I'm currently reading the Baby Rudin and I've trouble with understanding the proof of the theorem 2.41.

Theorem 2.41. If a set $E$ in $R^k$ has one of these three properties, it has the other two:

$(a)$ $E$ is closed and bounded;

$(b)$ $E$ is compact;

$(c)$ Every infinite subset of $E$ has a limit point in $E$.

In (c)->(a), the proof construct a set $S$. Assuming $E$ is not bounded, then $E$ contains points $x_n$ with $\left|x_n\right|>n(n=1,2,3,...)$. The proof then states "The set $S$ consisting of these points $X_n$ is infinite and clearly has no limit point in $R^k$, hence has none in $E$. "

My question is why $S$ has no limit point in $R^k$?


Suppose that $S$ has a limit point $x$. Note that there exists an integer $N > |x|$. It follows that whenever $n>N$, we have $$ |x_n - x| \geq |x_n| - |x| \geq N - |x| $$ So, the open ball around $x$ of radius $(N - |x|)$ contains at most finitely many points $x_n$, which means that $x$ cannot be a limit point after all.

  • $\begingroup$ Thanks! That makes sense! $\endgroup$ – Johnny Ji Jun 28 '17 at 15:12

Suppose that $S$ has a limit point $l$. Then there is a subsequence $(x_{n_k})$ of $(x_n)$ , which converges to $l$. Hence $(x_{n_k})$ is bounded. But from

$ |x_{n_k}|>n_k$ for all all $k$,

we get a contradiction.

  • $\begingroup$ You mean there is a $r_k$ for each $x_{n_k}$ that $\left|x_{n_k}-l\right|<r_k$, so $x_{n_k}$ is bounded, right? $\endgroup$ – Johnny Ji Jun 28 '17 at 15:20
  • 1
    $\begingroup$ @JohnnyJi no; at least, your statement isn't a proof of what we want unless you say something else about $r_k$. We can conclude that $x_{n_k}$ is bounded as follows: there exists a $K$ such that $k>K$ implies $|x_{n_k} - l| < 1$, which means that for all $k>K$, we have $$ |x_{n_k}| \leq |l| + |x_{n_k} - l| < |l| + 1 $$ which means that $x_{n_k}$ is bounded. $\endgroup$ – Omnomnomnom Jun 28 '17 at 15:31
  • $\begingroup$ Great! Thanks very much! $\endgroup$ – Johnny Ji Jun 28 '17 at 17:39

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.