Proving that the range of a function is dense in its codomain Let $l_q := \{ x = (x_i)_i \in \mathbb{C}^\mathbb{N}/\|x\|_q := (\sum_{i = 1}^{\infty}|x_i|^q)^{1/q}  < \infty \}$
and $c_o := \{ x = (x_i)_i \in \mathbb{C}^\mathbb{N}/ \lim\limits_{i}x_i = 0, \|x\|_\infty := \sup\limits_{i}|x_i| \}$.
Let $X = l_q$ or $c_0$.
How would I show that the range of $\phi : X \rightarrow X$, $\, \phi((x_i)_i) := (e^{x_i} - 1)_i$ is dense in the codomain?
From what I understand of density, this problem is the same as showing that there are points in ran($\phi$) arbitrarily close to each point in $X$. So for every $x \in X$ and $\epsilon > 0$, we can find a $y \in$ ran($\phi$) such that $||x - y|| < \epsilon$, where $||\cdot||$ is the corresponding norm.
 A: Range of $\phi$ is dense. Let $c_{00} \subseteq X$ be the subspace of finite sequences. Note that in both cases this is a dense subspace - for any $x \in X$ we have 
$$ (x_1,\ldots,x_n,0,\ldots) \xrightarrow{n \to \infty} (x_1,x_2,\ldots) \quad \text{in } X. $$
Choose arbitrary $(a_1,\ldots,a_n,0,\ldots) \in c_{00}$. For $i=1,\ldots,n$ let $b_i \in \mathbb{C}$ be any number such that $e^{b_i} = 1+a_i$. Then $b = (b_1,\ldots,b_n,0,\ldots) \in c_{00}$ satisfies $\phi(b) = a$, so $a$ lies in the range of $\phi$. Since the range of $\phi$ contains $c_{00}$, it is dense in $X$. 
$\phi$ is well-defined and continuous. Fix $R > 0$. The $\mathbb{C} \ni z \mapsto e^z-1 \in \mathbb{C}$ is Lipschitz on the disk $B(0,R) \subseteq \mathbb{C}$ with constant $e^R$. If $x,y \in B(0,R) \subseteq X$ (i.e. $\|x\|_X, \|y\|_X \le R$), then in particular $|x_n|,|y_n| \le R$ for all $n \in \mathbb{N}$ (this follows from the formula for $\| \cdot \|_X$), hence 
$$ |\phi(x)_n - \phi(y)_n| \le e^R |x_n-y_n|. $$
Then the formula for $\| \cdot \|_X$ implies 
$$ \| \phi(x) - \phi(y) \|_X \le e^R \|x-y\|_X. $$
If one takes $y=0$, the above reasoning shows that $\phi(y)$ is indeed an element of $X$. Thus $\phi$ is well-defined and continuous. 
