Proof that multilinear rank of tensor less of equal than border rank Let's consider the tensor $t \in \mathbb{R}^{d_1 \times d_2 \times d_3}$. It is well-know statement, that multilinear rank of tensor less or equal than border rank, which means that if we consider multilinear rank $rk_{mult} = (rk_1, rk_2, rk_3)$ than $rk_i(t) \le brk(t)$.
I have the proof of this simple fact, but I can not understand some notions in it. The proof is:
Let $t_n \to t$, $rk(t_n) = brk(t)$. Let's write the canonical decomposition for every $t_n: t_n = \sum a_i^n \otimes b_i^n \otimes c_i^n$. Then $t_1^n = \sum a_i^n \otimes (b_i^n \otimes c_i^n)$ converges to $t_1$.  Then because $t_1^n$ is a matrix we have that $rk(t_1) \le rk(t_1^n) \le rk(t_n) = brk(t)$.
I can not understand one thing. Here is written that $t_1^n = \sum a_i^n \otimes (b_i^n \otimes c_i^n)$. But $t_1^n$ is the first flattering, it is a matrix it equals $t_1(x) = \sum (a_i x^T)b_i \otimes c_i$, how it can be the Kronecker product of 3 vectors? Is there any typo or I misunderstanding something? If here is a typo, how I can fix it?
Thanks for the help!

The confusion cames from the fact that the symbol "$\otimes $" can be interpreted as the tensor product, the outer product, and the Kronecker product. So, the  expression $t=\sum a_i\otimes b_i\otimes c_i$ can be interpreted as a tensor (abstract notion), three-way array and a vector, respectively.
In you case, $t^n$ is interpreted as a matrix (flattening or unfolding): 
$t^n=\sum a_i^n\otimes (b_i^n\otimes c_i^n)$ should be read as
$t^n=\sum a_i^n (b_i^n\otimes c_i^n)^T$, i.e. $t_n$ is a sum of rank-$1$ $d_1\times d_2d_3$ matrices.
