. $\Box$

In particular, if $A$ and $B$ are fuzzy Lie ideals of a Lie algebra $L$, then
we obtain a special case of Shaqaqha’s result [32, Theorem 3.5].

##  5\. Direct Product of Intuitionistic fuzzy Lie $n$-subalgebras

Let $A=(\mu_{A},\lambda_{A})$ and $B=(\mu_{B},\lambda_{B})$ be an
intuitionistic fuzzy subsets of $L_{1}$ and $L_{2}$, respectively. Then the
generalized cartesian product $A\times B$ is defined to be $A\times
B=(\mu_{A}\times\mu_{B},\lambda_{A},\lambda_{B})$ where

$\mu_{A}\times\mu_{B}:L\times
L\rightarrow[0,1];~{}(x,y)\mapsto\mu_{A}(x)\wedge\mu_{B}(y),$

and

$\lambda_{A}\times\lambda_{B}:L\times
L\rightarrow[0,1];~{}(x,y)\mapsto\lambda_{A}(x)\vee\lambda_{B}(y).$

We have the following result.

######  Theorem 5.1.

Let $A=(\mu_{A},\lambda_{A})$ and $B=(\mu_{B},\lambda_{B})$ be an
intuitionistic fuzzy $n$-Lie subalgebras of $L_{1}$ and $L_{2}$, respectively.
Then $A\times B$ is an intuitionistic fuzzy $n$-Lie subalgebra of $L_{1}\times
L_{2}$.

Proof. Note that $A\times B$ is an intuitionistic fuzzy subset of $L_{1}\times
L_{2}$. Indeed if $x\in L_{1}$ and $y\in L_{2}$ such that
$\mu_{A}(x)\leq\mu_{B}(y)$ and $\lambda_{B}(y)\geq\lambda_{B}(x)$, then

$(\mu_{A}\times\mu_{B})(x,y)+(\lambda_{A}\times\lambda_{B})(x,y)=\mu_{A}(x)+\lambda_{B}(y)\leq\mu_{B}(y)+\lambda_{B}(y)\leq
1.$

Similarly, if $\mu_{B}(y)\leq\mu_{A}(x)$ and
$\lambda_{A}(x)\geq\lambda_{B}(y)$. Let $(x_{1},y_{1}),(x_{2},y_{2})\in
L_{1}\times L_{2}$. Then the proofs for
$(\mu_{A}\times\mu_{B})((x_{1},y_{1})+(x_{2},y_{2}))\geq(\mu_{A}\times\mu_{B})((x_{1},y_{1}))\wedge(\mu_{A}\times\mu_{B})((x_{2},y_{2}))$,
$(\lambda_{A}\times\lambda_{B})((x_{1},y_{1})+(x_{2},y_{2}))\leq\lambda_{A}\times\lambda_{B}((x_{1},y_{1}))\vee\lambda_{A}\times\lambda_{B}((x_{2},y_{2}))$,
$(\mu_{A}\times\mu_{B})(c(x_{1},y_{1}))\geq(\mu_{A}\times\mu_{B})((x_{1},y_{1}))$,
and
$(\lambda_{A}\times\lambda_{B})(c(x_{1},y_{1}))\leq(\lambda_{A}\times\lambda_{B})((x_{1},y_{1}))$
are similar to the proof of [11, Lemma 2.1].
For $(x_{1},y_{1}),(x_{2},y_{2}),\ldots,(x_{n},y_{n})\in L\times L$, we have

$\displaystyle\mu_{A}([(x_{1},y_{1}),(x_{2},y_{2}),\ldots,(x_{n},y_{n})])$
$\displaystyle=$
$\displaystyle(\mu_{A}\times\mu_{B})([x_{1},x_{2},\ldots,x_{n}],[y_{1},y_{2},\ldots,y_{n}])$
$\displaystyle=$
$\displaystyle\mu_{A}([x_{1},x_{2},\ldots,x_{n}])\wedge\mu_{B}([y_{1},y_{2},\ldots,y_{n}])$
$\displaystyle\geq$
$\displaystyle(\mu_{A}(x_{1})\wedge\cdots\wedge\mu_{A}(x_{n}))\wedge(\mu_{B}(y_{1})\wedge\cdots\wedge\mu_{B}(y_{n}))$
$\displaystyle=$
$\displaystyle(\mu_{A}\times\mu_{B})((x_{1},y_{1}))\wedge\cdots\wedge(\mu_{A}\times\mu_{B})((x_{n},y_{2})),$

and also

$\displaystyle\lambda_{A}([(x_{1},y_{1}),(x_{2},y_{2}),\ldots,(x_{n},y_{n})])$
$\displaystyle=$
$\displaystyle(\lambda_{A}\times\lambda_{B})([x_{1},x_{2},\ldots,x_{n}],[y_{1},y_{2},\ldots,y_{n}])$
$\displaystyle=$
$\displaystyle\lambda_{A}([x_{1},x_{2},\ldots,x_{n}])\vee\lambda_{B}([y_{1},y_{2},\ldots,y_{n}])$
$\displaystyle\leq$
$\displaystyle(\lambda_{A}(x_{1})\vee\cdots\vee\lambda_{A}(x_{n}))\vee(\lambda_{B}(y_{1})\vee\cdots\vee\lambda_{B}(y_{n}))$
$\displaystyle=$
$\displaystyle(\lambda_{A}\times\lambda_{B})((x_{1},y_{1}))\vee\cdots\vee(\lambda_{A}\times\lambda_{B})((x_{n},y_{2})).$

$\Box$

However the direct product of two intuitionistic fuzzy Lie ideals of $n$-Lie
algebras $L_{1}$ and $L_{2}$ is not nesaccary to be an intuitionistic fuzzy
Lie ideal of the $n$-Lie algebra $L_{1}\times L_{2}$.

##  6\. On Lie Algebra Homomorphism of Intuitionistic Fuzzy $n$-Lie Algebras

Let $L_{1}$ and $L_{2}$ be $n$-Lie algebras, $A=(\mu_{A},\lambda_{A})$ be an
intuitionistic subset of $L_{1}$, and $f:L_{1}\rightarrow L_{2}$ be a
function. Then the intuitionistic fuzzy subset $f(A)$ of $f(L_{1})$ defined by
$f(A)=(\mu_{f(A)},\lambda_{f(A)})$ where $\mu_{f(A)}(y)=\sup_{x\in
f^{-1}(y)}\\{\mu_{A}(x)\\}$ and $\lambda_{f(A)}(y)=\inf_{x\in
f^{-1}(y)}\\{\lambda_{A}(x)\\}$ for $y\in f(L_{1})$ is called the image of $A$
under $f$. Similarly, if $B=(\mu_{B},\lambda_{B})$ is an intuitionistic fuzzy
subset of $L_{2}$, then the intuitionistic fuzzy set on $L_{1}$ is
$f^{-1}(B)=(\mu_{f^{-1}(B)},\lambda_{f^{-1}(B)})$ where

$\mu_{f^{-1}(B)}(x)=\mu_{B}(f(x))~{}\mathrm{and}~{}\lambda_{f^{-1}(B)}(x)=\lambda_{B}(f(x))$

(see for example [32]). The proofs of the following two results are omitted
because they are routine and parallel to the corresponding results on
intuitionistic fuzzy Lie algebras ([3, 33]).

######  Theorem 6.1.

Let $\varphi:L_{1}\rightarrow L_{2}$ be a $n$-Lie algebra homomorphism. If
$B=(\mu_{B},~{}\lambda_{B})$ is an intuitionistic fuzzy $n$-Lie subalgebra
(resp. ideal) on $L_{2}$, then the intuitionistic fuzzy set $\varphi^{-1}(B)$
is an intuitionistic fuzzy $n$-Lie subalgebra (rep. ideal) of $L_{1}$.

######  Theorem 6.2.

Let $\varphi:L_{1}\rightarrow L_{2}$ be a $n$-Lie algebra homomorphism. If
$A=(\mu_{A},~{}\lambda_{A})$ is an intuitionistic fuzzy $n$-Lie subalgebra
(resp. ideal) on $L_{1}$, then the intuitionistic fuzzy set $\varphi(A)$ is an
intuitionistic fuzzy $n$-Lie subalgebra (resp. ideal) of
$\mathrm{im}(\varphi)$.

##  7\. Intuitionistic Fuzzy Quotient $n$-Lie Algebras

Let $L$ be an $n$-Lie algebra and $I$ be an ideal of $L$. Then the factor
space $L/I=\\{x+I~{}:~{}x\in I\\}$ acquires an $n$-Lie algebra structure
(called a quotient $n$-Lie algebra) by setting

$[x_{1}+I,x_{2}+I,\ldots,x_{n}+I]=[x_{1},~{}x_{2},\ldots,x_{n}]+I$

for $x_{1},x_{2},\ldots,x_{n}\in L$. In this paper we define and study the
intuitionistic Fuzzy quotient $n$-Lie algebra by an intuitionistic fuzzy
$n$-Lie ideal.

######  Definition 7.1.

Let $A=(\mu_{A},\lambda_{A})$ be an intuitionistic fuzzy $n$-Lie ideal of an
$n$-Lie algebra $L$. Then for $x\in L$, the intuitionistic fuzzy subset

$x+A=(x+\mu_{A},x+\lambda_{A})$

where

$x+\mu_{A}:L\rightarrow[0,~{}1];~{}y\mapsto\mu_{A}(y-x)$

and

$x+\lambda_{A}:L\rightarrow[0,~{}1];~{}y\mapsto\lambda_{A}(y-x)$

is called a coset (determined by $x,\mu_{A}$, and $\lambda_{A}$) of the
intuitionistic fuzzy $n$-Lie ideal $A$.

The case where $n=2$ was introduced and studied by Chen [12] in 2010. The set
of all cosets of an intuitionistic fuzzy $n$-Lie ideal will be denoted by
$L/A$. The following lemma proves that a coset may have many different labels.

######  Lemma 1.

Let $A=(\mu_{A},~{}\lambda_{A})$ be an intuitionistic fuzzy $n$-Lie ideal of
an $n$-Lie algebra $L$, and let $x,y\in L$. The following statements are
equivalent:

  * (i)

$x+A=y+A$,

  * (ii)

$\mu_{A}(x-y)=\mu_{A}(0)$ and $\lambda_{A}(x-y)=\lambda_{A}(0)$.

Proof. If $x,~{}y\in L$ with $x+A=y+A$, then $x+\mu_{A}=y+\mu_{A}$ and
$x+\lambda_{A}=y+\lambda_{A}$. Consider $x+\mu_{A}=y+\mu_{A}$, then evaluating
both sides for $x$ implies that $\mu_{A}(0)=\mu_{A}(x-y)$. Similarly
$\lambda_{A}(x-y)=\lambda_{A}(0)$. Conversely, let $\mu_{A}(x-y)=\mu_{A}(0)$
and $\lambda_{A}(x-y)=\lambda_{A}(0)$. Then for any $z\in L$, we have
$(y+\mu_{A})(z)=\mu_{A}(z-y)\geq\mu_{A}(z-x)\wedge\mu_{A}(x-y)=\mu_{A}(z-x)\wedge\mu_{A}(0)=\mu_{A}(z-x)=(x+\mu_{A})(z)$.
Thus $y+\mu_{A}\geq x+\mu_{A}$. Also
$\mu_{A}(y-x)=\mu_{A}(-(x-y))=\mu_{A}(x-y)=\mu_{A}(0)$. So we can prove that
$x+\mu_{A}\geq y+\mu_{A}$ in the same way as above. Hence
$x+\mu_{A}=y+\mu_{A}$. By almost the same aegument we can prove that
$x+\lambda_{A}=y+\lambda_{A}$. $\Box$

######  Theorem 7.1.

Let $A=(\mu_{A},~{}\lambda_{A})$ be an intuitionistic fuzzy $n$-Lie ideal of
$L$ and $L/A$ be the set of all cosets of $L$ on $A$. Then the set $L/A$ is an
$n$-Lie algebra under the following operations:

  * (i)

$(x+A)+(y+A)=(x+y)+A$ for all $x,y\in L$,

  * (ii)

$\alpha(x+A)=\alpha x+A$ for all $x\in L$ and $\alpha\in F$,

  * (iii)

$[(x_{1}+A),(x_{2}+A),\ldots,(x_{n}+A)]=[x_{1},x_{2},\ldots,x_{n}]+A$ for all
$x_{1},x_{2},\ldots,x_{n}\in L$.

Proof. First, we show that the operations are well defined. Let $x,~{}y,~{}u$
and $v$ be elements in $L$ such that $x+A=y+A$ and $u+A=v+A$. Consequently
$\mu_{A}(x+u-(y+v))=\mu_{A}((x-y)+(u-v))\geq\mu_{A}(x-y)\wedge\mu_{A}(u-v)=\mu_{A}(0)$
(because $\mu_{A}(x-y)=\mu_{A}(0)~{}\mathrm{and}~{}\mu_{A}(u-v)=\mu_{A}(0)$).
As $\mu_{A}(0)\geq\mu_{A}(x+u-(y+v))$, we have $\mu_{A}(x+u-y-v)=\mu_{A}(0)$.
Almost the same argument one can obtain that
$\lambda_{A}(x+u-y-v)=\lambda_{A}(0)$ Therefore $(x+u)+A=(y+v)+A$. Also,
$\mu_{A}(\alpha x-\alpha y)\geq\mu_{A}(x-y)=\mu_{A}(0)$ and
$\lambda_{A}(\alpha x-\alpha y)\leq\lambda_{A}(x-y)=\lambda_{A}(0)$. Hence
$\alpha(x+A)=\alpha(y+A)$. If $x_{1},x_{2},\ldots,x_{n}\in L$ such that
$x_{i}+A=y_{i}+A$ ($i=1,\ldots,n$), then

$\displaystyle\mu_{A}([x_{1},x_{2},\ldots,x_{n}]-[y_{1},y_{2},\ldots,y_{n}])$
$\displaystyle=$ $\displaystyle\mu_{A}([x_{1}-y_{1},x_{2},\ldots,x_{n}]$
$\displaystyle+[y_{1},x_{2},\ldots,x_{n}]-[y_{1},y_{2},\ldots,y_{n}])$
$\displaystyle\geq$ $\displaystyle\mu_{A}([x_{1}-y_{1},x_{2},\ldots,x_{n}])$
$\displaystyle\wedge\mu_{A}([y_{1},x_{2},\ldots,x_{n}]-[y_{1},y_{2},\ldots,y_{n}])$
$\displaystyle\geq$
$\displaystyle\mu_{A}(0)\wedge\mu_{A}([y_{1},x_{2},\ldots,x_{n}]-[y_{1},y_{2},\ldots,y_{n}])$

Now,

$\displaystyle\mu_{A}([y_{1},x_{2},\ldots,x_{n}]-[y_{1},y_{2},\ldots,y_{n}])$
$\displaystyle=$ $\displaystyle\mu_{A}([y_{1},x_{2}-y_{2},x_{3},\ldots,x_{n}]$
$\displaystyle+[y_{1},y_{2},x_{3},\ldots,x_{2}]-[y_{1},y_{2},y_{3},\ldots,y_{n}])$
$\displaystyle\geq$
$\displaystyle\mu_{A}(y_{1})_{\vee}\mu_{A}(x_{2}-y_{2})\vee\mu_{A}(x_{3})\vee\cdots\mu_{A}(x_{n})$
$\displaystyle\wedge\mu_{A}([y_{1},y_{2},x_{3},\ldots,x_{2}]-[y_{1},y_{2},y_{3},\ldots,y_{n}])$
$\displaystyle\geq$
$\displaystyle\mu_{A}(0)\wedge\mu_{A}([y_{1},y_{2},x_{3},\ldots,x_{2}]-[y_{1},y_{2},y_{3},\ldots,y_{n}]).$

So

$\mu_{A}([x_{1},x_{2},\ldots,x_{n}]-[y_{1},y_{2},\ldots,y_{n}])\geq\mu_{A}(0)\wedge\mu_{A}(0)\wedge\mu_{A}([y_{1},y_{2},x_{3},\ldots,x_{2}]-[y_{1},y_{2},y_{3},\ldots,y_{n}]).$

Continuing in the same way, we obtain

$\mu_{A}([x_{1},x_{2},\ldots,x_{n}]-[y_{1},y_{2},\ldots,y_{n}])\geq\mu_{A}(0)\wedge\mu_{A}(0)\wedge\cdots\wedge\mu_{A}(0)~{}(n~{}\mathrm{times}).$

Hence
$\mu_{A}([x_{1},x_{2},\ldots,x_{n}]-[y_{1},y_{2},\ldots,y_{n}])=\mu_{A}(0)$.
Using the same method one can prove that
$\lambda_{A}([x_{1},x_{2},\ldots,x_{n}]-[y_{1},y_{2},\ldots,y_{n}])=\lambda_{A}(0)$.
Therefore $[x_{1},x_{2},\ldots,x_{n}]+A=[y_{1},y_{2},\ldots,y_{n}]+A$. Now it
is straightforward to see that the product on $L/A$ is an $n$-linear map
satisfying the Filippov identity. $\Box$

The $n$-Lie algebra constructed in Theorem 7.1 is called intuitionistic fuzzy
quotient $n$-Lie algebra of $L$ by $A$.

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