May I please receive help proving the following? Thank you

$\def\R{{\mathbb R}} \def\p{{\bf p}}$ Let $E\subseteq\R^n$.
(i) Prove if $E$ satisfies the Heine-Borel Property, then $E$ is bounded.

$\textbf{Proof:}$ Consider a compact subset $E\subseteq \R^n$. Now, let $G= \{(-n,n) | n\in \R^n\}$ be an open cover of $E$. However, $E$ is compact. So, $G$ must have a finite subcover as $$G' = \{(-n_1,n_1), (-n_2,n_2), \dots (-n_k, n_k)\}$$ which also covers $E$.

Let us define a maximum as $m = \max\{(-n_1,n_1), (-n_2,n_2), \dots (-n_k, n_k)\}$. So, $E \subseteq (-m,m)$ and means that the maximum open set in the finite subcover will contain the set $E$. So we can say $|e| < m$ for all $e$ in $E$. Thus, we can say that $E$ is bounded.

(ii) Prove if $E$ satisfies the Heine-Borel Property, then $E$ is closed.

$\textbf{Proof:}$ Consider a compact set $E\subseteq\R^n$. Now, assume a point $\p$ that is not in $E$ and instead is in the complement of $E$ as $\p\in E^\complement$. Recall, a point, or a singleton set, can be represented as the intersection of closed balls, or $\displaystyle{\p = \bigcap_{r>0} \overline{B_r} (\p)}$ where $r$ is radius of the closed ball.

As $E$ is compact, there is a finite subcover $\displaystyle{\bigcup_{a \in F} \mathcal{O}_a}$, for every open cover $\mathcal{O} = \{\mathcal{O}_a | a \in A\}$ in $E$, that covers $E$. So, $\displaystyle{E_k \subseteq \bigcup_{a\in F} \mathcal{O}_a}$, $E$ lies in the finite union of open balls.

Now, consider $\p \in E^\complement$. The singleton set $\{\p\}$ can be represented as the intersection of closed balls, or $\displaystyle{\bigcap_{r>0} \overline{B_r}(p)}$ where $r$ is radius of the closed ball. Use De Morgan's law as $\displaystyle{A^\complement = (\cup B)^\complement = \cap(B^\complement)}$ where $A$ and $B$ are any sets. Thus, $\p \in E$.

Now, consider $E$. $E$ can be covered by complements of closed balls of $\p$. Let us consider the complements so we get $$E^\complement = \left( \bigcap_{r>0} \overline{B_r}(\p) \right)^\complement = \bigcup_{r>0}\left(\overline{B_r}(\p) \right)^\complement.$$ We know that a complement of a closed set gives an open set. So, it can be seen that $\p^\complement$ is the union of finite open sets. Thus, $E^\complement$ is open. Complement of an open set is closed. So, we get $E$ is a closed set.

  • $\begingroup$ A comment on notation: We usually use the term Heine-Borel property to refer to a property of metric spaces. A space satisfies the Heine-Borel property if closed and bounded sets are compact. I have not seen it used to refer to compact sets before. $\endgroup$ – K.Power Apr 2 '20 at 4:16
  • $\begingroup$ @K.Power $A$ has the Heine-Borel property iff every open cover of $A$ has a finite subcover. The Heine-Borel theorem says that in $\Bbb R^n$ in the usual metric $A$ has the Heine-Borel property iff $A$ is closed and bounded. In a general metric space we have to replace the last property by "complete and totally bounded" instead. $\endgroup$ – Henno Brandsma Apr 2 '20 at 6:26
  • $\begingroup$ @HennoBrandsma That is just the definition of compactness though? I have never before seen the Heine-Borel property used in place of compact before. I have always seen it used in the context of metric spaces, as defined in the Wikipedia article. For example the statement: "no infinite dimensional Banach space has the Heine-Borel property" $\endgroup$ – K.Power Apr 2 '20 at 12:14
  • $\begingroup$ @K.Power That's a derived property from the aforementioned Heine-Borel theorem: a space has the Heine-Borel property iff the Heine-Borel theorem holds in it. The property is an historical name given before there was an agreed upon notion of compactness. In the early 20th century people in different countries called different spaces "compact" (or "bicompact"). See the "Handbook of the history of general topology" , e.g. Some used it only for countable covers, others for limit point compact, others again for sequentially compact (the Bolzano-Weierstraß property) etc. $\endgroup$ – Henno Brandsma Apr 2 '20 at 13:17
  • $\begingroup$ @HennoBrandsma I did not know that! Thank you for teaching me something new. $\endgroup$ – K.Power Apr 2 '20 at 14:18

You were close on the closed-part: If $E$ is not closed, it has a limit point $p$ that is not in $E$. Write $\{p\} = \bigcap_n D(p, \frac1n)$ the intersection of closed balls, just as you did.

It follows that $E \subseteq X\setminus \{p\} \subseteq \bigcup_n D(p, \frac1n)^\complement$ by de Morgan, so we have a countable open cover of $E$, which should have a finite subcover $D(p, \frac{1}{n_1})^\complement,\ldots,D(p, \frac{1}{n_k})^\complement$. Let $N > \max(n_1,\ldots, n_k)$ so that $\frac{1}{N}$ is smaller than all used radii in the finite cover. Then, as $p$ is a limit point, we have some $x \in B(p, \frac{1}{N}) \cap E$. But then $x \in B(x, \frac{1}{N}) \subseteq D(p, \frac{1}{n_i})$ for all $i$ and so is not covered by the supposedly finite subcover, contradiction. QED.

As an alternative: let $p \notin E$. Then for each $x \in E$ we take $O_x:=B(x, r_x)$ with $r_x=\frac{d(x,p)}{2} >0$. Then $\{O_x: x \in E\}$ is an open cover of $E$, so has a finite subcover $O_{x_1},\ldots, O_{x_n}$. Then setting $r=\min(r_{x_1},\ldots,r_{x_n})>0$ we find (check this!) that $B(p,r) \cap E = \emptyset$, showing that $E$ is closed, as $p \notin E$ was arbitrary.

For the boundedness, it's easier to take $p \in E$ and as the cover all balls $B(p,n), n \in \Bbb N$ which has a finite subcover. The ball with largest radius contains all of $E$ and so $E$ is bounded (e.g.the diameter is $\le$ two times that radius, or whatever your definition of a bounded subset is). Your notation $(-n,n): n \in \Bbb R^n$ makes no sense. Maybe you meant $(-m,m)^n \subseteq \Bbb R^n$, which would be OK (and $m$ runs to infinity).

  • $\begingroup$ Thank you Henno for the explanation. Yes, I mean $(-m,m)^n\subseteq \mathbb{R}^n$ but everything else in the bounded portion is properly indicated. Apologies for the notation error. $\endgroup$ – rudinsimons12 Apr 2 '20 at 13:28

The Heine-Borel theorem is that a set in $\mathbb{R}^N$ is compact iff it is closed and bounded. Presumably you want a proof that if $E$ is compact then it must be bounded, because if $E$ is assumed to be closed and bounded you would be done?

Suppose $E$ is compact but unbounded. If the set is unbounded, there is a sequence $\{x_n\}$ in $E$ that has no convergent subsequence in $E$ (start with an open ball of radius $r_1$ around the origin large enough to include at least one point of $E$ and pick a point of $E$, then pick a point of $E$ from the ball of radius $r_2> r_1$ large enough to include additional points of $E$ complement the ball of radius $r_1$, and so on --- because $E$ is unbounded you will never run out of such balls and this sequence diverges.). But this contradicts the Bolzano-Weierstrass theorem, which is that compact sets in $\mathbb{R}^N$ are sequentially compact, and all sequences in compact sets have convergent subsequences. Therefore $E$ must be compact.

  • $\begingroup$ Thank you, Renard :) $\endgroup$ – rudinsimons12 Apr 2 '20 at 13:32

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