Questions about coercive functions and its implications Given this definition: A function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is $coercive$ if 
$$\lim_{||x||\rightarrow\infty}f(x) = \infty.$$
Explicitly, this means that for any $M>0$ there is an $R>0$ such that $||x||\geq R$ implies $f(x)\geq M$.
I understand this definition, it is saying if the norm of a vector in $\mathbb{R}^n$ becomes large then the functions values lie outside of $M$. 
My questions are:
1) Does this mean that for $||x||<R$ implies $f(x)<M$? 
2) This may seem trivial to most but I was told that $||x||\leq R$  is a compact set in $\mathbb{R}^n$. I am not seeing this because of my first question. My perspective is that if $||x||\geq R$ implies $f(x)\geq M$ that the $||x||<R$ implies $f(x)<M$. Which is not closed.
3) If $||x||\leq R$ is indeed a compact set how does it guarantee that $||x||=R$ will not have a function value $f(x)>M$?
If anyone could help me clear my confusion I would greatly appreciate it!
Thank you for any help and comments. 
 A: *

*No. The condition only tells you information about the behavior of $f(x)$ when $\|x\|$ is large.
For example: If $f$ is coercive and you choose $M$ and an $R$ corresponding to $M$ like you described, then define $g: \mathbb{R}^n \to \mathbb{R}$ by
$$g(x) = \begin{cases}f(x)&\text{ if $\|x\| \geq R$}\\
M &\text{ if $\|x\| < R$}.\end{cases}$$
Then $g$ is coercive but $g(x)$ is never less than $M$.

*The Heine-Borel theorem says that $K \subset \mathbb{R}^n$ is compact if and only if $K$ is closed and bounded. Certainly $B = \{x : \|x\| \leq R\}$ contains its limit points (i.e., is closed), and by definition $B$ is bounded. So yes, $B$ is compact.

*If I'm understanding your point of confusion, the quick answer is that, in your interpretation of the meaning of the limit, there are infinitely many possible choices for $R$ but there's never a minimal one. (If you replace "$\|x\| \geq R$" with $\|x\| > R$ there is, though.) If you want to choose $M$ and $R$ so that $f(x) > M$ whenever $\|x\| \geq R$, then you can select any $R$ such that $R > R' := \inf\{S \mid \forall x, \|x\| \geq S \implies f(x) > M\}$ but you can't set $R = R'$ (since there may be $x$ with $\|x\| = R'$ such that $f(x) = M$).
For example, $f(x) = x^2$ is coercive. For $M = 100$, I know that for any $R > 10$ it's true that $f(x) > 100$ if $|x| \geq R$, but the claim that $f(x) > 100$ if $|x| \geq R$ is false if I set $R = 10$. Make sense?

In response to your comment$\ldots$
For any compact $K \subset \mathbb{R}^n$, a continuous function $f: K \to \mathbb{R}$ attains its infimum and supremum (i.e., has a minimum and maximum) on K. (This is one of the many versions of the Extreme Value Theorem, and of course it also implies a continuous $f:\mathbb{R}^n\to\mathbb{R}$ has a minimum and maximum on each compact subset of $\mathbb{R}^n$.) The same is true of lower semi-continuous functions, except that they only attain their infima on compact sets. As a rule, if you're function $f$ is continuous or satisfies a property similar to continuity, it's going to be easier to study maxima and minima of $f$ on compacta.
So that's one thing that definitely motivates restricting to a compact subset of $\mathbb{R}^n$ for part of the proof here. The other motivation is that, while $f$ being coercive doesn't tell you anything about the behavior of $f$ on the balls $B_r = \{x:\|x\| \leq r\}$ directly, it is a statement about their complements; a function $f: \mathbb{R}^n \to \mathbb{R}$ is coercive iff for all $M > 0$ there exists $R = R_M > 0$ such that $x \in \mathbb{R}^n\setminus B_R \implies f(x) > M$. Note that this isn't exactly the same as your restatement of the definition, but it's equivalent. The only practical difference is that the way I've stated the definition allows me to pick $R=R'$ (from point 3 above). (Also, there was a typo in the way I defined $R'$ in point 3 that I've now corrected.)
You can see how the proof should form just using these two facts. If $M \in \mathbb{R}$ and $R'$ is the infimum in point 3 corresponding to $M$, I can take $R = R'$ and observe that $f(x)$ must be less than or equal to $M$ for some $x \in B_R$ (otherwise $R'$ wouldn't be the infimum). Since $f(x) > M$ for all $x \in \mathbb{R}^n\setminus B_R$, this tells me that $\inf\{f(x) : x \in \mathbb{R}^n\} = \inf\{f(x) : x \in B_R\}$. Now I can use the fact that $f$ is lower semi-continuous and the compactness of $B_R$ to conclude $\inf f$ exists and equals $f(x)$ for some $x \in B_R$.
