# What is the smallest number of hyperplanes covering $\ell_2$?

For a Banach space $$X\ne \{0\}$$, let $$\mathrm{cov}_H(X)$$ be the smallest number of hyperplanes covering $$X$$.

By a hyperplane in a Banach space I understand any closed affine subspace of codimension 1.

By induction it can be shown that every Banach space $$X$$ of finite positive dimension has $$\mathrm{cov}_H(X)=\mathfrak c$$.

Problem 1. Is it true that $$\mathrm{cov}_H(X)=\mathfrak c$$ for any infinite-dimensional separable Banach space $$X$$?

In particular,

Problem 2. Is $$\mathrm{cov}_H(\ell_2)=\mathfrak c$$?

Remark 1. For any non-separable Hilbert space $$X$$ we have $$\mathrm{cov}_H(X)=\omega_1$$.

• better $\mathrm{Hcov}(X)$? $\mathrm{cov}_H$ looks like something depending on a variable $H$. – YCor Apr 29 at 20:46
• By the way you could mention the trivial bounds $\aleph_0<\mathrm{Hcov}(X)\le\mathfrak{c}$ for every $X\neq 0$. – YCor Apr 29 at 20:49
• Erm. Take some spanning sequence of unit vectors $x_j$ and consider the curve $C(r)=\sum_jr^j x_j$, $0<r<1/2$. It intersects every closed hyperplane by at most finitely many points and countable times less than continuum is still less than continuum (this requires a very weak form of AC, just the possibility to represent any infinite set as a union of disjoint countable sets), so it seems like the answer is "yes" to both. Am I missing anything? – fedja Apr 29 at 23:13
• It took me a few minutes to figure out why $C(r)$ only intersects every closed hyperplane at finitely many points, so to save time for anyone as slow as me: a hyperplane consists of those points $x$ satisfying $f(x) = c$ for some $f \in X^*$, $c \in \mathbb{R}$. Now $f(C(r)) = \sum_j r^j f(x_j)$ which is a power series in $r$ with radius of convergence at least $1$. Since $x_j$ has dense span, it is not the zero series. In particular, it is a nonzero analytic function of $r$, and so by the isolated zeros theorem, the equation $f(C(r))=c$ has only finitely many solutions in $[0,1/2]$. – Nate Eldredge Apr 30 at 3:17
• (Note that the coefficients $f(x_j)$ of the power series are bounded by $\|f\|$.) – Nate Eldredge Apr 30 at 3:18