Let $ c_0 = \{ x = (x_n)_{n \in \mathbb N} \in l^\infty : \lim_{n \to \infty} x_n = 0\}$. Show that $c_0$ is a Banach space with the norm $\rVert \cdot \lVert_\infty$
I am capable of showing the space where the limit of $x_n$ exists is normed linear space but am having trouble with showing that the limit of Cauchy sequences must converge to 0.
Let $(x^{(n)})_{n \in \mathbb N}$ be a Cauchy sequence in $c_0$ such that $x^{n} = (x^n_1, x^n_2,…)$.
Fix $k \in \mathbb N$ consider the sequence $(x^n_k)_{n \in \mathbb N}$ in $\mathbb F$. For any $n,m \in \mathbb N$
$\lvert x^n_k – x^m_k \rvert \le \sup_{k \in \mathbb N} \lvert x^n_k – x^m_k \rvert = \lVert x^n – x^m \rVert_\infty \lt \epsilon $ (1)
Thus $x^n_k$ is Cauchy in $\mathbb F$ and so has limit $y_k$ such that $y = (y_1,y_2,…)$ and y is the limit of $x^n$
To show that such a y exists we look at the value of $\lvert y_n – y_m \rvert \le \lvert y_n – x^N_n \rvert + \lvert x^N_n – x^N_m \rvert + \lvert x^N_m – y_m \rvert \lt \epsilon$ for all $n,m \ge N$ (2)
The middle expression on RHS of (2) is $\lt \epsilon/3$ by (1)
The other two are also $\lt \epsilon/3$ follow from $x^N_k$ being Cauchy and converging to $y_k$
This shows that $\lim_{n \to \infty} y_n$ exists but we still have not shown that $y \in c_0$.
I know that to show y tends to 0 i should show that $\lvert y_k \rvert \lt \epsilon$ for $k \ge N$
This is where I am stuck. Perhaps
$\lvert y_k \rvert = \lvert \lim_{n \to \infty} x^n_k \rvert$ and then we can take the limit function outside the absolute value sign by continuity?
Then we might say due to it being a Cauchy sequence $x^n_k \lt \epsilon$. I know this last bit isn't at all convincing so I could do with some help.
Best Answer
Suppose $x^k \in c_0$ and $x^k \to x$. Let $\epsilon>0$ and pick $N$ such that $\|x^k-x\|_\infty < {1 \over 2 } \epsilon$ for $k \ge N$. Since $x^N \in c_0$, there is some $N'$ such that $|x_i^N| < {1 \over 2 } \epsilon$ for $i \ge N'$. Then $|x_i| \le |x_i^N| +|x_i-x_i^N| \le |x_i^N| +\|x-x^N\|_\infty <\epsilon$. Hence $x_i \to 0$ and so $x \in c_0$.
Hence $c_0$ is a closed subspace of $l_\infty$.
It follows that $c_0$ is Banach since $l_\infty$ is Banach (any Cauchy sequence in $c_0$ is Cauchy in $l_\infty$ hence converges to some point an closedness shows that this point lies in $c_0$).