I have a very difficult excercise. I see now that it became too much text for someone to might go through it, if you can please help me, but don't want to read all, can you please then only answer my question in bold, and check that the lemma at the end is correct? Also if you see a simple way of doing this, without so much text, please tell me how, the excercise is only one sentence, so I do not think it should be so much work.
Let $V \ne \{\textbf{0}\}$ be a normed vector space. Show that V is complete if and only if the unit sphere $S = \{x \in V: \|x\| = 1\}$ is complete.
Doesnt this hold, even if $V = \{\textbf{0}\}$? Because $V = \{\textbf{0}\}$ is complete, and since the empty space does not contain any Cauchy-sequences, every Cauchy sequence in the empty space converges?
In the solution, I have to use a lemma, it is the lemma I am most unsure about.
$\rightarrow$
Assume that V is complete. Then every Caucy-sequence in S converges, since S is closed, S is complete.
$\leftarrow$
Assume that S is complete, we must show that every Cauchy-sequence in V converges. Let $\{v_n\}$, be a Cauchy sequence in V. By the lemma below, either $\{v_n\}$ converges to $\textbf{0}$ or there is an $M>0$ and an $N$, so that if $n \ge N$, $\|v_n\|>M$. In the first case the Caucy-sequence converges and we are done, so assume that we are in the second case.
Work with the new sequence for $n \ge N$, $\{\frac{v_n}{\|v_n\|}\}$, since we are in the second case, we never divide by zero. This is a sequence in S, and it is Cauchy since, given an $\epsilon$, there is an $N_1$ so that if $n,m \ge N_1$, $\|v_n-v_m\|<\epsilon/(2M)$. So if $m,n \ge max(N,N_1):$ $\|\frac{v_n}{\|v_n\|}-\frac{v_m}{\|v_m\|}\|\le \|\frac{v_n}{\|v_n\|}\|+\|\frac{v_m}{\|v_m\|}\|<\frac{\epsilon/(2M)}{M}+\frac{\epsilon/(2M)}{M}=\epsilon$.
Since S is complete, $\{\frac{v_n}{\|v_n\|}\}$ converges to a vector $u$. Since $\|\|v_n\|-\|v_m\|\|\le\|v_n-v_m\|$, the sequence of the norms are also Cauchy. Since $\mathbb{R}$ is complete, $\{\|v_n\|\}$, converges to a number $b$. But then $\{v_n\}$ must converge to $bu$. Because given an $\epsilon$(b can't be zero).:
$\|v_n-bu\|=b\|\frac{v_n}{b}-\frac{v_n}{\|v_n\|}+\frac{v_n}{\|v_n\|}-u\|\le b\|v_n\|*|\frac{1}{b}-\frac{1}{\|v_n\|}|+b*\|\frac{v_n}{\|v_n\|}-u\|$. The first expression we can get as small as we want because a Cauchy sequence in $\mathbb{R}$ is bounded, and because of convergence properties of the real numbers. The second part because we showed that the sequence must converge because it is in S.
Now comes the Lemma:
Lemma: Let V be a vector space, and $\{v_n\}$ a Cauchy sequence in V. Either $\{v_n\}$ converges to $\textbf{0}$, or there is an $M > 0$, and an $N$, so that if $n \ge N$, $\|v_n\| > M$.
We prove this by assuming we are not in the first case, and show that we must be in the second case. Start my choosing a number $L>0$, since $\{v_n\}$ is Cauchy, there is an $N$, so that if $m,n \ge N \rightarrow \|v_n-v_m\|<L/2$. Assume first that $\|v_N\|> L/2$. Then for all $n \ge N$, $\|v_n\|>\|v_N\|-L/2=M>0$, and we are done. But it might be that $\|v_N\|\le L/2$, then choose $L_2=L/2$, and find an $N_2$ for $L_2/2$, if this doesnt work choose $L_3 = L/3$, and find an $N_3$.
Now, this procedure must stop, because if it did not stop, we can show that the sequence must converge to $\textbf{0}$. Because assume that the procedure never stopped, and choose an $\epsilon$, we can find an $N$ such that $n \ge N \rightarrow \|v_N-v_n\| < \epsilon/2$(divide L by a big enough N), since the procedure never stopped, we have that $\|v_N\|\le\epsilon/2$. But for any $n \ge N$, we then have:
$\|v_n-\textbf{0}\|=\|v_n-v_N+v_N\|\le \|v_n-v_N\|+\|v_N\|<\epsilon/2+\epsilon/2=\epsilon.$
It is clear that the completeness of the whole space implies the completeness of the unit sphere. Suppose we have that the unit sphere is complete, we want to show that the whole space is also complete.
$\{x_n\}$, not necessarily on the unit sphere, is a Cauchy Sequence, so is $\{\|x_n\|\}$. We have that $\|x_n\|\to a$. If $a=0$, i.e. $\|x_n\|\to 0$ which means that $x_n \to 0$ in $V$. If $a>0$, then we can find a $N$ large enough such that:
$\|x_n\|>0 ,\;\forall n>N$
Starting from $N$, we can define $y_n = \frac{x_n}{\|x_n\|}$. It is easy to check that $y_n$ is a Cauchy sequence on the unit sphere. And thus, we get $y_n \to y$. One can easily check that $a\cdot y$ is the limit of $\{x_n\}$.