Denote $c_0 = \{ (\xi_n)_{n \in \mathbb{N}}: \lim_{n \rightarrow \infty}\xi_n=0 \}$ and $c = \{ (\xi_n)_{n \in \mathbb{N}}: \lim_{n \rightarrow \infty}\xi_n \text{ exists}\}$. Both spaces are equipped with supremum norm $\| (\xi_n)\| = \sup_{n \in \mathbb{N}}|\xi_n|.$
Question: Show that $c_0$ is a closed subspace of $c$
The question above is taken from the book 'An introductory Course in Functional Analysis' by Bower and Kalton, page $26$, exercise $2.1$.
My attempt:
I use the fact that $A$ is closed if and only if for every Cauchy sequence $(x_n)$ in $A$ converges to $x$, we have $x \in A.$
Let $(\xi_n^{(i)})_{i \in \mathbb{N}}$ be a sequence in $c_0$ such that it converges to
$\xi = (\xi_n)_{n \in \mathbb{N}}$. 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
$\xi_n$, there exists $I$ such that $\| \xi_n^{(I)} – \xi_n
\|_{\infty} < \frac{\varepsilon}{2}.$ In particular, for all $n \in
\mathbb{N},$ we have $|\xi_n^{(I)} – \xi_n| < \frac{\varepsilon}{2}.$
Since $(\xi_n^{(i)})_{n \in \mathbb{N}}$ converges to $0,$ there
exists $N$ such that for all $n \geq N,$ we have $\| \xi_n^{(i)}
\|_{\infty} < \frac{\varepsilon}{2}.$ Therefore, for $n \geq N,$ we
have $$|\xi_n| \leq |\xi_n – \xi_n^{(I)}| + |\xi_n^{(I)}| \leq
\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$
Is my proof correct?
Best Answer
If you know that the limit operator $\lim$ is a continuous functional on $c$ then $c_0=\ker \lim$ is a closed subspace.