Theorem 2.41 in Baby Rudin

Question

I am reading Theorem 2.41 in Baby Rudin. Rudin proves that in $R^n$ every infinite subset $E$ has a limit point of $E$ implies that $E$ is closed. I understand what he wants to do. For references his proof is here:

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 three 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.
$\qquad$ If $E$ is not closed, then there is a point ${\bf x}_0\in {\bf R}^k$ which is a limit point of $E$ but not a point of $E$. For $n=1,2,3,...,$ there are points ${\bf x}_n\in E$ such that $|{\bf x}_n-{\bf x}_0|<1/n$. Let $S$ be the set of these points ${\bf x}_n$ Then $S$ is infinite (otherwise $|{\bf x}_n-{\bf x}_0|$ would have a constant positive value, for infinitely many $n$), $S$ has ${\bf x}_0$ as a limit point, and $S$ has no other limit point in ${\bf R}^k$. For if $\mathbf y\in{\bf R}^k$, ${\bf y}\neq{\bf x}_0$, then $$\begin{align}|{\bf x}_n-{\bf y}| & \geq |{\bf x}_0-{\bf y}| - |{\bf x}_n-{\bf x}_0| \\ & \geq |{\bf x}_0-{\bf y}|-\dfrac1n\geq \dfrac12|{\bf x}_0-{\bf y}|\end{align}$$ for all but finitely many $n$; this shows that $\bf y$ is not a limit point of $S$ (Theorem $2.20$).
$\qquad$ Thus $S$ has no limit point in $E$; hence $E$ must be closed if $(c)$ holds.

Note that Theorem 2.20 is that if $p$ is a limit point of $E$, then every neighborhood of $p$ contains infinitely many points of $E$.

I have troubled with that "this shows that $y$ is not a limit point of $E$ (Theorem 2.20)." I can't understand how to use Theorem 2.20. I think use the contrapositive of Theorem 2.20, but I don't think above inequality gives us a finite neighborhood.

Also, I read Baby Rudin theorem 2.41 . But I couldn't understand the answer that "We have shown that $\vert x_n - y \vert$ isn't getting smaller and smaller as it should be if $y$ were a limit point of $S$".

Answer 1

Let $\epsilon=\frac12|x_0-y|$, and let $B(y,\epsilon)=\left\{x\in\Bbb R^k:|x-y|<\epsilon\right\}$, the open ball of radius $\epsilon$ centred at $y$. We know that $|x_n-y|\ge\epsilon$ for all but finitely many $n$, so there are only finitely many $n$ such that $|x_n-y|<\epsilon$. That is, there are only finitely many $n$ such that $x_n\in B(y,\epsilon)$, and $B(y,\epsilon)$ is therefore a nbhd of $y$ that does not contain infinitely many points of $S$. By the contrapositive of Theorem $2.20$, therefore, $y$ is not a limit point of $S$.

The language that you quote from the linked answer is a bit sloppy, as it makes it sound as if the points $x_n$ ought to be a sequence converging to $y$. That’s not true, but it is true that if $y$ were a limit point of $S$, there would be points of $S$ as close to $y$ as we like. Rudin shows that this isn’t true: there are only finitely many points of $S$ closer to $y$ than $x_0$ is, and if we let $f$ be the distance from $y$ to the closest of these points, then no point of $S$ is within $\frac{r}2$ of $y$.

Answer 2

Actually, it is saying that a sequence of points in $\Bbb R^k$, it converges, has only one limit point. The way that Rudin has chosen is to show that if the sequence converges to one point then its members can not approach another point other than the limit point.

To see this in other ways, suppose $y$ is another limit point of a sequence $\{x_n\}$, then : $$\forall \epsilon > 0, \exists M_1 , M_2 \in \Bbb N ,\forall n_1 \geq M_1 \hspace{0.2cm} \wedge \forall n_2 \geq M_2 (|y - x_{{n_{1}}}| \leq \epsilon \wedge |x_0 - x_{n_{2}}| \leq \epsilon).$$

Now let $M = \max(M_1,M_2)$. So for every $ n \geq M$ we have $|y-x_{0}| \leq \epsilon.$ thus $ y = x_0$.