Editing Intermediate value theorem (section)

==Proof==
=== Proof version A===
<!-- This section is linked from [[Continuity property]] -->
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.

===Proof version B===
We will only prove the case of <math>f(a)<u<f(b)</math>, as the <math>f(a)>u>f(b)</math> case is similar.<ref>Slightly modified version of {{cite book |title=Understanding Analysis|first=Stephen|last=Abbot|publisher=Springer | year=2015|page=123}}</ref>

Define <math>g(x)=f(x)-u</math> which is equivalent to <math>f(x)=g(x)+u</math> and lets us rewrite <math>f(a)<u<f(b)</math> as <math>g(a)<0<g(b)</math>, and we have to prove, that <math>g(c)=0</math>  for some <math>c\in[a,b]</math>, which is more intuitive. We further define the set <math>S=\{x\in[a,b]:g(x)\leq 0\}</math>. Because <math>g(a)<0</math> we know, that <math>a\in S</math> so, that <math>S</math> is not empty. Moreover, as <math>S\subseteq[a,b]</math>, we know that <math>S</math> is bounded and non-empty, so by Completeness, the [[supremum]] <math>c=\sup(S)</math> exists.

There are 3 cases for the value of <math>g(c)</math>, those being <math>g(c)<0,g(c)>0</math> and <math>g(c)=0</math>. For contradiction, let us assume, that <math>g(c)<0</math>. Then, by the definition of continuity, for <math>\epsilon=0-g(c)</math>, there exists a <math>\delta>0</math> such that <math>x\in(c-\delta,c+\delta)</math> implies, that <math>|g(x)-g(c)|<-g(c)</math>, which is equivalent to <math>g(x)<0</math>. If we just chose <math>x=c+\frac{\delta}{N}</math>, where <math>N>\frac{\delta}{b-c}+1</math>, then as <math>1 < N</math>, <math>x<c+\delta</math>, from which we get <math>g(x)<0</math> and <math>c<x<b</math>, so <math>x\in S</math>. It follows that <math>x</math> is an upper bound for <math>S</math>.  However, <math>x>c</math>, contradicting the '''upper bound''' property of the ''least upper bound'' <math>c</math>, so <math>g(c)\geq 0</math>. Assume then, that <math>g(c)>0</math>. We similarly chose <math>\epsilon=g(c)-0</math> and know, that there exists a <math>\delta>0</math> such that <math>x\in(c-\delta,c+\delta)</math> implies <math>|g(x)-g(c)|<g(c)</math>. We can rewrite this as <math>-g(c)<g(x)-g(c)<g(c)</math> which implies, that <math>g(x)>0</math>. If we now chose <math>x=c-\frac{\delta}{2}</math>, then <math>g(x)>0</math> and <math>a<x<c</math>. It follows that <math>x</math> is an upper bound for <math>S</math>. However, <math>x<c</math>, which contradict the '''least''' property of the ''least upper bound'' <math>c</math>, which means, that <math>g(c)>0</math> is impossible. If we combine both results, we get that <math>g(c)=0</math> or <math>f(c)=u</math> is the only remaining possibility.

'''Remark:''' The intermediate value theorem can also be proved using the methods of [[non-standard analysis]], which places "intuitive" arguments involving infinitesimals on a rigorous{{Clarify|reason=The placement and phrasing of this remark may suggest that the classical proof is somehow "intuitive" and not rigorous, which is not the case.|date=January 2023}}  footing.<ref>{{cite arXiv |last=Sanders|first=Sam | eprint=1704.00281 | title=Nonstandard Analysis and Constructivism!|date=2017|class=math.LO}}</ref>