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.


1 Answer 1


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.

  • $\begingroup$ 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... $\endgroup$ Oct 27, 2022 at 4:24
  • $\begingroup$ @SassatelliGiulio I initially did the same with $\ell^1$. $\endgroup$ Oct 27, 2022 at 4:27

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