Proof of the special case of the Intermediate Value Theorem:
Let $f$ be a continuous function on $[a, b]$ and suppose that:
$$ f(a) < 0 < f(b) $$
Then there exists a number $c$ in $(a, b)$ such that $f(c) = 0$.
Consider the following proof:
First, define $[a_0, b_0] = [a, b]$ and let $p = 1/2(a_0+b_0)$, the midpoint of $[a_0, b_0]$.
If $f(p) = 0$, then the proof is complete. Otherwise define:$[a_1, b_1] = [a_0, p]$ if $f(p) > 0$, or $[a_1, b_1] = [p, b_0]$ if $f(p) < 0$.
In either case we have:
- $[a_1, b_1] \subseteq [a_0, b_0]$
- $b_1 - a_1 = 1/2(b_0 - a_0)$
- $f(a_1) < 0 < f(b_1)$
Now repeat this process, bisecting $[a_1, b_1]$ to obtain $[a_2, b_2]$, etc. If at any stage, we find $f(p) = 0$, the proof is done.
Otherwise we obtain a sequence of closed intervals $[a_n, b_n]$ for $n=1,2,...$ with properties
- $[a_{n+1}, b_{n+1}] \subseteq [a_n, b_n]$
- $b_n - a_n = (1/2)^n(b_0 - a_0)$
- $f(a_n) < 0 < f(b_n)$
Property 1 implies that $(a_n)$ is increasing and bounded above by $b_0$. Hence by the Monotone Convergence Theorem $(a_n)$ is convergent. Let $\lim _{ n \to \infty } a_n = c$. ...
The proof then goes on to say that the limit of $b_n$ must also be $c$ etc.
But I'm already somewhat stuck. How can they claim that "Property 1 implies that $(a_n)$ is increasing ..."?
I understand we are repeatedly bisecting the current interval to obtain another interval half the size. Any interval along the way is bounded above by $b_0$, that's also clear. But why does the lower value of the new interval necessarily increase?
My thinking: When I'm bisecting, I find a midpoint $p$. If the midpoint is larger than $0$, then I continue the bisecting with $[a_n, p]$ and if that repeatedly happens (choosing to continue with the interval on the "left side" of the midpoint), then the sequence of lower bounds $(a_n)$ wouldn't ever increase, it always could stay at $a_0$. Is that not right?
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