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.


1 Answer 1


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

  • $\begingroup$ Alright, thank you! $\endgroup$
    – Yeo
    May 19, 2019 at 10:50

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