$\newcommand{\Vs}{V^{\!\star}}\newcommand{\I}{\mathcal{I}}\newcommand{\L}{\mathcal{L}}$The following is an excerpt from Guillemin and Haine's book on Differential Forms. Note that $\L^k(V)$ denotes the set of $k$-tensors over $V$.
Definition. A decomposable $k$-tensor $\ell_1\otimes\cdots\otimes\ell_k$, with $\ell_1,\dots,\ell_k\in\Vs$ is redundant if for some index $i$ we have $\ell_i=\ell_{i+1}$. Let $\I^k(V)\subset\L^k(V)$ be the linear span of the set of redundant $k$-tensors.
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Proposition. If $T\in \I^r(V)$ and $T'\in\L^s(V)$ then $T\otimes T'$ and $T'\otimes T$ are in $\I^{r+s}(V)$.
Proof. We can assume that $T$ and $T'$ are decomposable, i.e., $T=\ell_1\otimes\dotsm\otimes \ell_r$ and $T'=\ell_1'\otimes\dotsm\otimes\ell_s'$ and that $T$ is redundant: $\ell_i=\ell_{i+1}$. Then $$T\otimes T'=\ell_1\otimes\dotsm\otimes\ell_{i-1}\otimes\ell_i\otimes\ell_i\otimes\dotsm\otimes\ell_r\otimes\ell_1'\otimes\dotsm\otimes\ell_s'$$ is redundant and hence in $\I^{r+s}$. The argument for $T'\otimes T$ is similar. $\square$
The proof of the proposition is rather clear to me, except for the first part: why can we assume that $T'$ is decomposable? Of course we can assume $T$ to be decomposable since the notion of redundancy is one defined for decomposable tensors, but why can we do so for $T'$ since (unless I'm horribly mistaken) there are plenty of tensors which are not decomposable? What am I missing here?
Thanks in advance!
Elements of $\mathcal{L}^s(V)$ are sums of decomposable tensors, while elements of $\mathcal{I}^r(V)$ are sums of decomposable redundant tensors. Neither $T$ nor $T'$ are necessarily decomposable, but the proof for general tensors reduces to decomposable tensors by linearity. If $T = \sum_i t_i$ and $T' = \sum_j t_j'$ where $t_i$ and $t_j'$ are decomposable, then $T\otimes T' = \sum_{i,j} t_i\otimes t_j'$. If $t_i\otimes t_j'$ is redundant for every $i$ and $j$, then $T\otimes T' \in \mathcal{I}^{r+s}(V)$ by definition.