0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Alright, thank you! $\endgroup$ – S Y May 19 '19 at 10:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.