Calculus/Extreme value theorem

This is very rough, and needs to be seriously edited

Introduction

The Mean Value Theorem(MVT) says that continuous functions have maximiums and minimiums on a closed and bounded interval.

For example consider the function f(x)=1/x. On the interval [1,2] f(x) is continuous with a maximium value of 1, and a minimum value of 1/2. However, if we consider f(x) on the interval [0,1], f(x) isn't continuous at 0 where f(x)=1/0 so the theorem doesn't apply. However, f(x) is continuous on the interval (0,1), f(x) has no maximum value because as x approaches 0, f(x) approaches infinity.

Preliminary Concepts

Before formally stating the theorem, or proving it, we must give a few preliminary results from real analysis.

Interval

An interval is a set of contiguous on the real line. A closed interval denoted [a,b] is the set of points x, such that $a\le x\le b$. An open interval denote (a,b) is the set of points x, such that $a<x<b$. A half-open interval can be denoted (a,b] or [a,b) which is the just open interval plus the point next to the bracket.

Compact

Let a cover of a set S be a collection of open sets, whose union contains S. We say that a cover is finite if and only if it is collection of a finite number of sets. A set is compact if and only if every cover of the set has a finite subcover.

The Heine-Borel theorem states that a subset of R is compact if and only if it is closed and bounded. The theorem can be proved by considering the endpoints of the intervals. If the set doesn't include the boundaries, there infinite sets that approach the boundary, but no set contains all the points up to the boundary. If the intervals are all closed, some set will have to include the boundary.

Formal Theorem

If f(x) is a continuous function over the interval [a,b] then there exists c,d such that

$c\le f(y)\le d$ holds strictly over the interval [a,b].

Proof

The theorem can easily be translated to saying compact subsets of the reals map to compact subsets of the reals. This is true, because the inverse mapping sends all covers back to the interval.