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It is shown here that an infinite dimensional Banach space cannot have a countable basis. I understand the usage of Baire, but don't understand why the sets $M_n = $span{${e_1, ..., e_n}$} are closed, and have an empty interior. Can someone explain this?

paxtibimarce
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    Finite-dimensional normed spaces are isomorphic to $\mathbb{K}^d$ for some $d$ (the ground field you use) and those are Banach spaces. Hence, they are complete. Now pick a sequence $(x_m)_{m\geq 1}\subseteq M_n$ that converges in the ambient space. Then this sequence is Cauchy and by completeness of $M_n$ the limit is in $M_n$. Thus, $M_n$ is closed in the ambient Banach space. To see that they have empty interior it is useful to think of some lower dimensional subspace of $\mathbb{R}^3$. If you pick a ball, there is always a direction where you are out of the lower dimensional space. – Severin Schraven Jul 24 '21 at 09:31
  • @SeverinSchraven Thanks. I think I got it. If we assume by contradiction it contains some open ball, we can add to its center $e_{n+1}$ multiplied by a scalar smaller than the radius and reach contradiction to the span. – paxtibimarce Jul 24 '21 at 09:44
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    That is exactly the argument :) – Severin Schraven Jul 24 '21 at 13:31

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The map $f_i$ that sends $x$ to the coefficient of $e_i$ in the basis expansion wrt $\{e_1,e_2,\ldots\}$ is a linear and continuous map from $X$ to $\Bbb R$.

And $M_n = \bigcap_{i > n} f_i^{-1}[\{0\}]$ is thus closed as an intersection of closed sets (inverse images of closed sets under continuous maps etc.)

And if $B(x, r)$ is any open ball ($r>0$) around some $x \in M_n$ then $x' = x+\frac{r}{2}e_{n+1}$ is in $B(x,r) \cap M_n^\complement$ showing that no such ball is a subset of $M_n$ and $\operatorname{int}(M_n)=\emptyset$, as claimed.

Henno Brandsma
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