The expression of the title seems to assert that $\Bbb R$ is a subset of $\Bbb R^n$ but I cant see how this make sense, it is surely some kind of convention that I dont know.
I must suppose that $\Bbb R$ is the set of $x\in\Bbb R^n$ such that $x=\langle x,0,\ldots,0\rangle$ or something like this?
Can someone clarify this question? Thank you.
On
As was mentioned in the comments, if you take the embedding where you map $\mathbb{R}$ to one of the canonical coordinates of $\mathbb{R}^n$, the image is a closed set. However, it's important to note that not all embeddings of $\mathbb{R}$ into $\mathbb{R}^n$ have this property!
A good example is embedding $\mathbb{R}$ into $\mathbb{R}^2$ (and by extension $\mathbb{R}^n$ where $n \geq 2$) as the unit circle about the origin missing one point. Since the missing point is a limit point, the set is not closed.
The easiest way to see this is possible is to use $f(x) = 2 \tan^{-1}(x)$ to map the entire real line to the interval $(-\pi, \pi)$ bijectively, and then interpreting this as an angular coordinate, fixing the radial coordinate to $1$. The final embedding $f : \mathbb{R} \to \mathbb{R}^2$ would then be:
$$ f(x) = \langle \cos(2 \tan^{-1}(x)), \sin(2 \tan^{-1}(x)) \rangle $$
in rectangular coordinates, and the missing point would be $\langle -1, 0 \rangle$.
Ok, from the context of the book the meaning of "$\Bbb R$ is closed in $\Bbb R^n$" it is what I supposed it would be. Let define the projection
$$\pi: \Bbb K^n\to \Bbb K^{n-j},\; \langle a_1,a_2,\ldots,a_n\rangle\mapsto \langle a_{j+1},\ldots,a_n\rangle$$
then $\pi^{-1}(0)=\Bbb K^{j}$, under this "embedding" of $\Bbb K^j$ in $\Bbb K^n$ we have that
$$\Bbb K^j\text{ in }\Bbb K^n:=\Bbb K^j\times\underbrace{\{0\}\times\{0\}\times\cdots\times\{0\}}_{n-j\text{ times}}$$
P.S.: I opened the question with the hope that this convention could be standard, this is the reason why in first place I dont provided more details, sorry for the inconveniences.