TITLE: Why is an open interval not a compact set?
QUESTION [13 upvotes]: I learned that every compact set is closed and bounded; and also that an open set is usually not compact. 
How to show that a concrete open set, for example the interval $(0,1)$, is not compact? I tried to show that $(0,1)$ has no finite sub cover.

REPLY [33 votes]: 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.