Q:

Questions concerning a proof that $\mathcal{D}$ is dense in $\mathcal{S}$.

I am currently working through 
this lecture notes
and on page 164, there it is said

The space of $\mathcal{D}(\mathbb{R}^n)$ of smooth complex-valued functions with compact support is contained in the Schwartz space $\mathcal{S}(\mathbb{R}^n)$. If $f_k \to f$ in $\mathcal{D}$, then $f_k \to f$ in $\mathcal{S}$, so $\mathcal{D}$ is continuously embedded in $\mathcal{S}$. Furthermore, if $f\in \mathcal{S}$, and $\eta \in C_c^{\infty}(\mathbb{R}^n)$ is a cutoff function with $\eta_k(x) = \eta(x/k)$, then $\eta_k f \to f$ in $\mathcal{S}$ as $k \to \infty$, so $\mathcal{D}$ is dense in $\mathcal{S}$.

I don't understand the arguments in this paragraph, for a subset $\mathcal D$ of $\mathcal S$ to be dense in $\mathcal S$ for every element $s$ of $\mathcal S$ I need to find a sequence in $\mathcal D$ which converges to $s$, but there just stands that $\eta_k f \to f$ in $\mathcal{S}$, but what i need is a sequence in $\mathcal{D}$ not in $\mathcal{S}$, so why does it follow from this that $\mathcal{D}$ is dense in $\mathcal{S}$?

A:

The sequence $\{\eta_k\}$ is contained in $\mathcal D(\Bbb R^n)$ and we can check that the product of a function in the Schwartz space with a test function is a test function (it's smooth because the product of two smooth functions is smooth, and it has a compact support because it's contained in the support of the test function). 
We have $\mathcal D(\Bbb R^n)\subset \mathcal S(\Bbb R^n)$, and what we want to see is that it's a dense subset for the topology of $\mathcal S(\Bbb R^n)$. So what we have to show is that 
$$\sup_{x\in \Bbb R^n}|x^p\partial^{\alpha}(\eta_kf-f)(x)|\to 0$$
for all integer $p$ and all $\alpha\in\Bbb N^n$. 

