Q:

Show $\{x\}$ is a bounded sequence

Was reviewing the lecture notes for the test. My math gets rusty after a few years away from the college, so this might be a dumb question. The following statement, the bold part, I didn't know why we can reach such a conclusion given the information.
Let $m:= inf\{f(x):x \in \mathbb{R}^n\}$ and take a sequence $\{x_k\}$ such that $f(x_k) \rightarrow m$ as $k \rightarrow \infty$. Since $f$ is coercive, $\{x_k\}$ must be bounded; otherwise it has a subsequence $\{x_{k_j}\}$ with $\|x_{k_j}\| \rightarrow \infty$ as $j \rightarrow \infty$, and hence $m=lim_{j \rightarrow \infty} f(x_{k_j})= + \infty$, a controdiction.
Though the explanation has been given in following sentence, but I didn't get it, what's the contradiction here?
Note: A function $f$: $\mathbb{R}^n \rightarrow \mathbb{R}$ is coercive if $f(x) \rightarrow +\infty$ as $|x| \rightarrow \infty$.

A:

We have $m:= inf\{f(x):x \in \mathbb{R}^n\}$ 
If $m= +\infty$ , we would have $ + \infty \le f(x)$  for all $x$. But this is absurd.

