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Let $c=\lbrace x=\lbrace x_n\rbrace ,n\in \mathbb{N}: \exists \, \text{lim}_{n\to \infty}x_n\rbrace$ and $c_0=\lbrace x=\lbrace x_n\rbrace ,n\in \mathbb{N}: \text{lim}_{n\to \infty}x_n=0\rbrace$. I want to show that $c$ anc $c_0$ are Banach (I have to show that both are complete metric spaces).

Is it enough to show that $\text{lim}_{n\to \infty}x_n$ belongs to space?

These are my thoughts:

Let: $$x^{(n)}=\left( x^{(n)}_j\right)_{j=1}^\infty =(x^{(n)}_1,x^{(n)}_2,\dots )$$ be a Cauchy sequence in $c$.

Lets define the $\infty$-norm $||\cdot||_\infty$: $$||(x_j)_{j\geqslant 0}||_\infty = \text{sup}_{j\geqslant 0} |x_k|$$

Let $\varepsilon >0$, there exists $n_0\in \mathbb{N}$ such that: $$||x_j^{(n)}-x_k^{(m)}||<\varepsilon,\,\,\,\forall n,m >n_0$$ $$\text{sup}_{j\geqslant 0} |x_j^{(n)}-x_j^{(m)}|<\varepsilon,\,\,\,\forall n,m >n_0$$ $$|x_j^{(n)}-x_j^{(m)}|<\varepsilon,\,\,\,\forall n,m >n_0, \,\,\,\forall j\geqslant 0$$

Now I have to prove that $$\text{lim}_{n\to\infty} x^{(n)}= x$$ and $$x\in c.$$

Unfortunately I don't know how to show that $x\in c$.

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  • $\begingroup$ Show that every Cauchy sequence converges in $c_0$. If $x^{(n)}=(x^{(n)}_j)_{j=1}^\infty$ is Cauchy in $c_0$ then pick a subsequence $(x^{(n_{k,1})})_{k=1}^\infty$ with $|x^{(n_{k,1})}-x_1|<k^{-1}$ for each $k$ and some $x_1\in\mathbb{R}$. Now find a further subsequence $(x^{(n_{k,2})})_{k=1}^\infty$ with $|x^{(n_{k,2})}_2-x_2|<k^{-2}$ for some $x_2\in\mathbb{R}$. Etc. Set $x=(x_j)_{j=1}^\infty$ so that the diagonal sequence $(x^{(n_{k,k})})_{k=1}^\infty$ satisfies $|x^{(n_{k,k})}_j-x_j|<k^{-1}$ and hence $\|x^{(n_{k,k})}-x\|<k^{-1}$. This is the idea for $c_0$, and $c$ is similar. $\endgroup$ – Ben W Apr 5 '16 at 5:12
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    $\begingroup$ Although, it might be easier (and more interesting) to prove that $c$ is Banach by simply showing that the map $(x_1,x_2,x_3,\cdots)\mapsto(L,x_1-L,x_2-L,x_3-L,\cdots)$ is a linear isomorphism between $c$ and $c_0$. $\endgroup$ – Ben W Apr 5 '16 at 5:15
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Let $(X. \vert \cdot \vert)$ be some Banach space. Let $(x^{(n)})_{n\in \mathbb{N}} \subseteq c$ be a Cauchy sequence, where $x^{(n)}=(x_k^{(n)})_{k\in \mathbb{N}} \subseteq X$ is a convergent sequence. We define

$$ x_{\infty}^{(n)}:= \lim_{k\rightarrow \infty} x_k^{(n)}.$$

We need to construct $x=(x_k)_{k\in \mathbb{N}}\in c$ such that $x^{(n)} \rightarrow x$ in the supremum norm. As usual we construct a candidate and verify that indeed this is the desired limit.

We prove first that the pointwise limit of each coordinate exists. Let $\epsilon >0$ and choose some $N\in \mathbb{N}$ such that for all $m,n\geq N$ holds

$$ \Vert x^{(m)} - x^{(n)} \Vert <\epsilon.$$

Fix some $k\in \mathbb{N}$, then we get

$$ \vert x_k^{(m)} - x_k^{(n)} \vert \leq \Vert x^{(m)} - x^{(n)} \Vert <\epsilon.$$

Hence, $(x_k^{(l)})_{l\in \mathbb{N}} \subseteq X$ forms a Cauchy sequence in $X$. As $X$ is a Banach space there exists $x_k := \lim_{l \rightarrow \infty} x_k^{(l)}$. We found our candidate, namely $x:=(x_k)_{k\in \mathbb{N}}$. We need to check that $x\in c$. For this we note that for $m,n\geq N$ (where $N$ is chosen as above) we get

$$ \vert x_{\infty}^{(m)} - x_{\infty}^{(n)} \vert = \lim_{k\rightarrow \infty} \underbrace{\vert x_k^{(m)} - x_k^{(n)} \vert}_{< \epsilon, \text{by choice of }N} \leq \epsilon.$$

Hence, $(x_{\infty}^{(l)})_{l\in \mathbb{N}}\subseteq X$ forms a Cauchy sequence in $X$. Again, as $X$ is a Banach space, the limit exist and we call it $x_{\infty}:= \lim_{l\rightarrow \infty} x_{\infty}^{(l)}$. This is our candidate for the limit of $(x_k)_{k\in \mathbb{N}}$.

Let $\epsilon >0$, choose $\tilde{N}\in \mathbb{N}$ such that for all $\tilde{n}\geq \tilde{N}$ holds $\vert x_{\infty}^{(\tilde{n})} -x_{\infty} \vert <\epsilon/3$. Choose $\tilde{K}\in \mathbb{N}$ such that for all $\tilde{k}\geq \tilde{K}$ holds $\vert x_k -x_k^{(\tilde{n})}\vert, \vert x_k^{\tilde{n}} - x_{\infty}^{(\tilde{n})} \vert < \epsilon/3$ (convince yourself, that can we choose such $\tilde{N}, \tilde{K}$). Then holds for every $\tilde{k}\geq \tilde{K}$

$$ \vert x_{\tilde{k}} - x_{\infty} \vert \leq \vert x_{\tilde{k}} - x_{\tilde{k}}^{(\tilde{n})} \vert + \vert x_{\tilde{k}}^{(\tilde{n})} - x_{\infty}^{(\tilde{n})} \vert + \vert x_{\infty}^{(\tilde{n})} - x_{\infty} \vert < \epsilon.$$

Hence, $\lim_{\tilde{k} \rightarrow \infty} x_{\tilde{k}}=x_{\infty}$ and therefore $x\in c$.

We are left to show, that $x^{(n)}\rightarrow x$ in the supremum norm. Choose again $N\in \mathbb{N}$ such that for all $n,m\geq N$ holds $\Vert x^{(m)} - x^{(n)} \Vert < \epsilon/2$. Let $k\in \mathbb{N}$. Choose $m_k\geq N$ such that $\vert x_k - x_k^{(m_k)} \vert <\epsilon/2$, then holds for every $n\geq N$

$$ \vert x_k - x_k^{(n)} \vert \leq \vert x_k -x_k^{(m_k)} \vert + \vert x_k^{(m_k)} -x_k^{(n)} \vert \leq \epsilon/2 + \Vert x^{(m_k)} - x^{(n)} \Vert < \epsilon.$$

Hence, for $n\geq N$ holds

$$ \Vert x -x^{(n)} \Vert = \sup_{k\in \mathbb{N}} \underbrace{\vert x_k - x_k^{(n)}}_{<\epsilon} \vert <\epsilon.$$

This shows that $(c, \Vert \cdot \Vert)$ forms a Banach space. Redoing precisely the same argument would give you that $(c_0, \Vert \cdot \Vert)$ forms a Banach space as well. However, there is an easier way. All we need to show is that $c_0$ is a closed subspace of $c$. This can be shown in the following way:

We define

$$ L: c \rightarrow X, \ L\left((x_k)_{k\in \mathbb{N}} \right):= \lim_{k\rightarrow \infty} x_k.$$

This is a linear functional. In fact it is bounded. Let $x=(x_k)_{k\in \mathbb{N}}\in c$, then

$$ \vert L(x) \vert = \vert \lim_{k\rightarrow \infty} x_k \vert = \lim_{k\rightarrow \infty} \underbrace{\vert x_k \vert}_{\leq \sup_{k\in \mathbb{N}} \vert x_k \vert = \Vert x \Vert} \leq \Vert x \Vert.$$

This mean, $c_0 = L^{-1}(\{ 0\})$ (i.e. the kernel of a continuous linear functional) and therefore $c_0$ is closed subspace of a Banach space and hence a Banach space itself.

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  • $\begingroup$ That helped me alot. I would like to thank you for proof. $\endgroup$ – Rico Apr 30 '16 at 8:07

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