Baby Rudin Theorem 2.41 boundedness

Question

I'm currently reading the POMA of Rudin and I don't understand the proof of boundedness of the theorem 2.41 (p. 40) of the book.

It wants to prove that if $E \subset \mathbb{R}^k$ we have that every infinite subset of $E$ has a limit point in $E$ implies that $E$ is bounded and closed. I have some troubles in understanding what the set $S$ is...

Is $S$ the $\mathbb{R}^k$ complement of the open ball centered around 0 of radius 1? If someone could explain me its proof this would be very nice,

Thanks

From the book: "If $E$ is not bounded, then $E$ contains points $\textbf{x}_n$, with $|\textbf{x}_n| > n (n=1, 2, 3)$. The set $S$ consisting of these points $\textbf{x}_n$ is infinite etc." — yarmenti, Aug 31 '15 at 14:14

score 1 · Accepted Answer · answered Aug 31 '15 at 14:02

1

Because $E$ is unbounded, for each positive integer $n$ there is an $x_n\in E$ such that $|x_n|>n$.

$$S=\{x_1,x_2,\ldots,x_n,\ldots\}.$$

Does this clear things up?

answered Aug 31 '15 at 14:02

Tim Raczkowski

20,046

Ok, so you construct the sequence $S$ as you mentionned (which is discrete and infinite). As it is discrete, we can find for a given $p \in \mathbb{R}^k$ a neighborhood which does not contain any element of $S$ and thus it is finished? – yarmenti Aug 31 '15 at 14:04
Yes, that's correct. – Tim Raczkowski Aug 31 '15 at 14:05
Sorry I edited my previous comment – yarmenti Aug 31 '15 at 14:10

Baby Rudin Theorem 2.41 boundedness

1 Answers1