Prove that if $f(x)$ is continuous on $[a,b]$, then $f(x)$ takes on a minimum value on $[a,b]$.

The Extreme-Value Theorem states that if $f(x)$ is continuous on $[a,b]$, then $f(x)$ takes on both maximum and minimum values on $[a,b]$. We prove here the case for the minimum value.

Since $f(x)$ is continuous on $[a,b]$, we know that it is bounded on $[a,b]$. Since it is bounded, the Axiom of Completeness tells us we have a greatest lower bound that we denote by

<Sign in to see all the formulas>

Our goal is to show that there exists an <Sign in to see all the formulas> such that <Sign in to see all the formulas>.

We introduce the function

<Sign in to see all the formulas>

and note that if <Sign in to see all the formulas> then the function is continuous on $[a,b]$. (It is continuous because it is a function composition <Sign in to see all the formulas> where <Sign in to see all the formulas>.)

However, we know that for every <Sign in to see all the formulas> there exists <Sign in to see all the formulas> such that <Sign in to see all the formulas>. Therefore, for every <Sign in to see all the formulas> there exists a $z$ such that

<Sign in to see all the formulas>

Therefore, $g(x)$ is unbounded.

Now we will prove that there exists an <Sign in to see all the formulas> such that <Sign in to see all the formulas> by contradiction.

If <Sign in to see all the formulas> for all $x$ in $[a,b]$ then $g(x)$ is a continuous function. Because it is continuous it is bounded. However, we have just shown that $g(x)$ is unbounded. We have reached the contradiction that when <Sign in to see all the formulas> for all $x$ in $[a,b]$ then $g(x)$ is both bounded and unbounded. Therefore, there must exist an <Sign in to see all the formulas> in $[a,b]$ such that <Sign in to see all the formulas>.