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Consider a complete normed vector space $V$ and its subset $C:=\{x\in V:\lVert x\rVert=2\}$. Prove that $C$ is also complete.

Here's the part I'm not quite sure about. If we consider a Cauchy sequence $\{x_n\}$ in $C$, we will know that it converges in $C$ because $\lim\limits_{n\to\infty} \lVert c_n\lVert = 2$ for any $\{c_n\}\in C$. Thus $C$ is complete anyway. Why do we need to know that $V$ is complete, other than for the fact that Cauchy sequences are convergent in $V$?

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  • $\begingroup$ Knowing that $\|c_n\| \to 2$ does not imply that there is some $c$ such that $c_n \to c$. $\endgroup$ – copper.hat Oct 13 '16 at 5:57
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If $V$ is complete and $C\subset V$ is closed, then $C$ is also complete.

If $x_n$ is Cauchy in $C$ then it is Cauchy in $V$ and hence there is some $x \in V$ such that $x_n \to x$. Since $x_n \in C$ and $C$ is closed, we have $x \in C$, hence $C$ is complete.

We use completeness of $V$ to know that there exists some $x$ to which the sequence converges.

To illustrate what happens if $V$ is not complete, let $V$ be the continuous functions on $[-1,1]$ with the norm $\|x\|^2 = \int_{-1}^1 x(t)^2 dt$. Let $y_n(t) = {2 \over \pi} \arctan (nt)$, and $x_n = 2{y_n \over \|y_n\| }$

Then $x_n$ is Cauchy but does not converge to a continuous function (this takes a little work, but should be clear from a picture), hence $V$ is not complete. Since $x_n \in C$, clearly it does not converge in $C$ either.

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  • $\begingroup$ If $x_n \in C$ and $C$ is closed, why does this imply that $x$ must be in $C$ and not outside of $C$? $\endgroup$ – sequence Oct 13 '16 at 6:15
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    $\begingroup$ If $x \notin C$, then since $C$ is closed there is some open ball containing $x$ that does not intersect $C$, but this contradicts $x_n \to x$. Hence $x \in C$. $\endgroup$ – copper.hat Oct 13 '16 at 6:17
  • $\begingroup$ Sorry, why can't $x_n\to x$ if an open ball does not intersect $C$? $\endgroup$ – sequence Oct 13 '16 at 6:26
  • $\begingroup$ Let $C$ be a closed and bounded set. If we suppose that a sequence $\{x_n\}$ in $C$ converges outside of $C$, then it will have some points $x_n \not\in C$, and we can still have an open ball around $x$, not intersecting $C$. @copper.hat $\endgroup$ – sequence Oct 13 '16 at 23:09
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    $\begingroup$ @sequence: I don't really follow your comment stream here. You need to show that $C$ is complete. It follows from the fact that $C$ is a closed subset of a complete space. Boundedness has nothing to do with it. $\endgroup$ – copper.hat Oct 14 '16 at 2:40
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You may need to study an example to see what goes wrong: Take for $V$ the set $C^1([0,1])$ of continuously differentable functions on $[0,1]$ and use the norm $$\|x\|=\sup_{0\leq t\leq 1} |x(t)|$$ then the set $C=\{x\in V: \|x\|=2\}$ is not going to be complete. You may e.g. try to find a Cauchy sequence (for this norm) of functions in $C$ that converges pointwise to a continuous function but which is not differentiable at some point.

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