# Baby Rudin Proof for Theorem 2.41 - help with first part

Below is the first part of the proof. I cannot seem to visualise this out. What does $$|{\bf x}_n|>n$$ actually mean in a diagram?

2.41 $$\ \$$ Theorem $$\ \$$ If a set $$E$$ in $${\bf R}^k$$ has one of the following three properties, then it has the other two:

$$\quad(a)\ \$$ $$E$$ is closed and bounded.
$$\quad(b)\ \$$ $$E$$ is compact.
$$\quad(c)\ \$$ Every infinite subset of $$E$$ has a limit point in $$E$$.

Proof $$\ \$$ If $$(a)$$ holds, then $$E\subset I$$ for some $$k$$-cell $$I$$, and $$(b)$$ follows from Theorems $$2.40$$ and $$2.35$$. Theorem $$2.37$$ shows that $$(b)$$ implies $$(c)$$. It remains to be shown that $$(c)$$ implies $$(a)$$.
$$\qquad$$ If $$E$$ is not bounded, then $$E$$ contains points $${\bf x}_n$$ with $$|{\bf x}_n|>n\qquad(n=1,2,3,...).$$ The set $$S$$ consisting of these points $${\bf x}_n$$ is infinite and clearly has no limit point in $${\bf R}^k$$, hence has none in $$E$$. Thus $$(c)$$ implies that $$E$$ is bounded.

The radial distance of the point $${\bf x}_n$$ from the origin is $$> n$$. This is what $$|{\bf x}_n|>$$ means.