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

  • 1
    $\begingroup$ How did you define compact? The usual definition is the one in terms of open covers. $\endgroup$ – 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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – 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 $\{\mathbb{R}\backslash (x-1/n,x+1/n):n\in\mathbb{N}\}$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.