How to verify if a set admits a finite sub covering? 
A metric space $X$
is called compact iff each of its open cover has a finite subcover.
For example $[0,1)$ is not compact. I understand what open covers are;
for example set $A = \{(-1,1)\}$ may be one of open coverings of $[0,1)$.

How do I check if set $A$  admits a finite sub covering? I understand what open covers are.
Attempt:
Let $O$ be the set of open intervals of the form $(-1/2, 1-1/t)$. $O$ is an open cover but it does not admit a finite subcover. My question is related with understanding of how can  we say $O$ doesn't admit a finite subcover.
I find it difficult to apply the open covering definition of compactness. If someone can explain this definition with examples it would be good for me.
 A: Here's the first task.

Let $U_n = (0, 1 - \frac1n)$ for $n \in \mathbf N$. Show that the open covering $\{U_n\}$ of $(0, 1)$ has no finite subcover.

Some steps you could follow:


*

*A finite subcover is of the form $\{U_n\}_{n \in S}$ for some finite subset $S$ of $\mathbf N$.

*If $S$ is non-empty then let $N$ be the largest element of $S$. Then $U_n \subset U_N$ for all $n \in S$, so it is enough to show that $U_N$ does not contain all of $(0, 1)$.

*$1 - \frac1N$ is an element of $(0, 1) \setminus U_N$.


I can't tell you how to understand the definition in a few words, but here are some thoughts. It's often enough to understand a special case of something, while keeping in mind that exceptions exist. Here, the familiar case is that of $\mathbf R$. More generally, in a metric space compactness is equivalent to sequential compactness. This might be a more visceral notion, and it might help to see the proof of equivalence in section 2 of this handout of Brian Conrad's.
For me, this justifies thinking of non-compact sets as those in which points can run off, either to a point outside of the set or "to infinity". See also this (closed) question at MO. Try thinking about why you can't construct an example like the one you gave for $[0, 1]$.
As opposed to applying the open covering definition directly, it's often enough to remember the fundamental fact that a subset of $\mathbf R^n$ is compact if and only if it is closed and bounded, together with standard theorems on compactness.
