Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
We wouldn't need the words "Banach space" if this were true :) –  Dylan Moreland May 13 '12 at 14:57
1  
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
add comment

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.

share|improve this answer
add comment

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. $\|x\| := \sum_i |x_i|$.

share|improve this answer
2  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.