Intermediate Value Theorem proof Question












1












$begingroup$


There is a certain implication $overset{(*)}{Longrightarrow}$ of the proof that I do not understand.




Intermediatevaluetheorem
:$[a,b]subseteqmathbb{R},f:[a,b]rightarrowmathbb{R}
text{ continuous, }f(a)leq y_0leq f(b) Longrightarrow
exists_{xin[a,b]}f(x)=y_0$




Proof:



$$text{Let }M:={xin[a,b]:f(x)geq y_0}$$



$$bin MLongrightarrow Mneqemptysettext{ and }aleq x,forall _{xin M}Longrightarrow existsinf M=:x_0$$



$$Longrightarrowexists_{(x_{n})_{ninmathbb{N}}}x_nrightarrow x_0,x_nin M,forall_{ninmathbb{N}}$$



$$overset{ftext{ is continuous}}Longrightarrow f(x_n)rightarrow f(x_0)overset{x_ngeq y_0,forall_{ninmathbb{N}}}{Longrightarrow}f(x_0)geq y_0Longrightarrow x_0=max M$$



$$text{Suppose }f(x_0)>y_0Longrightarrow exists_{epsilon_0>0}epsilon_0=f(x_0)-y_0$$



$$overset{ftext{ is continuous}}{Longrightarrow}exists_{delta_0>0}forall_{xin[a,b]}|x-x_0|<delta_0Longrightarrow |f(x)-f(x_0)|<epsilon_0$$



$$overset{(*)}{Longrightarrow}f(x)>f(x_0)-epsilon_0=y_0text{ contradicition to }x_0=inf M$$



$(*)$ Is only true if the $x$ I am looking for is smaller than $x_0$ How do I know that such a $x$ exists.



I.e How can I make sure that ${(x_0-delta_0,x_0)cap[a,b]}=:Ssubset M$ is not empty?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can take $delta$ small enough so that $(x -delta, x+ delta) subset [a,b]$. You can always take smaller $delta$ in general, as if $delta_1 < delta_2$, we have that $|x - a| < delta_1 implies |x- a| < delta_2$
    $endgroup$
    – rubikscube09
    Jan 25 at 21:35










  • $begingroup$
    I do not know how exact your copy of the proof is, but as the version that you have presented here is written very poorly.
    $endgroup$
    – Carsten S
    Jan 26 at 14:19










  • $begingroup$
    The original proof was in German and there was more text involved...
    $endgroup$
    – RM777
    Jan 26 at 16:29
















1












$begingroup$


There is a certain implication $overset{(*)}{Longrightarrow}$ of the proof that I do not understand.




Intermediatevaluetheorem
:$[a,b]subseteqmathbb{R},f:[a,b]rightarrowmathbb{R}
text{ continuous, }f(a)leq y_0leq f(b) Longrightarrow
exists_{xin[a,b]}f(x)=y_0$




Proof:



$$text{Let }M:={xin[a,b]:f(x)geq y_0}$$



$$bin MLongrightarrow Mneqemptysettext{ and }aleq x,forall _{xin M}Longrightarrow existsinf M=:x_0$$



$$Longrightarrowexists_{(x_{n})_{ninmathbb{N}}}x_nrightarrow x_0,x_nin M,forall_{ninmathbb{N}}$$



$$overset{ftext{ is continuous}}Longrightarrow f(x_n)rightarrow f(x_0)overset{x_ngeq y_0,forall_{ninmathbb{N}}}{Longrightarrow}f(x_0)geq y_0Longrightarrow x_0=max M$$



$$text{Suppose }f(x_0)>y_0Longrightarrow exists_{epsilon_0>0}epsilon_0=f(x_0)-y_0$$



$$overset{ftext{ is continuous}}{Longrightarrow}exists_{delta_0>0}forall_{xin[a,b]}|x-x_0|<delta_0Longrightarrow |f(x)-f(x_0)|<epsilon_0$$



$$overset{(*)}{Longrightarrow}f(x)>f(x_0)-epsilon_0=y_0text{ contradicition to }x_0=inf M$$



$(*)$ Is only true if the $x$ I am looking for is smaller than $x_0$ How do I know that such a $x$ exists.



I.e How can I make sure that ${(x_0-delta_0,x_0)cap[a,b]}=:Ssubset M$ is not empty?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can take $delta$ small enough so that $(x -delta, x+ delta) subset [a,b]$. You can always take smaller $delta$ in general, as if $delta_1 < delta_2$, we have that $|x - a| < delta_1 implies |x- a| < delta_2$
    $endgroup$
    – rubikscube09
    Jan 25 at 21:35










  • $begingroup$
    I do not know how exact your copy of the proof is, but as the version that you have presented here is written very poorly.
    $endgroup$
    – Carsten S
    Jan 26 at 14:19










  • $begingroup$
    The original proof was in German and there was more text involved...
    $endgroup$
    – RM777
    Jan 26 at 16:29














1












1








1





$begingroup$


There is a certain implication $overset{(*)}{Longrightarrow}$ of the proof that I do not understand.




Intermediatevaluetheorem
:$[a,b]subseteqmathbb{R},f:[a,b]rightarrowmathbb{R}
text{ continuous, }f(a)leq y_0leq f(b) Longrightarrow
exists_{xin[a,b]}f(x)=y_0$




Proof:



$$text{Let }M:={xin[a,b]:f(x)geq y_0}$$



$$bin MLongrightarrow Mneqemptysettext{ and }aleq x,forall _{xin M}Longrightarrow existsinf M=:x_0$$



$$Longrightarrowexists_{(x_{n})_{ninmathbb{N}}}x_nrightarrow x_0,x_nin M,forall_{ninmathbb{N}}$$



$$overset{ftext{ is continuous}}Longrightarrow f(x_n)rightarrow f(x_0)overset{x_ngeq y_0,forall_{ninmathbb{N}}}{Longrightarrow}f(x_0)geq y_0Longrightarrow x_0=max M$$



$$text{Suppose }f(x_0)>y_0Longrightarrow exists_{epsilon_0>0}epsilon_0=f(x_0)-y_0$$



$$overset{ftext{ is continuous}}{Longrightarrow}exists_{delta_0>0}forall_{xin[a,b]}|x-x_0|<delta_0Longrightarrow |f(x)-f(x_0)|<epsilon_0$$



$$overset{(*)}{Longrightarrow}f(x)>f(x_0)-epsilon_0=y_0text{ contradicition to }x_0=inf M$$



$(*)$ Is only true if the $x$ I am looking for is smaller than $x_0$ How do I know that such a $x$ exists.



I.e How can I make sure that ${(x_0-delta_0,x_0)cap[a,b]}=:Ssubset M$ is not empty?






real-analysis







share|cite|improve this question











$endgroup$




There is a certain implication $overset{(*)}{Longrightarrow}$ of the proof that I do not understand.




Intermediatevaluetheorem
:$[a,b]subseteqmathbb{R},f:[a,b]rightarrowmathbb{R}
text{ continuous, }f(a)leq y_0leq f(b) Longrightarrow
exists_{xin[a,b]}f(x)=y_0$




Proof:



$$text{Let }M:={xin[a,b]:f(x)geq y_0}$$



$$bin MLongrightarrow Mneqemptysettext{ and }aleq x,forall _{xin M}Longrightarrow existsinf M=:x_0$$



$$Longrightarrowexists_{(x_{n})_{ninmathbb{N}}}x_nrightarrow x_0,x_nin M,forall_{ninmathbb{N}}$$



$$overset{ftext{ is continuous}}Longrightarrow f(x_n)rightarrow f(x_0)overset{x_ngeq y_0,forall_{ninmathbb{N}}}{Longrightarrow}f(x_0)geq y_0Longrightarrow x_0=max M$$



$$text{Suppose }f(x_0)>y_0Longrightarrow exists_{epsilon_0>0}epsilon_0=f(x_0)-y_0$$



$$overset{ftext{ is continuous}}{Longrightarrow}exists_{delta_0>0}forall_{xin[a,b]}|x-x_0|<delta_0Longrightarrow |f(x)-f(x_0)|<epsilon_0$$



$$overset{(*)}{Longrightarrow}f(x)>f(x_0)-epsilon_0=y_0text{ contradicition to }x_0=inf M$$



$(*)$ Is only true if the $x$ I am looking for is smaller than $x_0$ How do I know that such a $x$ exists.



I.e How can I make sure that ${(x_0-delta_0,x_0)cap[a,b]}=:Ssubset M$ is not empty?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 22:05









Bernard

122k741116




122k741116










asked Jan 25 at 21:17









RM777RM777

38312




38312












  • $begingroup$
    You can take $delta$ small enough so that $(x -delta, x+ delta) subset [a,b]$. You can always take smaller $delta$ in general, as if $delta_1 < delta_2$, we have that $|x - a| < delta_1 implies |x- a| < delta_2$
    $endgroup$
    – rubikscube09
    Jan 25 at 21:35










  • $begingroup$
    I do not know how exact your copy of the proof is, but as the version that you have presented here is written very poorly.
    $endgroup$
    – Carsten S
    Jan 26 at 14:19










  • $begingroup$
    The original proof was in German and there was more text involved...
    $endgroup$
    – RM777
    Jan 26 at 16:29


















  • $begingroup$
    You can take $delta$ small enough so that $(x -delta, x+ delta) subset [a,b]$. You can always take smaller $delta$ in general, as if $delta_1 < delta_2$, we have that $|x - a| < delta_1 implies |x- a| < delta_2$
    $endgroup$
    – rubikscube09
    Jan 25 at 21:35










  • $begingroup$
    I do not know how exact your copy of the proof is, but as the version that you have presented here is written very poorly.
    $endgroup$
    – Carsten S
    Jan 26 at 14:19










  • $begingroup$
    The original proof was in German and there was more text involved...
    $endgroup$
    – RM777
    Jan 26 at 16:29
















$begingroup$
You can take $delta$ small enough so that $(x -delta, x+ delta) subset [a,b]$. You can always take smaller $delta$ in general, as if $delta_1 < delta_2$, we have that $|x - a| < delta_1 implies |x- a| < delta_2$
$endgroup$
– rubikscube09
Jan 25 at 21:35




$begingroup$
You can take $delta$ small enough so that $(x -delta, x+ delta) subset [a,b]$. You can always take smaller $delta$ in general, as if $delta_1 < delta_2$, we have that $|x - a| < delta_1 implies |x- a| < delta_2$
$endgroup$
– rubikscube09
Jan 25 at 21:35












$begingroup$
I do not know how exact your copy of the proof is, but as the version that you have presented here is written very poorly.
$endgroup$
– Carsten S
Jan 26 at 14:19




$begingroup$
I do not know how exact your copy of the proof is, but as the version that you have presented here is written very poorly.
$endgroup$
– Carsten S
Jan 26 at 14:19












$begingroup$
The original proof was in German and there was more text involved...
$endgroup$
– RM777
Jan 26 at 16:29




$begingroup$
The original proof was in German and there was more text involved...
$endgroup$
– RM777
Jan 26 at 16:29










1 Answer
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$begingroup$

Take $x=x_0-delta /2$ in the argument. Note that $|x-x_0|=delta /2 <delta$ so we get $|f(x)-f(x_0)| <epsilon_0$ and hence $f(x) >f(x_0)-epsilon_0 >y_0$. This is a contradiction since $x <x_0$.






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    $begingroup$

    Take $x=x_0-delta /2$ in the argument. Note that $|x-x_0|=delta /2 <delta$ so we get $|f(x)-f(x_0)| <epsilon_0$ and hence $f(x) >f(x_0)-epsilon_0 >y_0$. This is a contradiction since $x <x_0$.






    share|cite|improve this answer









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      $begingroup$

      Take $x=x_0-delta /2$ in the argument. Note that $|x-x_0|=delta /2 <delta$ so we get $|f(x)-f(x_0)| <epsilon_0$ and hence $f(x) >f(x_0)-epsilon_0 >y_0$. This is a contradiction since $x <x_0$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Take $x=x_0-delta /2$ in the argument. Note that $|x-x_0|=delta /2 <delta$ so we get $|f(x)-f(x_0)| <epsilon_0$ and hence $f(x) >f(x_0)-epsilon_0 >y_0$. This is a contradiction since $x <x_0$.






        share|cite|improve this answer









        $endgroup$



        Take $x=x_0-delta /2$ in the argument. Note that $|x-x_0|=delta /2 <delta$ so we get $|f(x)-f(x_0)| <epsilon_0$ and hence $f(x) >f(x_0)-epsilon_0 >y_0$. This is a contradiction since $x <x_0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 26 at 12:12









        Kavi Rama MurthyKavi Rama Murthy

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