Why is an open interval not a compact set?

I think the following may be a source of confusion: the statement "$(0,1)$ has no finite sub cover" doesn't make any sense. You first have to choose a cover of $(0,1)$ by open sets. Then this may or may not have a finite sub-cover.

If $(0,1)$ were compact, any such cover would (by the definition of compact in terms of open covers) have to have a finite subcover. In fact, $(0,1)$ is not compact, and so what this means is that we can find some cover of $(0,1)$ by open sets which does not have a finite subcover.

Omnomnomnom gives an example of such a cover in their answer, and there are lots of others; here's one: the cover $\{U_2, \ldots, U_n , \ldots\}$ where $U_n = (1/n, 1-1/n).$

Here is a cover which is finite, and hence does have a finite subcover (namely, itself): $\{(0,1)\}$.

Here is another: $\{(0,1/2), (1/3,1)\}$.

Here is an infinite cover which admits a finite subcover $\{(0,1), (0,1/2), \ldots, (0,1/n) , \ldots \}$.

Hopefully these examples help to clarify what the definition of compactness in terms of finite subcovers is about.

To be sure, there exist sets that are open and closed and bounded. For example, if we take the space $X = [0,1] \cup [2,3] \cup [4,5]$ under the typical topology of $\mathbb{R}$, then $[2,3]$ is a closed, open, compact proper subset of $X$.

As for open intervals, consider the example of $X = (0,1)$. Defining $U_n = (1/n,1)$, we note that $\{U_1,U_2,\dots\}$ is an open cover of $X$ (since $X \subset \bigcup_{n=1}^\infty U_n$) that has no finite subcover.

(Assuming you're talking about $\Bbb{R}^1$)

Consider the open interval $(a,b)$. Let $d=|b-a|/2$, and let $c=(a+b)/2$, the midpoint of $(a,b)$. Take the collection of open intervals: $\{(c-\frac{n}{n+1}d,c+\frac{n}{n+1}d)\}_{n=1}^\infty$. This collection covers $(a,b)$ (its union equals $(a,b)$), but no finite sub-collection covers $(a,b)$.

For what it's worth:

A subset of the euclidean space $\mathbb R^n$ is compact if and only if it is closed and bounded. This is a possible definition of compactness of sets like these.

Regarding the concepts of open, closed, bounded: You will have to look up their definitions. Some examples of subsets of $\mathbb R$:

• The empty set is open, closed and bounded.
• The set $\mathbb R$ is open, closed and not bounded.
• The interval $(0,1)$ is open, not closed, bounded.
• The interval $(0,\infty)$ is open, not closed, not bounded.
• The interval $[0,1]$ is not open, closed, bounded.
• The interval $[0,\infty)$ is not open, closed, not bounded.
• The interval $[0,1)$ is not open, not closed, bounded.
• The set $\{0\}\cup (1,\infty)$ is not open, not closed, not bounded.