Show by an example that a proper closed subset of $\mathbb{R}$ in the sandard (order) topology need not be compact.

Question

In my mind, I can think of below example which seems to work.

If $(X,T) = \mathbb{R}$, and $A = (0,\infty)$, then as far as I know it comes in the standard (order) topology of $\mathbb{R}$, but what I dont know is whether it will be a proper closed subset of $\mathbb{R}$. If it works, then it will be perfect b/c for all positive integer $i$, if I let $P_i$ be the open interval $(0,i)$, then clearly $A \subset \cup$ $O_i$, where $i = 1$ to $\infty$,

however there doesnt exist $i_1, i_2,......, i_n$, such that $A \subset (0,i_1),(0,i_2),....,(0,i_n)$, therefore by the definition of compactness we can see that we dont have a finite sub-cover, so its not compact.

Kindly check my proof, let me know if there is anything wrong. Also, give it better style and notation if required.

Answer 1

The set $(0,\infty)$ is not closed. But $[0,\infty)$ will work. For instance, the sequence $1,2,3,\ldots$ has no convergent subsequence. Or you can say that $\{[0,n)\mid n\in\mathbb N\}$ is an open cover with no finite subcover.

Answer 2

Noting that any bounded closed subset of $\mathbb{R}$ is compact, so we need to think of unbounded closed subsets like: $$[a,\infty), \mathbb{N}, \ldots$$

Answer 3

Let $A = [0, \infty)$, then its complement $A^c = (-\infty, 0)$ is an open interval, and hence $A^c$ is an open set. Thus, by definition, $A$ is a closed set.

To show it is not compact, consider the following open cover: $\{(-n,n)| \, \, n \in \Bbb{N}\}$. This open cover in fact covers all of $\Bbb{R}$, so in particular covers $A$. Clearly, there is no finite subcover. Hence, $A$ is not compact.

Answer 4

By Heine-Borel, a compact subset of $\Bbb R^n$ is closed and bounded.

So take any closed, unbounded set.