Editing Extreme value theorem (section)

===Proof of the boundedness theorem===
{{Math theorem
|name=Boundedness Theorem
|If <math>f(x)</math> is continuous on <math>[a,b],</math> then it is bounded on <math>[a,b].</math>}}
{{Math proof|
Suppose the function <math>f</math> is not bounded above on the interval <math>[a,b]</math>. Pick a [[sequence]] <math>(x_n)_{n \in \mathbb{N}}</math> such that <math>x_n \in [a,b]</math> and <math>f(x_n)>n</math>. Because <math>[a,b]</math> is bounded, the [[Bolzano–Weierstrass theorem]] implies that there exists a convergent subsequence <math>(x_{n_k})_{k \in \mathbb{N}}</math> of <math>({x_n})</math>. Denote its limit by <math>x</math>. As <math>[a,b]</math> is closed, it contains <math>x</math>.  Because <math>f</math> is continuous at <math>x</math>, we know that <math>f(x_{{n}_{k}})</math> converges to the real number <math>f(x)</math> (as <math>f</math> is [[sequentially continuous]] at <math>x</math>). But <math>f(x_{{n}_{k}}) > n_k \geq k </math> for every <math>k</math>, which implies that <math>f(x_{{n}_{k}})</math> diverges to [[Extended real number line|<math>+ \infty </math>]], a contradiction. Therefore, <math>f</math> is bounded above on <math>[a,b]</math>.&nbsp;[[Q.E.D.|∎]]}}
{{Math proof
|title=Alternative proof
|Consider the set <math>B</math> of points <math>p</math> in <math>[a,b]</math> such that <math>f(x)</math> is bounded on <math>[a,p]</math>.  We note that <math>a</math> is one such point, for <math>f(x)</math> is bounded on <math>[a,a]</math> by the value <math>f(a)</math>.  If <math>e > a</math> is another point, then all points between <math>a</math> and <math>e</math> also belong to <math>B</math>.  In other words <math>B</math> is an interval closed at its left end by <math>a</math>.

Now <math>f</math> is continuous on the right at <math>a</math>, hence there exists <math>\delta>0</math> such that <math>|f(x) - f(a)| < 1</math> for all <math>x</math> in <math>[a,a+\delta]</math>. Thus <math>f</math> is bounded by <math>f(a) - 1</math> and <math>f(a)+1</math> on the interval <math>[a,a+\delta]</math> so that all these points belong to <math>B</math>.

So far, we know that <math>B</math> is an interval of non-zero length, closed at its left end by <math>a</math>.

Next, <math>B</math> is bounded above by <math>b</math>.  Hence the set <math>B</math> has a supremum in <math>[a,b]</math> ; let us call it <math>s</math>. From the non-zero length of <math>B</math> we can deduce that <math>s > a</math>.

Suppose <math>s<b</math>.  Now <math>f</math> is continuous at <math>s</math>, hence there exists <math>\delta>0</math> such that <math>|f(x) - f(s)| < 1</math> for all <math>x</math> in <math>[s-\delta,s+\delta]</math> so that <math>f</math> is bounded on this interval.  But it follows from the supremacy of <math>s</math> that there exists a point belonging to <math>B</math>, <math>e</math> say, which is greater than <math>s-\delta/2</math>.  Thus <math>f</math> is bounded on <math>[a,e]</math> which overlaps <math>[s-\delta,s+\delta]</math> so that <math>f</math> is bounded on <math>[a,s+\delta]</math>.  This however contradicts the supremacy of <math>s</math>.

We must therefore have <math>s=b</math>.  Now <math>f</math> is continuous on the left at <math>s</math>, hence there exists <math>\delta>0</math> such that <math>|f(x) - f(s)| < 1</math> for all <math>x</math> in <math>[s-\delta,s]</math> so that <math>f</math> is bounded on this interval.  But it follows from the supremacy of <math>s</math> that there exists a point belonging to <math>B</math>, <math>e</math> say, which is greater than <math>s-\delta/2</math>.  Thus <math>f</math> is bounded on <math>[a,e]</math> which overlaps <math>[s-\delta,s]</math> so that <math>f</math> is bounded on <math>[a,s]</math>.&nbsp;&nbsp; 
[[Q.E.D.|∎]]}}