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Let $V$ be a vector space such that there is a $v \in V$ with $\|v\|_V = \infty$. Can you conclude from this that $V$ is not complete, i.e. that there is a Cauchy sequence in $V$ which does not converge in $V$?

The question came up when my lecture notes stated (without proof) that the space $C((0,1))$ of continuous functions on the open interval $(0,1)$ are not complete with the sup-norm $\|\cdot\|_\infty$. I assume this is a concrete instance of the above problem, since there are $f \in C((0,1))$ with $\|f\|_\infty = \infty$, e.g. $f(x) = \frac{1}{x}$.

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  • $\begingroup$ A norm on a real or complex vector space, by definition, is non-negative real-valued. It can't take the value $\infty$. It doesn't make sense to say that $C((0,1))$ is not complete with respect to the supremum norm, because the supremum doesn't define a norm on this vector space. $\endgroup$ – Keenan Kidwell Dec 12 '15 at 2:51
  • $\begingroup$ It's a kind of sloppy language. In a normed space all vectors have real (finite) norm. If you find what appears to be a vector of infinite norm, it's therefore not in your space. What you're really saying is that you have found a sequence of vectors of unbounded norm, which is the definition of "not complete". $\endgroup$ – ziggurism Dec 12 '15 at 2:54
  • $\begingroup$ @ziggurism: A sequence of vectors of unbounded norm has nothing to do with not being complete. In fact, it still makes sense to talk about Cauchy sequences with respect to a "norm" that is sometimes infinite, and it turns out that in this case every Cauchy sequence does converge. $\endgroup$ – Eric Wofsey Dec 12 '15 at 2:58
  • $\begingroup$ @EricWofsey: yes, upon second thought my comment is wrong. Not sure what I was thinking. Thanks for the correction. $\endgroup$ – ziggurism Dec 12 '15 at 3:40
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Actually, for any reasonable definition of what "complete" means when you have a norm that is allowed to be infinite (e.g., every Cauchy sequence converges), $C((0,1))$ is complete with respect to the sup norm (the proof is exactly the same as the proof that $C([0,1])$ is complete). All that was probably meant by the remark in your lecture notes is that since $\|\cdot\|_\infty$ is not even a norm on $C((0,1))$ in the usual sense (since the usual definition of a norm requires it to always be finite), it does not make it into a complete normed space.

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  • $\begingroup$ Wait how does allowing norm to be infinite force every Cauchy sequence to converge? $\endgroup$ – ziggurism Dec 12 '15 at 3:41
  • $\begingroup$ It doesn't force it; it just happens that in this case ($C((0,1))$ with the sup "norm") every Cauchy sequence does converge. $\endgroup$ – Eric Wofsey Dec 12 '15 at 3:44
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It seems the following. Define a norm $\|\cdot\|$ on a vector space $X=\Bbb R^2$ as follows. For a vector $x=(x_1,x_2)\in V$ put $\|x\|=|x_1|$, if $x_2=0$ and $\|x\|=\infty$, if $x_2\ne 0$. It is easy to check that the space $(V,\|\cdot\|)$ is complete.

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