$[a,b]$ has an open cover then every point of this interval is covered in an open set. Since a is in an open set some neighborhood of a is a subset of that set. Say $[a,a+r_1) \subset O_i $ similarly $[a+r_1,a+r_1+r_2)\subset O_j$. Going like this we can reach a $r_n$ such that $b \in [a+r_1…r_n,a+r_1….r_{n+1})$ Radius of these neighborhoods are real numbers greater than zero. $M=min(r_1,r_2…r_{n+1}))$ It is obvious that $n+1 \le \frac{b-a}{M}$ Then n is clearly finite and even if these neighborhoods belong to different open sets number of open sets are finite then every open cover has a finite subcover. Then [a,b] is compact.
It is really easy to show that a compact subset of R is closed and bounded.
My argument seems to be correct but shouldn't this work for intervals (a,b) or [a,b)?
Some say that a sequence of real numbers doesn't add up to any number. But since [a,b] is closed a neighborhood of b is also in those open sets. What I mean is we can always add whatever is missing in this case.
Best Answer
The comments have already explained where your argument breaks down; here's an example of an open cover of a non-compact set, with no finite subcover, which hopefully makes things clear:
Consider the following sequence of open intervals: $U_n=(0, 1-{1\over n+1})$. Clearly $\{U_n: n\in\mathbb{N}\}$ is an open cover of $(0, 1)$; however, the corresponding $r$-values are $$r_1={1\over 2}, r_{n+1}={1\over n+1}-{1\over n+2}.$$ It's easy to see (and you should check this) that while $\lim_{n\rightarrow\infty}\sum^n_1r_i=1$, no finite sum of the $r_i$s equals or exceeds $1$. This is why $\{U_n\}$ has no finite subcover.