The largest topological copy of a Hilbert space contained in $\ell^1$

Let us consider $\ell^1$, the space of absolutely summable sequences in the space of complex numbers. Clearly every finite dimensional Hilbert space is topologically embedded into $\ell^1$.

Convention. For given Hilbert spaces $H$ and $K$, let us write $H\leq K$ if the Hilbertian dimension of $H$ is less than $K$.

Q. What is the Hilbertian dimension of the largest Hilbert space which can be topologically embedded into $\ell^1$?

• Your convention is not used in the question, so why include it? – LSpice May 9 '18 at 18:46
• Yes you are right. – Ali Bagheri May 9 '18 at 18:49
• relevant: mathoverflow.net/questions/79713/… – Pietro Majer May 9 '18 at 19:14
• ANSWER: $\ l^2\$ is TOPOLOGICALLY embeddable in $\ l^1$. – Wlod AA May 10 '18 at 4:10

No infinite dimensional reflexive space can be embedded into $\ell_1$, because every infinite dimensional closed subspace of $\ell_1$ has a non separable dual.
• @Pietro Majer What happens if we change the roles? I mean what is the Hilbertian dimension of the smallest Hilbert space contains $\ell^1$ up to topological vector spaces. – Ali Bagheri May 9 '18 at 18:55
• rmk: Of course if one only wants a homeo onto a subspace, not necessarily linear, then $\ell_2$ is already homeomorphic to $\ell_1$, via $\ell_2\ni x:=(x_j)_j\mapsto (|x_j|x_j)_j\in\ell_1$ – Pietro Majer May 9 '18 at 19:03
• @LSpice of course topological embedding (not necessarily linear) need not preserve completeness- e.g. R can be embedded in R as an open interval. But the question is only meaningful in the top. linear sense ($\ell_1$ and $\ell_2$ are already homeomorphic, as topological spaces) – Pietro Majer May 9 '18 at 19:37