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Intermediate value theorem
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===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>
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