How are manifolds embedded into an euclidean space? I am really new to differential geometry and topology and I am trying to understand how it is possible to embed a minfold into an euclidean space.
In particular, I am looking at projective spaces, but as I have already said I am really new, so I don't know if it is really realated.
So, my question is really short and it is:
How are manifolds embedded into an euclidean space?
If possible, I would like to have a more intuitive explanation of how this happens.
 A: A $n$-manifold is, by definition, a (second countable, Hausdorff) topological space $M$ equipped with a atlas $\{(U_i,f_i)\}$ where $\{U_i\}$ is an open cover of $M$ and $f_i:U_i\to V_i$ is a homeo from $U_i$ to an open sub-set of $\mathbb R^n$. If one is interested in  differentiable manifolds, then we have to add the requiremente that $f_j\circ f_i^{-1}$ is a diffeomorphisms from $f_i(U_i\cap U_j)$ to $f_j(U_i\cap U_j)$).
Now, suppose for simplicity that the manifold is compact, so we can always extract a finite altas from any atlas.
Given such a finite atlas, build a partition of unity $\{g_i\}$, subordinate to the finte cover $\{U_1,\dots, U_k\}$.
Then any map $f_ig_i$, which is a priori defined only on $U_i$, extends to the whole $M$, by setting it zero outside $U_i$.
Then the map $F:M\to \mathbb R^{kn+k}$ given by
$$F(x)=(f_1g_1(x),...,f_kg_k(x),g_1(x),...,g_k(x))$$
is an embedding.
A similar construction can be made also if $M$ is not compact. You need a locally finite atlas, which always exists, but whose existence is not trival.

EDIT: How to proof that $F$ is an embedding.
$F$ is clearly continuous. Let's see that it is injective. If $F(x)=F(y)$ then in particular $g_i(x)=g_i(y)$ for all $i$. Since $\{g_i\}$ is a partition of unity there is $i$ so that $g_i(x)\neq 0$. In particular $x,y$ both belong to $U_i$. But now from $f_ig_i(x)=f_ig_i(y)$ we deduce $f_i(x)=f_i(y)$, whence $x=y$ because $f_i$ is injective.
So $F:M\to F(M)$ is a bijective, continuous map from a compact set to a Hausdorff space, hence it is a homeo.
In case $M$ is a differentiable manifold, ane can also easily compute $dF$ and see that it always has maximal rank, so $F$ is in fact a smooth embedding.
A: For an intuitive picture, it may be helpful to start with submanifolds (https://en.wikipedia.org/wiki/Submanifold), i.e. subsets of Euclidean space defined by some equations.
Maybe the simplest example would be the unit circle in the plane: Let $x$ and $y$ be the coordinates of $\mathbb{R}^2$, then the circle $S^1$ can be defined by $$x^2+y^2=1\,.$$
In this case, it is straightforweard to choose a coordinate chart that covers $S^1$, except at a single point, and write $x=\cos\phi$, $y=\sin\phi$, where $\phi\in\left(0,2\pi\right)$. You can choose a second chart where $x=\cos\alpha$ and $y=\sin\alpha$, where $\alpha\in\left(-\pi,\pi\right)$. On the overlap of the coordinates, the coordinates are related by $\alpha=\phi$ or $\alpha=\phi-2\pi$.
