# Baby Rudin Theorem 2.14 (c) ->(a)

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.

• Thanks! That makes sense! – 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 $k$,

• 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? – Johnny Ji Jun 28 '17 at 15:20
• @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. – Omnomnomnom Jun 28 '17 at 15:31