Verify that any Cauchy sequence in $c_0$ (equipped with the supremum norm) converges to a limit in $c_0.$ Conclude that $c_0$ is a Banach space.

My attempt:

Let $(\xi_n^{(i)})_{i \in \mathbb{N}}$ be a Cauchy sequence in $c_0.$ Let $\varepsilon>0$ be given. Then there exists $I$ such that for all $i > j \geq I,$ we have $\| \xi_n^{(i)} - \xi_n^{(j)} \|_{\infty} < \varepsilon.$ It follows that for each $n \in \mathbb{N},$ we have $|\xi_n^{(i)} - \xi_n^{(j)}| \leq \| \xi_n^{(i)} - \xi_n^{(j)} \|_{\infty} < \varepsilon.$ Hence, $(\xi_n^{(i)})_{i \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$ for each $n \in \mathbb{N}$. Since $\mathbb{R}$ is complete, we have $\lim_{i \rightarrow \infty}\xi_n^{(i)} = \xi_n$ for each $n \in \mathbb{N}.$ Denote $\xi = (\xi_n)_{n \in \mathbb{N}}.$ Clearly $(\xi_n^{(i)})_{i \in \mathbb{N}}$ converges to $\xi.$ We wish to show that $\xi \in c_0,$ that is, $\lim_{n \rightarrow \infty}\xi_n = 0.$ Let $\varepsilon>0$ be given. Since $(\xi_n^{(i)})_{i \in \mathbb{N}}$ converges to $\xi,$ there exists $I$ such that $\| \xi_n^{(I)} - \xi_n \|_{\infty} < \frac{ \varepsilon}{2}.$ Since $\lim_{n \rightarrow \infty}\xi_n^{(I)} = 0,$ there exists $N$ such that for all $n \geq N,$ we have $\| \xi_n^{(I)} \|_{\infty} < \frac{\varepsilon}{2}.$ Hence, for any $n \geq N,$ we have $$\| \xi_n \|_{\infty} \leq \|\xi_n - \xi_n^{(I)} \|_{\infty} + \| \xi_n^{(I)}\|_{\infty} < \varepsilon.$$ Therefore, $c_0$ is a Banach space.

Is my proof correct?

  • $\begingroup$ yes, it is right. $\endgroup$ – Masacroso May 24 '17 at 7:57

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