Exercise 6 in section 50 in Topology textbook of Munkres. I am looking for a reference of the proof of the following thoerem in Munkres it's exericse 6 in section 50 page 315 and it goes as follows:
Let $X$ be a locally compact Hausdorff space with a countable basis, such that every compact subspace of $X$ has topological dimension at most $m$. Then $X$ is homeomorphic to to a closed subspace of $\mathbb{R}^{2m+1}$.
I am giving a sort of guidance assignments, but I got stuck on assignment b), it asks me to show that if $f(x)\to \infty$ as $x\to \infty$ then $f$ extends to a continuous map of one point compactifications. Conclude that if $f$ is injective as well then $f$ is homeomorphism of $X$ with a closed subspace of $\mathbb{R}^N$; where for $f: X \to \mathbb{R}^N$ is a continuous map, we say that $f(x)\to \infty$ as $x\to \infty$ if given $n$, there's a compact subspace $C$ of $X$ s.t $f(x)>n$ for $x\in X\setminus C$.
So my thoughts about this b) assignment:
I am stumpped, do you have hints or advice on how to prove it?
If you have an official reference even better, thanks.
 A: Well the answer is straight forward, and you don't need any reference as it is only several lines long (the Alexandrov compactification is indeed discussed in Munkres' if I recall correctly, probably in a previous section).
The one-point compactification of $\mathbb{R}^{m}$ is a suitable sphere (by stereographic projection), but it will be better for our purposes to think about it just as a set of $\mathbb{R}^{m}$ with the additional point at infinity.
Just define $\hat{f}$ to be $f$ on $X$ and $\hat{f}(\infty)=\infty$.
This is a continuous mapping, by immediate verification, the only thing left to prove is that the pre-image of a nhbd of $\infty$ in $\mathbb{R}^{m}$ is open in $\hat{X}$, but this is true by definition of nhbds of $\infty$ in the Alexandrov topology (I'm assuming you're using the actual definition of escape from compact sets, your definition doesn't make sense for function to general Euclidean spaces).
For the second part, if $f$ is injective, then $\hat{f}$ is injective as-well, so $\hat{f}$ is homeomorphism between $\hat{X}$ with its image by the usual argument of compact to Hausdorff bijections.
Hence we can identify the image of $X$ with $Y=\hat{f}(\hat{X})\setminus \{\infty\}$. So we just need to verify that $Y$ is closed. Let $y_{n}\to y_0$ be a sequence in $Y$, by pull-backing we get $\{x_{n}\}\subset \hat{X}$, by compactness we can assume $x_{n}$ converges to $x_{0} \in \hat{X}$, as $y_{0} \in Y$, we can assume that $\{y_{n}\}$ were inside a bounded set, hence $x_{n}$ are bounded and $x_0$ is bounded, and in particular, $x_0 \neq \infty$, therefore $y_0=f(x_0)$ is not $\infty$ as well.
