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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.

Feel free to contact me by leaving a comment or sending me a private message.

page revision: 30, last edited: 02 Mar 2009 14:20
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