# Is there a normed space $W$ such that $W=X\oplus Y$, with $X$ not closed and such that $W$ is homeomorphic to $X\times Y$?

If $$W$$ is a normed space, $$X,Y$$ are vector subspaces with the subspace topology, $$X$$ is not closed and $$X\oplus Y=W$$ in the algebraic sense, then the map $$X\times Y\to W$$, $$(x,y)\mapsto x+y$$ is bijective and continuous, but not a homeomorphism.

If $$W$$ is Banach, then the stronger conclusion holds that $$X\times Y$$ is not homeomorphic to $$W$$, as a consequence of the fact that Banach spaces cannot be homeomorphic to non-Banach normed spaces (e.g. here).

I believe that it should be possible to find a normed space $$W$$ with a non-closed subspace $$X$$ and another subspace $$Y$$ such that $$X\oplus Y=W$$ and $$X\times Y$$ is homeomorphic to $$W$$. However, I cannot find such an example.

Let $$W \subseteq \ell^2(\Bbb{R})$$ be the inner product space of real sequences that are eventually $$0$$, indexed by $$\Bbb{N}$$ where $$0 \in \Bbb{N}$$, equipped with the $$2$$-norm. Let $$e^n$$ be the $$n$$th standard basis vector of $$\ell^1(\Bbb{R})$$, for $$n \ge 0$$, i.e. $$e^n_i = \delta_{ni}$$ for all $$i \in \Bbb{N}$$.

Define $$X = \operatorname{span}\{e^0 + e^n / n : n \ge 1\}$$, and $$Y = \operatorname{span} \{e^0\}$$. Clearly, $$X + Y = W$$.

If $$X \cap Y$$ is non-trivial, being a subspace of one-dimensional $$Y$$, then $$X \cap Y = Y$$, hence $$e^0 \in X \cap Y \subseteq X$$. But, it's easy to see that $$e^0 \notin X$$, hence $$X \cap Y = \{0\}$$. That is, $$X \oplus Y = W$$.

Further, $$e^0 + e^n / n \to e^0$$ as $$n \to \infty$$, thus $$e^0 \in \overline{X} \setminus X$$, proving $$X$$ is not closed in $$W$$.

Now, we may equip $$X \times Y$$ with an inner product $$\langle (x_1, y_1), (x_2, y_2) \rangle = \langle x_1, x_2 \rangle + \langle y_1, y_2 \rangle$$. This inner product produces a norm compatible with the product topology on $$X \times Y$$. It has a countable basis $$(0, e^0), (e^0 + e^1, 0), (e^0 + e^2/2, 0), \ldots, (e^0+e^n/n,0), \ldots$$ Applying Gram-Schmidt yields a countable orthonormal basis for $$X \times Y$$. The same is true of $$W$$. Form a linear map that maps one orthonormal basis to the other. Such a map is a surjective linear isometry.

That is, it's possible for $$X \times Y$$ and $$W$$ to be isometrically isomorphic, not just homeomorphic.

• Look at me sitting for an hour on a similar idea in $\ell^\infty$, when I should have just realized that pre-Hilbert spaces of dimension $\aleph_0$ are isometrically isomorphic... Oct 27, 2022 at 4:24
• @SassatelliGiulio I initially did the same with $\ell^1$. Oct 27, 2022 at 4:27