About Michael Spivak's proof of Heine - Borel Theorem for the line in his book "Calculus on Manifolds"

Question

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 $x \in A$.

Answer 1

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 $A=[a,b]$.

Answer 2

My proof is the following:

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 $b$).

Since $\mathcal{O}$ is a cover of $[a, b]$, $a \in V$ for some $V$ in $\mathcal{O}$.

There exists $\delta > 0$ such that $a + \delta \leq b$ and $(a - \delta, a + \delta) \in V$.

Let $a^{'}$ be a real number such that $a < a^{'} < a + \delta \leq b$.

Then $[a, a^{'}] \subset (a - \delta, a + \delta) \subset V$.

$\therefore a^{'} \in A$.

$\therefore a < a^{'} \leq \alpha$.

$\therefore a < \alpha$.

Now we assume $\alpha \notin A$.

Since $\mathcal{O}$ is a cover of $[a, b]$ and $\alpha \leq b$, $\alpha \in U$ for some $U$ in $\mathcal{O}$.

Let $\epsilon > 0$ be a real number such that $a < \alpha - \epsilon$ and $(\alpha - \epsilon, \alpha + \epsilon) \in U$.

Then, there exists $\beta$ such that $\alpha - \epsilon < \beta \leq \alpha$ and $\beta \in A$ because $\alpha$ is the least upper bound of $A$.

Since $\alpha \notin A$ by our assumption, $\beta < \alpha$.

$[a, \beta]$ is covered by some finite number of open sets in $\mathcal{O}$ and $[\beta, \alpha]$ is covered by $U$ in $\mathcal{O}$.

So, $[a, \alpha]$ is covered by some finite number of open sets in $\mathcal{O}$.

$\therefore \alpha \in A$.

But this is a contradiction to our assumption that $\alpha \notin A$.

$\therefore \alpha \in A$.

Next, there exists $\gamma$ such that $\alpha < \gamma$ and $\gamma \in (\alpha - \epsilon, \alpha + \epsilon) \subset U$.

$[a, \alpha]$ is covered by some finite number of open sets in $\mathcal{O}$ and $[\alpha, \gamma]$ is covered by $U$ in $\mathcal{O}$.

So, $[a, \gamma]$ is covered by some finite number of open sets in $\mathcal{O}$.

If $\gamma \leq b$, then $\gamma \in A$.

This is a contradiction to the assumption that $\alpha$ is an upper bound of $A$.

So $b < \gamma$.

$[a, \gamma]$ is covered by some finite number of open sets in $\mathcal{O}$.

So, of course $[a, b] \subset [a, \gamma]$ is covered by some finite number of open sets in $\mathcal{O}$.