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As the title suggests, I would like to prove that the normed vector space $(\mathbb{R}^\infty , ||.||_{l^p})$ is not a Banach space, where $$\mathbb{R}^\infty :=\{ x:\mathbb{N} \rightarrow \mathbb{R} : \exists \ \bar{n} \in \mathbb{N} \ \ s.t. \ \ x(n)=0 \ \forall n>\bar{n} \}$$ and $$l^p:=\{x:\mathbb{N} \rightarrow \mathbb{R} : \sum_{n=1}^{\infty}|x(n)|^p< \infty \}.$$

As usual, I should start from a Cauchy sequence defined with respect to the distance induced by the considered norm.

Now I'm searching for a sequence of objects in $\mathbb{R}^\infty$ which converges to a sequence in $l^p \setminus \mathbb{R}^\infty$. Can someone give me some hint? I found the completeness of metric and normed spaces a very interesting topic, but I have the impression that the proof of the non-completeness, rather than the completeness, is always harder to achieve. Thank you all.

p.s.: Has the space $\mathbb{R}^\infty$, as defined previously, a particular name?

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It is the space of all realvalued sequences. – Stefan Nov 19 '12 at 16:55
$\mathbb R^\infty = c_c =$ "the space of compactly supported sequences = "the space of sequences that are eventually zero". – kahen Nov 19 '12 at 16:56
@Stefan: the space of all realvalued sequences is $\mathbb R^\omega$ or $\mathbb R^{\mathbb N}$. – kahen Nov 19 '12 at 16:58
up vote 4 down vote accepted

There are many different ways of proving that $\mathbb R^\infty$ is not a Banach space under any norm. The cleanest is probably to note that is has a countable basis, $(e_i)$, where $e_i(k) = \delta_{ik}$ (q.v. Kronecker delta), but a simple application of the Baire Category Theorem gives that no countably-infinite-dimensional vector space can be a Banach space: each finite-dimensional subspace is closed and nowhere dense.

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I didn't know the Baire Category thm. It's a useful cue for further study, thank you. – the_elder Nov 19 '12 at 17:09

Consider $e^n$ the sequence whose unique non-zero entry is the $n$-th (which is $1$), and the sequence $x_n=\sum_{j=1}^n2^{-j}e^j\in\Bbb R^{\infty}$. This sequence is Cauchy for the $\ell^p$ norm for all $p$, but doesn't converge to an element of $\Bbb R^{\infty}$ for the $\ell^p$ norm. Indeed, this converge implies componentwise convergence, so the potential limit is $\sum_{j=1}^{+\infty}2^{-j}e^j$, which is not finitely supported.

Actually, no norm can make $\Bbb R^\infty$ complete by Baire category theorem (but we don't need this argument if we work with a specified well-behaved norm).

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Very clear. Thank you. – the_elder Nov 19 '12 at 17:08

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