Subspace of convergent sequences in $l^{\infty}$: Some computations

Question

Let $l^{\infty}$ be the Banach space of sequences with finite sup norm and $c_0$ the subspace of $l^{\infty}$ which which contain sequences which converge to $0$. It can be shown $c_0$ is closed in $l^{\infty}$.

I found some notes which make the following two statements I'm unable to verify:

The distance between $c_0$ and the sequence $(1,1,1,...) \in l^{\infty}$ is $1$.

This statement confuses me, consider $(\frac{11}{n}) \in c_0$. Then $\|(\frac{11}{n}) - (1,1,1,...)\|_{\infty} = |11-1| = 10 \neq 1$

In fact, the distance between any $c_0$ and any bounded sequence $(a_n)$ with $a_n \in [0,2]$ is $1$.

Again consider $(\frac{11}{n}) \in c_0$, now $\|(\frac{11}{n}) - (a_n)\|_{\infty}$ seems like its anywhere between $9$ and $11$.

What am I missing?

Answer 1

By definition, the distance between sets is the smallest distance (infimum) between their elements. The distance $10$ what you got is not the smallest one. One approach is to prove first that the distance is at least $1$ (see what happens at infinity) and then to find a sequence that makes exactly $1$.

Answer 2

As already stated, the distance is defined as $$d((1,1, \ldots), c_0) = \inf_{(x_n) \in c_0}\|(1,1, \ldots) - (x_n)\|_\infty$$

Let $(x_n) \in c_0$ be arbitrary. We have

$$\|(1,1,\ldots) - (x_n)\|_\infty \ge |1-x_n| \ge 1-|x_n| \xrightarrow{n\to\infty} 1$$

so $d((1,1,\ldots), c_0) \ge 1$. On the other hand, for $(0,0, \ldots) \in c_0$ we have $$\|(1,1,\ldots) - (0,0,\ldots)\|_\infty = 1$$ so $d((1,1,\ldots), c_0) \le 1$. We conclude $d((1,1,\ldots), c_0) = 1$.

---

It is not true that $d((a_n), c_0) = 1$ for every bounded sequence with $a_n \in [0,2], \forall n \in \mathbb{N}$. Namely, for $(2,2,\ldots)$ and arbitrary $(x_n) \in c_0$ we have

$$\|(2,2, \ldots) - (x_n)\|_\infty \ge |2-x_n| \ge 2 - |x_n| \xrightarrow{n\to\infty} 2$$

so $d((2,2, \ldots), c_0) \ge 2$.