Is the Tensor Product Space Isomorphic to One of its Factors if They Are Infinite Dimensional? Let $X_1$ and $X_2$ be two infinite dimensional vector spacees and let $X_1 \otimes X_2$ be their tensor product space. My question is, is it possible to construct an isomorphism between $X_1 \otimes X_2$ and one of its facots $X_1$ or $X_2$ such that we have
$$
X_1 \cong X_1 \otimes X_2 \hspace{1cm} \text{or} \hspace{1cm} X_2 \cong X_1 \otimes X_2
$$
Don't mention something like $\mathbb{R} \otimes \mathbb{R}  \cong \mathbb{R} $ which is one dimensional and very trivial case that has nothing to do with the question here. I think if $X_1 = X_2 = c_{00}$, i.e. the  set of all sequences of only finitely many elements different from zero, then is possible to have
$$
c_{00} \cong  c_{00}\otimes  c_{00} \tag{*}
$$
by some mapping between basis in $c_{00}$ and the product basis in $ c_{00}\otimes  c_{00}$. Then for every bilinear map $\phi$ on $ c_{00} \times c_{00}$ there is a linear map $\tilde\phi$ on $ c_{00}$. But that is true only if (*) is true, so is it in general the case for the tensor product of infinite dimensional spaces ? and so is it a direct utilization of the linearisation of the bilinear maps ?
Another question is, if I chose $\ell^2 \otimes \ell^p$ spaces would
$$
\ell^p \cong \ell^p \otimes \ell^p 
$$
be true ? I think not.
 A: $c_{00}$ has countable dimension, so using the fact that the dimension of a tensor product is the product of the dimensions (which continues to hold in infinite dimensions) and that $|\mathbb{N} \times \mathbb{N}| = |\mathbb{N}|$ we get an abstract isomorphism $c_{00} \cong c_{00} \otimes c_{00}$, unconditionally.
Assuming the axiom of choice, Hamel bases always exist (this is in fact equivalent to AC), which means infinite-dimensional vector spaces have dimensions. Then we can show that $\ell^p$ has dimension the cardinality of the continuum $|\mathbb{R}|$, which squares to itself, so again we get an abstract isomorphism $\ell^p \otimes \ell^p \cong \ell^p$.
The dimension can be computed as follows. $\ell^p$ itself has the same cardinality as $\mathbb{R}$ (exercise), so $|\mathbb{R}|$ is an upper bound on its dimension. To give a lower bound it suffices to exhibit an $\mathbb{R}$'s worth of linearly independent elements of $\ell^p$. Explicitly if we work in $\ell^p(\mathbb{N})$ we can take
$$v_r(n) = e^{-rn}, r \in \mathbb{R}_{+}.$$
It's a nice exercise to prove that these sequences are linearly independent.
There is also a general argument using the Baire category theorem to show that an infinite-dimensional Banach space can't be countable-dimensional, but it doesn't follow that the dimension is at least $|\mathbb{R}|$ without the continuum hypothesis.
