Editing Intermediate value theorem (section)

=== Proof version A===
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The theorem may be proven as a consequence of the [[completeness (order theory)|completeness]] property of the real numbers as follows:<ref>Essentially follows {{cite book |title=Foundations of Analysis|first=Douglas A.|last=Clarke|publisher=Appleton-Century-Crofts | year=1971|page=284}}</ref>

We shall prove the first case, <math>f(a) < u < f(b)</math>. The second case is similar.

Let <math>S</math> be the set of all <math>x \in [a,b]</math> such that <math>f(x)<u</math>. Then <math>S</math> is non-empty since <math>a</math> is an element of <math>S</math>. Since <math>S</math> is non-empty and bounded above by <math>b</math>, by completeness, the [[supremum]] <math>c=\sup S</math> exists. That is, <math>c</math> is the smallest number that is greater than or equal to every member of <math>S</math>.

Note that, due to the continuity of <math>f</math> at <math>a</math>, we can keep <math>f(x)</math> within any <math>\varepsilon>0</math> of <math>f(a)</math> by keeping <math>x</math> sufficiently close to <math>a</math>. Since <math>f(a)<u</math> is a strict inequality, consider the implication when <math>\varepsilon</math> is the distance between <math>u</math> and <math>f(a)</math>. No <math>x</math> sufficiently close to <math>a</math> can then make <math>f(x)</math> greater than or equal to <math>u</math>, which means there are values greater than <math>a</math> in <math>S</math>. A more detailed proof goes like this:

Choose <math>\varepsilon=u-f(a)>0</math>. Then <math>\exists \delta>0</math> such that <math>\forall x \in [a,b]</math>, <math display="block">|x-a|<\delta \implies |f(x)-f(a)|<u-f(a) \implies f(x)<u.</math>Consider the interval <math>[a,\min(a+\delta,b))=I_1</math>. Notice that <math>I_1 \subseteq [a,b]</math> and every <math>x \in I_1</math> satisfies the condition <math>|x-a|<\delta</math>. Therefore for every <math>x \in I_1</math> we have <math>f(x)<u</math>. Hence <math>c</math> cannot be <math>a</math>.

Likewise, due to the continuity of <math>f</math> at <math>b</math>, we can keep <math>f(x)</math> within any <math>\varepsilon > 0</math> of <math>f(b)</math> by keeping <math>x</math> sufficiently close to <math>b</math>. Since <math>u<f(b)</math> is a strict inequality, consider the similar implication when <math>\varepsilon</math> is the distance between <math>u</math> and <math>f(b)</math>. Every <math>x</math> sufficiently close to <math>b</math> must then make <math>f(x)</math> greater than <math>u</math>, which means there are values smaller than <math>b</math> that are upper bounds of <math>S</math>. A more detailed proof goes like this:

Choose <math>\varepsilon=f(b)-u>0</math>. Then <math>\exists \delta>0</math> such that <math>\forall x \in [a,b]</math>, <math display="block">|x-b|<\delta \implies |f(x)-f(b)|<f(b)-u \implies f(x)>u.</math>Consider the interval <math>(\max(a,b-\delta),b]=I_2</math>. Notice that <math>I_2 \subseteq [a,b]</math> and every <math>x \in I_2</math> satisfies the condition <math>|x-b|<\delta</math>. Therefore for every <math>x \in I_2</math> we have <math>f(x)>u</math>. Hence <math>c</math> cannot be <math>b</math>.

With <math>c \neq a</math> and <math>c \neq b</math>, it must be the case <math>c \in (a,b)</math>. Now we claim that <math>f(c)=u</math>.

Fix some <math>\varepsilon > 0</math>. Since <math>f</math> is continuous at <math>c</math>, <math>\exists \delta_1>0</math> such that <math>\forall x \in [a,b]</math>, <math>|x-c|<\delta_1 \implies |f(x) - f(c)| < \varepsilon</math>.

Since <math>c \in (a,b)</math> and <math>(a,b)</math> is open, <math>\exists \delta_2>0</math> such that <math>(c-\delta_2,c+\delta_2) \subseteq (a,b)</math>. Set <math>\delta= \min(\delta_1,\delta_2)</math>. Then we have
<math display="block">f(x)-\varepsilon<f(c)<f(x)+\varepsilon</math>
for all <math>x\in(c-\delta,c+\delta)</math>. By the properties of the supremum, there exists some <math>a^*\in (c-\delta,c]</math> that is contained in <math>S</math>, and so
<math display="block">f(c)<f(a^*)+\varepsilon<u+\varepsilon.</math>
Picking <math>a^{**}\in(c,c+\delta)</math>, we know that <math>a^{**}\not\in S</math> because <math>c</math> is the supremum of <math>S</math>. This means that
<math display="block">f(c)>f(a^{**})-\varepsilon \geq u-\varepsilon.</math>
Both inequalities
<math display="block">u-\varepsilon<f(c)< u+\varepsilon</math>
are valid for all <math>\varepsilon > 0</math>, from which we deduce <math>f(c) = u</math> as the only possible value, as stated.