I am having trouble with the following problem.

Suppose $C = \{x ∈ V : \Vert x \Vert = 2\}$, a subset of the normed vector space ($V,\Vert \Vert$). Show that if V is complete then C is complete.

Possible Answer: Suppose that $C$ is a closed subset of $V$, and let $\{x_n\}$ be a Cauchy sequence in $C$. Since $V$ is complete there is some $L\in V$ such that $L$ is the limit of $\{x_n\}$.

We need to show that $x\in C$. By the definition of convergence for every $\epsilon>0$ there is some $N$ such that for all $n>N$, $\|L-x_n\|<\epsilon$. Therefore for every $\epsilon>0$ we can find other some point $x_0$ from $S$ such that $\|L-x_0\|<\varepsilon$. Since $C$ is closed this means that $L\in C$, as wanted. Therefore the subset C of a normed vector space V is complete in V as every Cauchy sequence in C converges to a vector in C.

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    $\begingroup$ Your title seems to be (if I read it correctly) the converse of what you're actually asking? $\endgroup$ – Ted Shifrin Oct 13 '16 at 17:07
  • $\begingroup$ Sorry. Let me correct that. $\endgroup$ – AzJ Oct 13 '16 at 17:09
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    $\begingroup$ The norm function is continuous, so $C$ is a closed subset of $V$. $\endgroup$ – Andreas Blass Oct 13 '16 at 17:20
  • $\begingroup$ @AndreasBlass Your comment is really helpful. But how does a continuous norm function mean C is closed. Note I have also put an a possible answer to my question. $\endgroup$ – AzJ Oct 13 '16 at 22:26
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    $\begingroup$ The pre-image of a closed set under a continuous function is always closed. Your $C$ is the pre-image under the norm function of the closed set $\{2\}$. $\endgroup$ – Andreas Blass Oct 13 '16 at 23:34

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