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I am studying the baire category theorem and trying to find a counterexample. This theorem says that a non-empty complete metric space can not be the countable union of nowhere-dense closed subsets

In particular, i'm trying to find a normed space that is the union of countably many closed nowhere-dense subsets. Obviously this means this set cannot be complete.

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    $\begingroup$ How about $\mathbb Q = \bigcup_{q \in \mathbb Q} \{q\}$? $\endgroup$
    – Stefan Mesken
    Feb 15, 2016 at 18:26
  • $\begingroup$ @Stefan what norm would we be using there? $\endgroup$
    – Sertii
    Feb 15, 2016 at 18:29
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    $\begingroup$ @user156715: Ordinary absolute value. But note that the field in question is $\Bbb Q$, not $\Bbb R$ or $\Bbb C$. $\endgroup$
    – Brian M. Scott
    Feb 15, 2016 at 18:31
  • $\begingroup$ Yes, I considered $\mathbb Q$ as a normed $\mathbb Q$-vector space - if that's what you were asking. $\endgroup$
    – Stefan Mesken
    Feb 15, 2016 at 18:31
  • $\begingroup$ @BrianM.Scott do you know of an example where the field is R or C? $\endgroup$
    – Sertii
    Feb 15, 2016 at 18:32

2 Answers 2

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For $n\in\Bbb Z^+$ let

$$A_n=\{f\in C[0,1]:|f(x)|\le n\text{ for all }x\in[0,1]\}\;,$$

and give $C[0,1]$ the $L^1$ norm.

Every open ball contains functions with very narrow, very tall spikes.

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Any normed vector space having countable dimension (over $\mathbb{C}$, for instance) will do. The finite-dimensional subspaces are closed and nowhere dense.

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  • $\begingroup$ I should clarify that by countable I mean countably infinite. $\endgroup$
    – carmichael561
    Feb 15, 2016 at 18:51

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