# Theorem 2.41 in Baby Rudin: Is this proof good enough? Can we generalise it?

Here is Theorem 2.41 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

If a set $E$ in $\mathbb{R}^k$ has one of the following three properties, then it has the other two:

(a) $E$ is closed and bounded.

(b) $E$ is compact.

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

The tricky part of the proof is proving that (c) implies (a). Here's my proof:

We need to show that (c) implies (a).

Suppose that every infinite subset of $E$ has a limit point in $E$. We show that then $E$ is both closed and bounded.

Suppose, if possible, that the set $E$ is unbounded. Then there exists a point $x_1 \in E$ such that $\vert x_1 \vert >1$, for otherwise the set $E$ would be contained in the unit closed ball about the origin in $\mathbb{R}^k$.

Using the same reasoning, we can find a point $x_2 \in E$ such that $$\vert x_2 \vert > 1 + \max \left( 2, \vert x_1 \vert \right).$$

Having chosen the point $x_{n-1}$ (where $n \geq 3$), we can choose a point $x_n \in E$ such that $$\vert x_n \vert > 1 + \max \left( n , \vert x_1 \vert, \ldots, \vert x_{n-1} \vert \right).$$ Otherwise, the set $E$ would be contained in a closed ball of radius equal to $1 + \max \left(n, \vert x_1 \vert, \ldots, \vert x_{n-1} \vert \right)$ and centered at the origin.

We have thus inductively chosen a sequence $\{x_n \}_{n \in \mathbb{N}}$ of distinct points of $E$ such that, for each $n \in \mathbb{N}$, we have $\vert x_n \vert > n$ and $\vert x_n \vert > \vert x_i \vert$ for all $i \in \{\ 1, \ldots, n-1 \ \}$.

Let us define the set $S$ as $$S \colon= \{ \ x_n \ \colon \ n \in \mathbb{N} \ \}.$$ This set $S$ is an infinite subset of $E$. We show that this set $S$ has no limit points in $\mathbb{R}^k$ and hence no limit points in $E$.

For any $m, n \in \mathbb{N}$ such that $n > m$, we have $$\vert x_n - x_m \vert \geq \vert x_n \vert - \vert x_m \vert \geq 1 + \max \left( n, \vert x_1 \vert, \ldots, \vert x_{n-1} \vert \right) - \vert x_m \vert \geq 1.$$ Thus it follows that, for any $m, n \in \mathbb{N}$ such that $n \neq m$, the inequality $\vert x_n - x_m \vert \geq 1$ holds.

So if some point $x \in \mathbb{R}^k$ were a limit point of $S$, then there would be infinitely many values $n \in \mathbb{N}$ such that $$\vert x_n - x \vert < \frac 1 4,$$ and for any two (distinct) such points $x_m$ and $x_n$ of $S$, we would have $$\vert x_m - x_n \vert \leq \vert x_m - x \vert + \vert x_n - x \vert < \frac 1 4 + \frac 1 4 = \frac 1 2,$$ which contradicts what we have shown above about the distance between any two distinct points of $S$.

So the set $S$, though an infinite subset of $E$, fails to have a limit point in $\mathbb{R}^k$ and hence in $E$.

Therefore, the set $E$ must be bounded.

Next, suppose that $E$ is not closed. Then $E$ has a limit point $x_0 \in \mathbb{R}^k - E$. Since $x_0$ is a limit point of $E$, every neighborhood of $x_0$ contains a point of $E$ distinct from the point $x_0$ itself (in fact infinitely many points of $E$).

Thus, there is a point $x_1 \in E$ such that $$0 < \vert x_1 - x_0 \vert < \frac 1 2.$$ Again there is a point $x_2 \in E$ such that $$0 < \vert x_2 - x_0 \vert < \min \left( \vert x_1 - x_0 \vert, \frac 1 3 \right).$$ Assuming that the point $x_{n-1}$ (where $n \geq 3$) has been chosen, we can choose a point $x_n \in E$ such that $$0 < \vert x_n - x_0 \vert < \min \left( \vert x_1 - x_0 \vert, \ldots, \vert x_{n-1} - x_0 \vert, \frac{1}{n+1} \right).$$

Thus we have recursively defined a sequence $\{x_n \}_{n\in\mathbb{N}}$ of points of $E$ for which $x_n \neq x_m$ for all $m, n \in \mathbb{N}$ such that $m \neq n$ and also $$0 < \vert x_n - x_0 \vert < \frac 1 n \ \mbox{ for all } \ n \in \mathbb{N}.$$ Let us define the set $S$ as follows: $$S \colon= \left\{ \ x_n \ \colon \ n \in \mathbb{N} \ \right\}.$$ This set $S$ is an infinite subset of $E$.

We show that $x_0$ is the only limit point of $S$. That is, we show that $x_0$ is a limit point of $S$ but no other point $y$ of $\mathbb{R}^k$ can be a limit point of $S$.

Let $\delta$ be any positive real number. Then, by the archimedean property of $\mathbb{R}$, we can find $n_\delta \in \mathbb{N}$ such that $$n_\delta > \frac 1 \delta,$$ and so, for all $n \in \mathbb{N}$ such that $n \geq n_\delta$, we have $$0 < \vert x_n - x_0 \vert < \frac{1}{n+1} < \frac 1 n \leq \frac 1 n_\delta < \delta,$$ which implies that $x_0$ is indeed a limit point of $S$.

Now if $y \in \mathbb{R}^k$ and $y \neq x_0$, then $\vert y - x_0 \vert > 0$. So we can find a positive integer $N$ such that $$N > \frac{2}{\vert y -x_0 \vert}.$$ So, for every $n\in \mathbb{N}$ such that $n \geq N$, we have $$0 < \vert x_n - x_0 \vert < \frac{1}{n} \leq \frac 1 N < \frac{\vert y - x_0 \vert}{2}$$ and hence, for every $n\in \mathbb{N}$ such that $n \geq N$, we have $$\vert x_n - y \vert \geq \vert y - x_0 \vert - \vert x_n - x_0 \vert > \vert y - x_0 \vert - \frac{\vert y - x_0 \vert}{2} > \frac{\vert y - x_0 \vert}{3}.$$ So if we take a positive real number $\epsilon$ such that $$0 < \epsilon < \frac{1}{2} \min \left( \vert x_1 - y \vert, \ldots, \vert x_N - y \vert, \frac{\vert y - x_0 \vert}{3} \right),$$ then there is no point of set $S$ that lies in the neighborhood of the point $y$ of radius $\epsilon$, other than the point $y$ itself if $y \in S$; that is, $$S \cap \left( N_\epsilon (y) - \{ y \} \right) = \emptyset,$$ which implies that the point $y$ cannot be a limit point of the set $S$.

But $y$ was any point of $\mathbb{R}^k$ other than the point $x_0$. Therefore, $x_0$ is the only limit point of the set $S$.

But $x_0 \not\in E$ by our assumption.

Thus we have found an infinite subset $S$ of $E$ such that no point of $E$ is a limit point of $S$. The only limit point of $S$, namely the point $x_0$, (which is also a limit point of the set $E$ by our assumption) does not belong to $E$.

So if every infinite subset of the set $E \subset \mathbb{R}^k$ were to have a limit point in $E$, then the set $E$ must be closed and bounded.

How good is the above proof? Is there any way I can improve it?

What if we replace $\mathbb{R}^k$ by an arbitrary metric space?

Would part (c) of Theorem 2.41 in Baby Rudin still imply (a)? Or vice versa?

• If every infinite subset of E has a limit point in E then every Cauchy sequence in E has a limit point in E, so E is closed. Note that in any metric space, (c) implies (a) and also (b) implies both (a) and (c). But (a) implies (c) iff the metric is complete. There are many complete metric spaces with some closed bounded non-compact subsets. – DanielWainfleet Jan 30 '16 at 22:56
• @user254665 at this point in Baby Rudin, Cauchy sequences have not been introduced. So we have to formulate our proof using only the machinary developed so far in the discussion. – Saaqib Mahmood Jan 31 '16 at 6:43
• OK. But it seems odd to me to be discussing limit points and closed sets in a metric space without using Cauchy sequences or convergent sequences. – DanielWainfleet Jan 31 '16 at 7:08
• I think I've spotted a problem with my proof. The problem arises from the following point on: So if we take a positive real number $\epsilon$ such that $$0 < \epsilon < \frac{1}{2} \min \left( \vert x_1 - y \vert, \ldots, \vert x_N - y \vert, \frac{\vert y - x_0 \vert}{3} \right),$$ – Saaqib Mahmood Mar 13 '16 at 8:05

Your proof is good, but too lengthy.

Suppose $E$ is not bounded. For each positive integer $n$, there is $x_n\in E$ such that $\|x_n\|>n$. I claim that $S=\{x_n:n>0\}$ is infinite. If it is finite, then $M=\max\{\|x_n\|:n>0\}$ exists; but if $n$ is the least integer greater than $M$, we have $\|x_n\|>n>M$: a contradiction.

The set $S$ cannot have a limit point, because every convergent sequence is bounded.

Suppose $E$ is not closed and consider $x$ in the closure of $E$ and $x\notin E$. Then there exists a sequence $(x_n)$ in $E$ converging to $x$. I claim that $S=\{x_n:n>0\}$ is infinite. If it is finite, then $m=\min\{\|x_n-x\|:n>0\}$ exists and $m>0$. Then, by definition of convergence, there exists $n>0$ such that $\|x_n-x\|<m/2$, a contradiction. Therefore $S$ has a limit point, but a convergent sequence in $\mathbb{R}^n$ has a unique limit point, which in this case is $x\notin E$. A contradiction. Therefore $E$ is closed.

This proof of course generalizes to every metric space. However (a)$\implies$(b) does not hold in general metric spaces. Consider an infinite set $X$ with the discrete metric $$d(x,y)=\begin{cases} 0 & \text{if x=y}\\[4px] 1 & \text{if x\ne y} \end{cases}$$ Then every subset of $X$ is closed and bounded, but no infinite subset of $X$ is compact.

I didn't read your proof ( I apologize), but here's my attempt:

Do this by contrapositive. First suppose $E$ is not closed. Then $E^c$ is not open, thus $\exists$ $x \in E^c$ such that $\forall$ $\epsilon>0$, $V_\epsilon(x)\nsubseteq E^c$, where $V_\epsilon(x)$ denotes an open neighborhood of radius $\epsilon$ centered at $x$. Then $\forall$ $n \in \mathbb{N}$ $\exists$ $y_n \in V_{\frac{1}{n}}(x) \cap E - \left\{x \right\}$. Consider $(y_n) \subseteq E$. $(y_n)$ is an infinite sequence in $E$ converging to $x \notin E$, thus, since limit points of real convergent sequences are unique, $(y_n)$ can have no limit point in $E$.

Now suppose $E$ is not bounded. Then $\forall$ $n \in \mathbb{N}$ $\exists$ $y_n \in E$ such that $|y_n| \geq n$. Consider $(y_n)$. Since $|y_n| \rightarrow \infty$, $(y_n)$ has no convergent subsequence (see here). Then $(y_n)$ has no limit points, since this would imply the existence of a convergent subsequence. Then trivially $(y_n)$ is an infinite subset of $E$ with no limit point in $E$.

The proof follows by contraposition.