I cannot understand the following Michael Spivak's proof of Heine – Borel Theorem for the line.
I think that we must show $a < \alpha$ first of all.
But he didn't prove that $a < \alpha$.
Is his proof correct?
The closed interval $[a, b]$ is compact.
Proof.
If $\mathcal{O}$ is an open cover of $[a, b]$,
let $A = \{x : a \leq x \leq b \text{ and } [a, x] \text{ is covered by some finite number of open sets in } \mathcal{O}\}.$
Note that $a \in A$ and that $A$ is clearly bounded above (by $b$).
We would like to show that $b \in A$.
This is done by proving two things about $\alpha =$ least upper bound of $A$ ; namely,
(1) $\alpha \in A$ and
(2) $b = \alpha$.
Since $\mathcal{O}$ is a cover, $\alpha \in U$ for some $U$ in $\mathcal{O}$.
Then all points in some interval to the left of $\alpha$ are also in $U$.
Since $\alpha$ is the least upper bound of $A$, there is an $x$ in this interval such that $x \in A$.
Best Answer
Clearly $a\in A$.
An element of one of the covering intervals is in $A$ if and only if all elements of that interval are in $A$.
In particular, the elements of an interval covering $a$ are in $A$.
Letting $\alpha$ be the LUB of $A$, it follows that $\alpha > a$.
Now consider an interval which covers $\alpha$.
If $\alpha\notin A$, then all the elements of an interval covering $\alpha$ which are slightly to the left of $\alpha$ would not be in $A$, contrary the choice of $\alpha$ as the LUB of $A$.
Hence $\alpha\in A$.
But if $\alpha < b$, then the elements of an interval covering $\alpha$ which are slightly to the right of $\alpha$ would be in $A$, again contrary the choice of $\alpha$ as the LUB of $A$.
Hence $\alpha = b$.
Thus, since $\alpha\in A$, we get $b\in A$, hence $A=[a,b]$.