Sequences with finite range are dense in $\ell^{\infty}$

Question

Let $m_{0}$ be the subspace of $\ell^{\infty}$ which consists of the sequences with finite range. Show that $m_{0}$ is dense in $\ell^{\infty}$.

Answer 1

I assume you are talking about $\ell^\infty = \ell^\infty(\mathbb{N})$.

Let $x = (x_n)_{n \in \mathbb{N}} \in \ell^\infty$. We need to show that for every $\varepsilon > 0$ there is a sequence $y = (y_n)_{n\in \mathbb{N}} \in \ell^\infty$ with finite range and $\| x - y\|_\infty \leq \varepsilon$.

So let $\varepsilon > 0$. You can give a sequence $y = (y_n)$ with $$ y(\mathbb{N}) \subset A := \{k \varepsilon + i l \varepsilon; k, l = -\lceil \|x\|_\infty/\varepsilon \rceil, \dots, -1, 0, 1, 2\, \dots, \lceil {\|x\|_\infty}/\varepsilon \rceil\},$$ where $\lceil \|x\|_\infty/\varepsilon \rceil$ denotes the smallest integer larger or equal than $\|x\|_\infty/\varepsilon$: Since any point in $\{ z \in \mathbb{C}; |z| \leq \| x\|_\infty\}$ (and therefore every point in $x(\mathbb{N})$) is at most $\frac{1}{\sqrt{2}}\varepsilon$ away from $A$, we may choose for every $n \in \mathbb{N}$ a $y_n \in A$ such that $|x_n - y_n| < \varepsilon$ and therefore have $\| x - y\|_\infty \leq \varepsilon$.