I am trying to understand the proof in my textbook for the Heine-Borel Theorem.
Theorem: Any closed interval $[a,b]$ is a compact subset of $\mathbb{R}$ (with the usual topology).
Proof:
Let $\mathcal{U}$ be a cover of $[a,b]$ by open subsets of $\mathbb{R}$.
Let $A$ denote the set of all $x\in[a,b]$ such that $[a,x]$ can be covered by a finite subcover taken from $\mathcal{U}$. Since $a\in A$ (we can certainly cover $[a,a]=\{a\}$ by one element of $\mathcal{U}$) the set $A$ is non-empty.
The set $A$ is also bounded above by $b$, so
The set $A$ is also bounded above by b, so we can set $c = \text{ sup } A$. Since
$a ≤ c ≤ b$, we must have $c ∈ U$ for some$ U ∈ \mathcal{U}$. Since $U$ is open, we
have $(c − δ, c + δ) ⊂ U$ for some $δ > 0.$
Since $c = \text{ sup }A$, there exists some $x ∈ A$ with $x > c − δ$. It follows
that
$$[a, c + δ) = [a, x] ∪ (c − δ, c + δ)
$$
can also be covered by a finite collection of sets from $\mathcal{U}$, since $[a, x]$ can
be and $(c − δ, c + δ) ⊂ U ∈ \mathcal{U}.$
Here is the part which I'm struggling to understand:
It follows (i) that $c = b$, for otherwise $c < b$ and this yields a cover of
$$[a, min(c + δ/2, b)]$$
Why if $c<b$ then this yields a cover of $[a,\text{min}(c+\delta/2,b)]$? I don't understand why would this happen?
by a finite number of sets from $\mathcal{U}$, contradicting the fact that $c = \text{ sup }A$;
Why would it contradict the fact that $c = \text{ sup }A$?
so (ii) a finite collection of sets from $\mathcal{U}$ cover$ [a, b + δ) ⊃ [a, b]$, which is
what we wanted.
Best Answer
We want to show that $c = b$. If $c < b$, then we've already picked a $\delta >0$ and $U \in \mathcal{U}$ so that $(c-\delta, c+\delta) \subseteq U$.
If now $c':=c+\frac{\delta}{2}$ (so that $c'> c$ and lies in $U$) would also obey $c' < b$, then $[a,c']$ is covered by finitely many members of $\mathcal{U}$ which contradicts the fact that $c$ is an upperbound for those numbers (as $c' \in A$ and so $c' \le \sup A = c < c'$). If $c' \ge b$ on the other hand, we would have that $b$ is already in $U$ and we're done: $[a,b]$ has a finite subcover and $b \in A$ and again this contradicts $b \le \sup A = c < b$.
So the $\min$ notation is a shorthand to cover these two similar cases. In both we go beyond $c$ to find a new, larger member of $A$ which is a contradiction.