# Heine-Borel Theorem implies compact proof

Let S be a nonempty subset of $\mathbb {R}$ such that H is any open covering of S, then S has an open covering $H^{\sim }$ comprised of ﬁnitely many open sets from H. Show that S is compact.

I got the easy end of the (Heine-Borel theorem) but i got race wheels going in circles here. Satisfying Heine-Borel is a stronger argument (in some spaces i believe) then being compact no?

anyway some butchers work.

i want the argue that S is bounded this should fall in realistically easily ( so it will be terrible) my set S can be covered by the union of finitely many open sets ie finitely many $B(r_1,x_{0})$,$B(r_2,x_{0})$,...,$B(r_n,x_{0})$ where $x_{0} \in S$ since there is finitely many of theses ball's if we $\sum_{0}^{n} r_{i}$ where $i \in \mathbb {N}$ this number must be finite. Since $S\in \mathbb {R}$ there exists a number closest to 0 take the absolute value of this closest number now define a new Ball=G($\sum_{0}^{n} r_{i})$+Absolute Value of number closets to 0) let this be R, thus we have $G(R,0)$ where $R < \infty$ since it is the sum of finite things this new open ball about the origin clearly contains all of S thus S is bounded by definition.

Still need to make Closed. if i can lean ont he fact that S is bounded ( which i don't think above is true) i can make a case argument. since S is bounded $\exists \sup S$ and $\inf S$ Eithier the $\sup S$ and $\inf S$ are limit points ( then we are done ) or they are isolated points

• How did you define compact? The usual definition is the one in terms of open covers. – Michael Greinecker May 9 '13 at 13:56

So I gather you want to show that $S$ is closed and bounded. I'll just work in any metric space $(X,d)$. For boundedness you have the right idea: take the cover $B(x_0, n)$, $n \in \mathbb{N}$, where $x_0$ is any point of $S$ (we can assume $S$ is non-empty or we would be done anyway). This covers $S$, because for any $x \in S$, $d(x, x_0)$ is some positive real number, and we can always find some $k$ with $d(x,x_0) < k$, and then $x \in B(x_0, k)$. Finitely many of these balls cover $S$ by assumption, say $B(x_0, n_1),\ldots,B(x_0, n_k)$. These open balls are nested (the one with the larger radius contains one with a smaller radius) so taking $N = \max(n_1,\ldots,n_k)$ we have $S \subset B(x_0, N)$, so $S$ is bounded.

To see $S$ is closed: suppose $S$ is not closed and $p$ is in the closure of $S$ but not in $S$. Recall that $S$ is closed iff it's equal to its closure $\overline{S}$, where a point $q$ is in $\overline{S}$ when every open ball $B(q,r), r>0$ intersects $S$. As we always have $S \subset \overline{S}$, $S$ is not closed iff there is some point in $\overline{S} \setminus S$, so some point $p$ that is close to $S$, in the sense that every ball around $p$ intersects $S$, but not in $S$ itself. So we are striving for a contradiction here.

Then the open sets $O_n = X \setminus \overline{B(p, \frac{1}{n})}$ cover $S$:

Let $x \in S$ be arbitrary, and as $p \notin S$, $d(x,p) > 0$. We can find some $n$ such that $0 < \frac{1}{n} < d(x,p)$, let $r = d(x,p) - \frac{1}{n}$ and then $B(x, r) \cap B(p,\frac{1}{n}) = \emptyset$, by the triangle inequality, and so $r$ witnesses that $x$ is not in $\overline{B(x, \frac{1}{n})}$ (alternatively we can use that $y \in \overline{B(x,\epsilon)}$ implies $d(y,x) \le \epsilon$ to see this), and so $x \in O_n$. This shows all $x$ in $S$ are covered by the $O_n$.

The $O_n$ have no finite subcover: note that the $O_n$ get larger with increasing $n$ (as the balls around $p$ get smaller and smaller). So if there is a finite subcover $O_{n_1},\ldots,O_{n_k}$ then again $O_N$ with $N = \max(n_1,\ldots,n_k)$ by itself covers $S$, so this would mean that $S \subset X \setminus \overline{B(p, \frac{1}{N})}$ for that $N$. But this implies that $B(p, \frac{1}{N})$ does not intersect $S$ and this means that $p$ is not in $\overline{S}$. This contradiction shows $S$ is closed.

Note that we only use countable covers, so we get the stronger statement that all countably compact (every countable cover has a finite subcover) subsets of $\mathbb{R}$ (any metric space will do, by the way) are closed and bounded.

• The closed argument i don't fully understand but that was a way better way to argue the bounded argument! i sort of understand what your saying in the second argument mostly what i don't get is how this violates closed? it feels like your making a balla round a point inside of S of some small radius then shifting inside that interval towards the point p but 1/n won't let u get there in finite n? ( will up vote later im our for the day) thx for the awesome answer btw. – Faust May 9 '13 at 14:21

To show tha the set is bounded, use the cover $\{(-n,n):n\in\mathbb{N}\}$.

To show it is closed, suppose there is a limit point $x$ that is not in the set and use the cover $\{\mathbb{R}\backslash (x-1/n,x+1/n):n\in\mathbb{N}\}$.