Draft in progress…
Intermediate Value Theorem
The Theorem
“Let $a<b$ be real numbers. Let $f(x)$ be a continuous function, $f:[a,b] \mapsto \mathbb{R}$. If $f(a)\leq 0$ and $f(b)\geq 0$ then there exists a $c \in [a,b]$ such that $f(c) = 0$.”
A Tunnelling-down Proof
The Set-up
Choices:
- Function $f$ is continuous (•)
- Function $f$ maps real numbers in $[a,b]$ to anywhere in $\mathbb{R}$ (•)
- Subset $X$ is the set of $x$ in $[a,b]$ for which $f(x)$ is less than $0$ (•).
Notation:
- By convention any member of set $[a,b]$ is a real number $x$ between $a$ and $b$ inclusive: $a\leq x \leq b$. (•)
The Steps
Let's name the lowest upper bound of $X$ as $x_0$ (•).
Because $f$ maps from $[0,1]$ to $\mathbb{R}$ we know that $f(x_0)$ is a real number. (•)
Therefore $f(x_0)$ must be greater than zero, less than zero, or equal to zero. (•)
If we assume $f(x_0)$ is less than zero or greater than zero then we get a contradiction. Therefore $f(x_0)$ must equal zero.
We have thus found a $c=x_0$ that satisfies $f(x)=0$ for some point $c$ in $[a,b]$. (•)
If you believe…
We know that $X$ will have various upper bounds, which are values that no element of $X$ ever exceeds.
For a set of real numbers, if there's an upper bound there will be a lowest upper bound, also known as a supremum. We'll use $x_0$ to label the lowest upper bound of $X$. (•)
Function $f$ is continuous. Therefore we can get as close to $f(x)$, but not equal it, if we get close enough to $x$.
If you assume $x_0$ is greater than zero you'll find a smaller number that's an upper bound.
If you assume $x_0$ is less than zero you'll find a largest number that's an upper bound.
Contradiction. So $x_0$ is neither positive nor negative. (•)
Therefore $x_0$ must be zero. (•)
If you believe…
$X$ is a non-empty set of real numbers, and therefore has a lowest upper bound. (•)
Any continuous function $f$ and positive number $\epsilon > 0$ will let you find a $\delta > 0$ that makes this guarantee:
Any $x_1$ within $\delta$ (plus or minus) of $x$ takes $f(x_1)$ to within $\epsilon$ (plus or minus) of $f(x)$.
If you assume $f(x_0)>0$ you can find some $\delta>0$ such that $f(x_0 - \delta) >0$ as well. (•)
So our assumption that $x_0$ is the lowest upper bound is contradicted. (•)
If you assume $f(x_0)<0$ you can find some $\delta>0$ such that $f(x_0 + \delta) <0$ as well. (•)
So our assumption that $x_0$ is an upper bound is contradicted. (•)
Sources
Anon. “The Intermediate Value Theorem.” MAS111 (“Convergence [and] Continuity”) class handout at the University of London. Undated, but from 2007/2008 course taught by Ilya Goldsheid. (PDF.)
Advisory
Please note that I am not a mathematician and so the presentation of proofs that I make may be deeply flawed. I'm using this writing process to figure out what I'm reading. Please consult more authoritative sources as well.
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