Set of sequences which converge to zero is a closed subspace of $l^\infty$

Question

Prove that:

$$c_0 = \{\{a_n\}_1^\infty:lim_{n\to\infty}a_n = 0\}$$

Is a closed subspace of $l^\infty$.

Answer 1

Hint:

Let $(u_n)$ be a sequence of elements of $c_0$ such that $u_n \to u$ in $\ell^{\infty}$.

Let $\varepsilon > 0$, then : $$\exists N \in \mathbb N, n \geq N \Rightarrow \|u_n-u\|_{\infty} < \varepsilon $$

In particular : $ \|u_N-u\|_{\infty} < \varepsilon/2 $

Since $u_N \in c_0$ we have:

$$\exists N_1 \in \Bbb N, \forall p \geq N_1 \quad|u_{N,p}| < \varepsilon / 2 $$

Let $N_2=\max(N,N_1)$, then

$$\forall p \geq N_2, |u_p| \leq |u_p-u_{N,p}| + |u_{N,p}| < \varepsilon /2 + \varepsilon /2 = \varepsilon $$

That shows that for all sequence $(u_n)$ of elements of $c_0$ if $u_n \to u \in \ell^{\infty}$ then $u \in c_0$. that means : $c_0$ is colsed set of $\ell^{\infty}$

Answer 2

Note that the space $c_0 \subset c \subset \ell_{\infty}$ is a, not dense, hyperplane (kernel) of the operator $T \colon c \to \mathbb{R}$ $$T(x) = \lim\limits_{n \to \infty} x_n$$

Answer 3

You can show that its complement is open.

Let $X = (c_o)^c = \{\{a_n\}_1^\infty: \{a_n\}_1^\infty \ \text{does not converge to 0}\}$.

That means that if $x \in X$ then there exists $\epsilon > 0$ such that for all $N \in \mathbb{N}$ there exists $n > N$ such that $|x_n - 0| = |x_n| > \epsilon$.

Now we'll find an $\epsilon$-neighborhood about x contained entirely within $X$. This will show that $X$ is open and that therefore $c_0$ is closed.

Let's take $B = \{y \in \ell^\infty : \|y - x\|_\infty = \sup_{i \in \mathbb{N}} |x_i - y_i|< \epsilon/2 \}$.

We know that for every $N \in \mathbb{N}$ there is some $n > N$ with $|x_n| > \epsilon$.

For any such $n$ we have

$|x_n| = |x_n - y_n + y_n| \leq |x_n - y_n| + |y_n|$, or

$|x_n| - |x_n - y_n| \leq |y_n|$. But the left hand side is at least $\epsilon - \epsilon/2 = \epsilon/2$, so $y$ can not converge to $0$.