 Border rank

Over $\mathbb{C}$, the border rank of a tensor $T$ is the least $r$ for which $T$ lies in the closure of tensors of tensor rank at most $r$. The usual Euclidean topology and Zariski topology give the same closure here. Hence $\underline{R}(T)\le R(T)$, but equality need not hold.^[ref]

Let $e_0,e_1$ be a basis of ${\mathbb{C}}^2$ and putFor nonzero $\epsilon$,Each approximant is a sum of two pure tensors, so the border rank is at most two. It is not one, since a matrix obtained by separating one factor from the other two has rank two.

Its exact tensor rank is three. The two tensor slices in the first factor span $\left\{\left(\begin{array}{cc}b& a\\ a& 0\end{array}\right):a,b\in \mathbb{C}\right\}$. Its determinant is $-a^2$, so this space has only one rank-one direction. A rank-two tensor with two independent slices would require their span to be generated by two independent rank-one matrices, a contradiction. The limit therefore permits a decomposition that is impossible exactly.