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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 $\mathrm{cov}_H(X)=\omega_1$.

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    $\begingroup$ better $\mathrm{Hcov}(X)$? $\mathrm{cov}_H$ looks like something depending on a variable $H$. $\endgroup$
    – YCor
    Commented Apr 29, 2019 at 20:46
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    $\begingroup$ By the way you could mention the trivial bounds $\aleph_0<\mathrm{Hcov}(X)\le\mathfrak{c}$ for every $X\neq 0$. $\endgroup$
    – YCor
    Commented Apr 29, 2019 at 20:49
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    $\begingroup$ 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? $\endgroup$
    – fedja
    Commented Apr 29, 2019 at 23:13
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    $\begingroup$ 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 $[0,1/2]$. $\endgroup$
    – Nate Eldredge
    Commented Apr 30, 2019 at 3:17
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    $\begingroup$ (Note that the coefficients $f(x_j)$ of the power series are bounded by $\|f\|$.) $\endgroup$
    – Nate Eldredge
    Commented Apr 30, 2019 at 3:18

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