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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.

A tensor slice is obtained by contracting one factor of a tensor with a linear functional. For $T\in A\otimes B\otimes C$ and $\alpha \in A^{\ast }$, the contraction $T(\alpha )\in B\otimes C$ is a two-factor tensor and can be displayed as a matrix after choosing bases. Coordinate slices are the contractions with members of a chosen dual basis; their individual matrices depend on that choice.^[ref]

For example, writeThe two first-factor slices are $I_2$ and $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Contracting with $a{e}^0+b{e}^1$ gives the matrix pencil $a{I}_2+bN$, whose determinant is $a^2$. A change of basis in $A$ replaces the generators of this pencil, but does not change its span.

The contraction map $A^{\ast }\to B\otimes C$ is intrinsic. If $T$ is a sum of $r$ pure tensors, its image is contained in the span of $r$ two-factor pure tensors. Its matrix rank therefore gives a lower bound for tensor rank. Studying slices converts a tensor-decomposition problem into restrictions on a space of matrices, although a single slice rank does not generally determine the tensor rank.