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Be $X$ a Banach space with countable basis. Suppose $(x_n)_{n \in \mathbb{N}} $ is a sequence in $X$ (allowing repetitions), such that

$$(\forall x \in X)(\exists M_x \subseteq \mathbb{N})\left(|M_x| \in \mathbb{N} \,\,\wedge \left(\forall m \in M_x)(\exists \lambda_m \in \mathbb{F} \right) \left(x=\sum_{m \in M_x} \lambda_m x_m \right)\right) $$ For each $k \in \mathbb{N}$, define $L_k :=$ sp$\{x_1, x_2, \ldots, x_k\}$.

It is supposed to be the case that there must exists at least one $k_0 \in \mathbb{N}$ such that the closure of $L_{k_0}$ has non-empty interior: $\left(\bar{L_{k_0}}\right)^{\circ} \neq \emptyset$.

I find the question oddly phrased, since each $L_k$ is a finite-dimensional subspace, and is therefore closed, so why speak of its closure? Is there something about the interior of the closure of a set that's useful here?

Having found such a $k_0$ then, how would this imply that $X = L_{k_0}$?

Suggestion

Let $x \in X \setminus \{0\}$. Note that since $L_{k_0}$ contains an open ball, say, $B_{\varepsilon}(a)$, $z := \frac{\varepsilon}{2}\frac{x}{\lVert x \rVert} + a \in B_{\varepsilon}(a) \subseteq L_{k_0}$, so $z \in L_{k_0}$. But then $x = \frac{2\lVert x\rVert}{\varepsilon}(z-a) \in L_{k_0} \oplus \text{sp}(a)$. This basically brings us back to what we want to prove, since we now need to have that $a \in L_{k_0}$... Is there a way to show that not only $L_{k_0}$ contains some open ball, but an open ball around 0?

Oops, of course $a \in L_{k_0}$, since $B_{\varepsilon}(a) \subseteq L_{k_0}$!

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    $\begingroup$ The closure is indeed redundant, but only because the $L_k$ are finite-dimensional. You do need closed sets for the Baire Category Theorem. $\endgroup$ – Mindlack Jan 21 at 17:09
  • $\begingroup$ Aaah yes! I would never have thought of that, thanks! $\endgroup$ – Jos van Nieuwman Jan 21 at 18:08

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