Question:
Show that the open rectangle $(a_1, b_1)×(a_2, b_2)$ is not compact in
$\mathbb{R}^2$ to the standard Euclidean metric by finding an open covering
that has no finite subcovering.
How to go about constructing this open covering that has no finite subcovering? What would be an example?
Additionally, why can't I argue directly that a set is compact $\Leftrightarrow$ it is closed and bounded, and since the open rectangle is not closed, it is not compact?
Best Answer
You can definitely argue directly by using the Heine-Borel theorem (since it is not closed, it cannot be compact). It’s just that the problem asks you to argue by showing directly it does not fulfill the definition of compact.
An example of an open cover which doesn’t have a finite subcover is the collection $\{U_n\}_{n\in\mathbb{Z}^+}$, where $$U_n = (a_1+1/n,b_1)\times(a_2,b_2).$$ Clearly, $\bigcup_{n\in\mathbb{Z}^+} U_n=(a_1,b_1)\times(a_2,b_2)$. On the other hand, observe that $U_n\subset U_{n+1}$ for all $n$. Therefore, for any finite subcover $\{U_{n_1},\ldots,U_{n_r}\}$, consider $m=\max\{n_1,\ldots,n_r\}$. We have that $\bigcup_{k=1}^{r} U_{n_k} = U_m\subsetneq (a_1,b_1)\times(a_2,b_2)$.