troubles to understand the embedding theorem

Question

I try to understand the proof of the embedding theorem:

"An $m-$dimensional differentiable manifold can be embedded as a closed subset of the Euclidean space $\Bbb R^n$, if $2m<n$."

The proof is given as follow: We know that we can find for a manifold a proper function $f:M^m \to \Bbb R^n$ for an $n>0$ and that we can approximate this by an injective immersion $g:M^m \to \Bbb R^n$, so that $\vert g - f \vert \leq 1$ and $A= \emptyset$. If $K \subset \Bbb R^n$ is compact, then $K \subset K(r)$ for some radius $r$, hence $g^{-1}(K)$ is closed in the compact set $f^{-1}(K\overline{(r+1)})$, hence compact. Therefore $g$ is proper, hence an embedding.

Now I have some questions in this proof:

1. Why we have that if $K \subset \Bbb R^n$ is compact, then $K \subset K(r)$ for some radius $r$?

2. Why it follows that $g^{-1}(K)$ is closed in the compact set $f^{-1}(K\overline{(r+1)})$?

3. Why we can now conclude that a proper is an embedding?

EDIT:

4. I guess, he claims that $g^{-1}(K) \subset f^{-1}(K\overline{(r+1)})$ to conclude that $g^{-1}(K)$ is compact. But Why is this inclusion right? I guess that it comes from the fact, that $\vert g - f \vert \leq 1$. But I am not sure.

Many thanks for your help!

Answer 1

1. Compact sets in $\mathbb{R}^n$ are bounded.

2. Preimage of a closed set by continuous $g$ is closed, and preimage of a compact set by proper $f$ is compact.

3. Proper injective immersion is exactly an embedding with closed image.

4. $g^{-1}(K)\subseteq f^{-1}(K(r+1))$ if and only if $[g(x)\in K\implies f(x)\in K(r+1)]$. But $g(x)\in K\implies g(x)\in K(r)$ i.e. $|g(x)|<r$ and by triangle inequality $|f(x)|\leq |g(x)|+|f(x)-g(x)|<r+1$, so indeed $[g(x)\in K\implies f(x)\in K(r+1)]$. Thus $g^{-1}(K)\subseteq f^{-1}(K(r+1))$ and $g^{-1}(K)\subset f^{-1}(\overline{K(r+1)})$.