Consider the set $S=\{x\in\mathbb{R}\,|\,0<x\leq1\}$. Give an example of an open cover of $S$ that has a finite subcover. Also, show that $S$ is not compact by giving an example of an open cover that does not admit a finite subcover. Justify your answer.
Can anyone help me with this? Please explain the intuition of finding it too.
Thank you so much!
Best Answer
A trivial cover with a finite subcover, is any cover that's finite to begin with, so $\{(0,2)\}$ (if we cover with open sets of $\Bbb R$) or $\{S\}$ if we cover by open sets of $S$ itself; it doesn't matter for the definition of compactness of a subspace which type we use).
One without a finite subcover: $\{(\frac{1}{n}, 1]\}_{n \in \Bbb N^+}$ or $\{(\frac{1}{n}, 2)\}_{n \in \Bbb N^+}$, depending on taste. In a finite subfamily a maximal $n$ is used and then $\frac{1}{n+1}$ for that $n$ cannot be covered. So no finite subcover exists.