Completeness of normed spaces

As earlier, I have received an answer from this site that Bolzano Weierstrass' theorem is true for finite dimensional normed spaces, but not for infinite dimensional spaces. This, in particular => all finite dim. normed spaces are complete(in the sense that every Cauchy sequence converges(w.r.t. norm)). However, is it true that every normed vector space is complete?

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We wouldn't need the words "Banach space" if this were true :) –  Dylan Moreland May 13 '12 at 14:57
There a lot of examples. math.stackexchange.com/questions/114070/… –  Norbert May 13 '12 at 15:04
Ignore my previous (non-)answer. Since I don't want to bamboozle you with function spaces I'm now trying to think of a vector space that is incomplete and is not a function space. –  Matt N. May 13 '12 at 15:31
This might be an easier example: math.stackexchange.com/questions/143857/… –  Michael Greinecker May 13 '12 at 15:34
@MichaelGreinecker I don't insist... –  Norbert May 13 '12 at 16:23
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2 Answers

Let $C[0,1]$ be the space of continuous, real-valued functions on $[0,1]$. Then $$\|f\|=\int_0^1 |f(x)|dx$$ defines a norm on this space. The sequence of functions defined by $$f_n(x) = \left\{ \begin{array}{rl} 0 & \text{if } x \leq 1/2,\\ 1 & \text{if } x \geq 1/2+1/n,\\ n(x-1/2) & \text{if } 1/2\leq x\leq 1/2+1/n. \end{array} \right.$$

is Cauchy, but does not converge. I have recycled the sequence from my answer here.

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Ok, here we go: take the one dimensional vector space over $\mathbb Q$ with the usual norm $|q|$. Then you can find a Cauchy sequence that converges to an irrational, hence the space is not complete.

Or: sequences that are non-zero only in finitely many places (over $\mathbb R$) and you can take the norm to be the $\ell^1$ norm, i.e. $\|x\| := \sum_i |x_i|$.

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This is not a vector space. –  Nate Eldredge May 13 '12 at 15:19
Dear Matt, can you just point me out what your "+" operation means in your vector space R or in your subspace (0,1)? Because in case it is the normal addition operation, then how can (0,1) be a subspace at all? Because 0.9 is in (0,1), but 0.9 + 0.9 = 1.8 > 1!! –  Somabha Mukherjee May 13 '12 at 15:22
@SomabhaMukherjee Sorry, didn't see the "vector" in your question. Let me correct my answer. –  Matt N. May 13 '12 at 15:24
@NateEldredge Thanks. I think I managed to fix my answer. –  Matt N. May 13 '12 at 15:47
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