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Defenition(to make sure we are talking about the same thing) :

A base $B=\{b_1,b_2,\dots \}$ to a normic space $X$ is a group of elements from $X$, that satisfies the following condition : $\forall x\in X$, $x$ can be written as a finite linear combination of elements from $B$.

I need to prove that a complete normic space has no countable base. What I tried : Let's assume we have a countable base : $B=\{b_1,b_2,\dots \}$ We would like to show that we can write $X$ as a countable union of nowhere dense set's, and then we will get a contridiction for Baire categoty theorome. I didn't find a way how to make these nowhere dense groups, I thought to let $A_i$ be the group of elements from $X$ which can be written as a linear combintaion of elements from $B$ with exactly $i$ elements, the union is indeed all the space, but I failed to show that it is nowhere dense.

Thanks

choco_addicted
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Amit
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2 Answers2

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You're almost right about the sets, even more easily:

Define $B_n$ to be the linear span of $\{b_1,\ldots,b_n\}$ for every $n$.

Then every $B_n$ is closed and has empty interior.

Closedness is clear, see this answer or this answer.

Nowhere denseness (i.e. empty interior in this case) follows from this answer.

Searching does help, you see...

Community
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Henno Brandsma
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Hint: consider these subspaces $F_n = Span(b_1,...,b_n)$

What can you say about them ?

Dark
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