Is the closure of $(0,+\infty) \cap (0, a)$ compact in $(0,+\infty)$?

Question

Consider the half-line $(0,+\infty)$ as a topological subspace of $\mathbb{R}$. Suppose I consider an open interval $(0,a)$ for some $a > 0$. The set $(0,+\infty) \cap (0,a)$ is open in $(0,+\infty)$. If I am not mistaken, its closure should be $(0,+\infty)\cap [0,a]$. The closed interval $[0,a]$ is compact in $\mathbb{R}$, so my question is: is the closure $(0,+\infty) \cap (0,a)$ compact in $(0,+\infty)$ as well?

Answer 1

First of all, a comment, I don't really know why you are writing the set as $(0,\infty)\cap (0,a)$, since this set is simply $(0,a)$.

Second, no, the set is not compact in $(0,\infty)$. There are several ways to prove this.

1. From definition of compactness (i. e. every open cover has a finite subcover), you can create a sequence of open intervals that become smaller and smaller as they approach $0$. This sequence will not have a finite subcover.
2. From sequences (in metric spaces, every bounded sequence has a convergent subsequence), you can take a sequence that converges to $0$ and there will be no convergent subsequence of it.