Editing Extreme value theorem (section)

===Proofs of the extreme value theorem===

{{Math proof
|title=Proof of the Extreme Value Theorem
|proof=By the boundedness theorem, ''f'' is bounded from above, hence, by the  [[Dedekind-complete]]ness of the real numbers, the least upper bound (supremum) ''M'' of ''f'' exists. It is necessary to find a point ''d'' in [''a'', ''b''] such that ''M'' = ''f''(''d''). Let ''n'' be a natural number. As ''M'' is the ''least'' upper bound, ''M'' – 1/''n'' is not an upper bound for ''f''. Therefore, there exists ''d<sub>n</sub>'' in [''a'', ''b''] so that ''M'' – 1/''n'' < ''f''(''d<sub>n</sub>''). This defines a sequence {''d<sub>n</sub>''}. Since ''M'' is an upper bound for ''f'', we have ''M'' – 1/''n'' < ''f''(''d<sub>n</sub>'') ≤ ''M'' for all ''n''.  Therefore, the sequence {''f''(''d<sub>n</sub>'')} converges to ''M''.

The [[Bolzano–Weierstrass theorem]] tells us that there exists a subsequence {<math>d_{n_k}</math>}, which converges to some ''d'' and, as [''a'', ''b''] is closed, ''d'' is in [''a'', ''b'']. Since ''f'' is continuous at ''d'', the sequence {''f''(<math>d_{n_k}</math>)} converges to ''f''(''d''). But {''f''(''d<sub>n<sub>k</sub></sub>'')} is a subsequence of {''f''(''d<sub>n</sub>'')} that converges to ''M'', so ''M'' = ''f''(''d''). Therefore, ''f'' attains its supremum ''M'' at ''d''.&nbsp;
[[Q.E.D.|∎]]}}

{{Math proof
|title=Alternative Proof of the Extreme Value Theorem
|proof=The set {{math|1= {''y'' &isin; '''R''' : ''y'' = ''f''(''x'') for some ''x'' &isin; [''a'',''b'']}<nowiki/>}} is a bounded set. Hence, its [[least upper bound]] exists by [[least upper bound property]] of the real numbers. Let {{math|1=''M'' = sup(''f''(''x''))}}&nbsp;on&nbsp;{{closed-closed|''a'', ''b''}}.  If there is no point ''x'' on [''a'',&nbsp;''b''] so that ''f''(''x'')&nbsp;=&nbsp;''M'', then
{{math|''f''(''x'') < ''M''}} on [''a'',&nbsp;''b''].  Therefore, {{math|1/(''M'' &minus; ''f''(''x''))}} is continuous on [''a'', ''b''].

However, to every positive number ''&epsilon;'', there is always some ''x'' in [''a'',&nbsp;''b''] such that {{math|''M'' &minus; ''f''(''x'') < ''&epsilon;''}} because ''M'' is the least upper bound. Hence, {{math|1/(''M'' &minus; ''f''(''x'')) > 1/''&epsilon;''}}, which means that {{math|1/(''M'' &minus; ''f''(''x''))}} is not bounded.  Since every continuous function on [''a'', ''b''] is bounded, this contradicts the conclusion that {{math|1/(''M'' &minus; ''f''(''x''))}} was continuous on [''a'',&nbsp;''b''].  Therefore, there must be a point ''x'' in [''a'',&nbsp;''b''] such that ''f''(''x'')&nbsp;=&nbsp;''M''.
[[Q.E.D.|∎]]}}