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Let $P_n$ be the vector space of polynomials of degree $\leq n$, with $n$ an integer. Now consider the norm $||x||=\max_i|a_i|$. Show tha $P_n$ is not a Banach space.

I'm learning functional analysis (from Kesavan's book) and I'm struggling with this argument. I know, by the Baire's theorem, that is a set is of first category, that is, a countable union of nowhere dense sets, then it cannot be a Banach space. The thing is, how can I prove the space $P_n$ is nowhere dense? I have no experience working with nowhere dense subsets. Is there any equivalent result to conclude $P_n$ is a countable union of nowhere dense vector spaces?

Byag
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    Are you sure you want the degree to be at most a fixed $n$? That space looks remarkably similar to $(\mathbb{R}^{n+1}, | \cdot |_\infty)$, which is a Banach space – Jose Avilez Oct 17 '23 at 02:34
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    Not just similar but isometric to.... – Moishe Kohan Oct 17 '23 at 02:35
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    It is a Banach space. Fun exercise: Prove that every finite-dimensional normed space is a Banach space. https://math.stackexchange.com/questions/168275/proof-that-every-finite-dimensional-normed-vector-space-is-complete – Severin Schraven Oct 17 '23 at 02:37
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    I strongly suspect the exercise actually asks (or was intended to ask) about the vector space of all polynomials, not just those of degree $\le n$. – Robert Israel Oct 17 '23 at 03:22
  • Each $P_n$ is closed in $C[0,1]$ with empty interior. Every proprer subspace has empty interior. – Kavi Rama Murthy Oct 17 '23 at 04:42

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