# Combination of simple tensors

I aksed this question on Math Stack Exchange 6 days ago, with no answer: https://math.stackexchange.com/q/4875445/1297919

Let $$X$$ and $$Y$$ two Banach spaces and let $$X\otimes Y$$ their tensor product. Let $$A(u)$$ be the collection of all finite sets of simple tensors $$\{x_1\otimes y_1,\dots ,x_n\otimes y_n\}$$ such that:

1. $$u=\sum_{i=1}^n x_i\otimes y_i,$$

2. There is no subset with at least two elements of $$\{x_1\otimes y_1,\dots ,x_n\otimes y_n\}$$ such that the sum of its elements is a simple tensor.

Is it true that for every $$u\in X\otimes Y$$ we have $$\max\{\text{card}\, s:s\in A(u)\}<\infty$$?

## 1 Answer

Let $$a_0,a_1,a_2\dots$$ be linearly independent elements of $$X$$ and $$b_0,b_1,b_2\dots$$ be linearly independent elements of $$Y$$. Let $$u=a_1\otimes b_1-a_0\otimes b_0$$, and let $$v_i=a_i\otimes (b_i-b_{i+1}),\quad w_i=(a_{i}-a_{i+1})\otimes b_{i+1},$$ $$x_i=a_i\otimes(b_i-b_0),\quad y_i=(a_i-a_0)\otimes b_0.$$ Then $$v_1+w_1+v_2+w_2+\dots+v_{n-1}+w_{n-1}+x_n+y_n= u$$ but no subset of $$\{v_1,w_1,v_2,w_2,\dots,v_{n-1},w_{n-1},x_n,y_n\}$$ of size at least two adds up to a simple tensor. So the answer to your question is no.

• Thanks! Do you think that it's possibile to have an example with this property where the sets $S_n$ of simple tensors that you use at stage $n$ have empty intersection, that is $\cap_n S_n=\emptyset$? Commented Mar 13 at 11:01
• Yes, a variation on this example should do that for you. Commented Mar 13 at 11:36