Condition (1): Any of its closed infinite-dimensional subspaces can contain X.

Condition (2): For a vector topology T, it is the case if and only if it can be extended on all of X that T, the only weaker Hausdorff topology on a subspace of X is a strictly weaker T.

Question 1: Is it the case that (1) is a consequence of (2)?

Answer: No in a word, and in more words, condition (1) Is not a consequence of condition (2).

Proof:

Always define a connecting plane or union of spaces, this is what induction in geometry is all about or what PT is about.

On the contrary, about direct embeddings/inclusions onto infinite dimensional closed subspaces.

A counterexample exists: let us take p dimension in space of sequences for p≠2 with ℓp⊕ℓ2 representation.

Condition (1) is false, hence, it is not a topology in exploration that can be embedded into an infinite dimension.

However, (2) satisfy as every weaker topology practically extend to any space.

Condition (2) extend to radial B-space and it is expected to be implemented because such space radially extend to any dimension.

Question 2: Let X be a reflexive space: does it at every point must be equipped with a separable Hilbert space as its centre?

Answer: Yes, a separable Hilbert space is equipped as centre, but on assumptions that X is reflexive and X satisfies both conditions (1) and (2).

Proof: The above proof clearly indicates that (i) there must be certain separation, meaning X is enough to consider super reflexive.

Indications also suggest that such properties violate containment, similar to (ii) and X must contain Radon-Nikodym property.

Thanks for the input.

If separable Hilbert space are contains dense countable closure sets then spaces are isomorphic to separable Hilbert space