Closure of finite span has non-empty interior

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 $$B_{\varepsilon}(a) \subseteq L_{k_0}$$!

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