# Showing $c_0$ is complete

Let $c_0$ be the vector space of all convergent sequences $x = \{ x_n \}_{n=1}^\infty$ in $\mathbb{R}$ $\$ with: $\$ $\lim_{n \to \infty} x_n = 0$. Let $\lVert x \rVert_{\infty} = \sup_{n \in \mathbb{N}} \lvert x_n \rvert$. Consider a Cauchy Sequence $\{ x_n \}_{n=1}^\infty$ in $c_0$ whose elements are denoted by $x_k = \{ x_{k,n} \}_{n=1}^\infty$ $\\$

1. show that for each fixed $n \in \mathbb{N}, \{ x_{k,n} \}_{n=1}^\infty$ is Cauchy in $\mathbb{R}$
2. Deduce that the limit $y_n = lim_{k \to \infty} x_{k,n}$ defines $y = \{ y_n \}_{n=1}^\infty$ with $\lVert y \rVert_{\infty} \lt \infty$
3. Show that $lim_{k \to \infty} \lVert x_k - y \rVert_{\infty} = 0$
4. Conclude that $y \in c_o$ and thus $c_0$ is complete.

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My attempt:

Visualization- $$\begin{pmatrix} x_{11} & x_{12} & \cdot & \cdot & \cdot & \cdot \\ x_{21} & x_{22} & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{pmatrix}$$

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So, $\{ x_{k,n} \}_{n=1}^\infty$ is a sequence of sequences like the above and each row $\to 0$.

Now, fixing $n \in \mathbb{N}$ lets me deal with an individual column of the above. Since $\{ x_{k} \}_{n=1}^\infty$ is a Cauchy sequence in $c_0$, for $\forall \epsilon \gt 0, \ \exists M \in \mathbb{N} \$ s.t. $\ m,m' \ge M \implies \lVert x_m - x_{m'} \rVert \lt \epsilon.$

So, $$m,m' \ge M \implies \lvert x_{m,n} - x_{m',n} \rvert \lt \epsilon; \ \ \forall n \ge 1$$
Therefore, when n is fixed, $\{ x_{k,n} \}_{n=1}^\infty$ is Cauchy in $\mathbb{R}$

Because $\{ x_{k,n} \}_{n=1}^\infty$ is Cauchy, the limit exists and $\lim_{k \to \infty} \{ x_{k,n} \}_{n=1}^\infty = y_n$

Now, there is a new sequence using the limit of each column involved, let's call it: $y = \{ y_n \}_{n=1}^\infty$ (Note: I'm drawing a blank on formal reason why $\lVert y \rVert_{\infty} \lt \infty$)

Letting $m' \to \infty$ in the above result, we get: $$m \ge M \implies \lvert x_{m,n} - y_n \rvert \lt \frac{\epsilon}{2}; \ \ \forall n \ge 1$$ Setting m = M, $\forall n \ge 1$, we have: $$\lvert y_n \rvert = \lvert y_n - x_{M,n} + x_{M,n} \le \lvert y_n - x_{M,n} \rvert + \lvert x_{M,n} \rvert \lt \frac{\epsilon}{2} + \lvert x_{M,n} \rvert$$ Since $x_M \in c_0$ we have $x_{M,n} \to 0$ as $n \to \infty$; $\ \exists N \in \mathbb{N}$ $\$ s.t. $\$ $n \ge N \implies \lvert x_{M,n} \rvert \lt \frac{\epsilon}{2}$

Hence: $$n \ge N \implies \lvert y_n \rvert \lt \frac{\epsilon}{2} + \lvert x_{M,n} \rvert \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ Therefore, $y \to 0$ as $n \to \infty$, thus $y \in c_0$ and $c_0$ is complete.

Also, $y$ is the limit of the $x_k$'s

And so: $$\lVert x_k - y \rVert_{\infty} = sup_{n \in \mathbb{N}} \lvert x_k - y \rvert \lt \epsilon$$ Which means: $$\lim_{k \to \infty} \lVert x_k - y \rVert_{\infty} = 0$$ $\ \$