Questions about the proof of the intermediate value theorem












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$begingroup$


Let$ f:[a,b]rightarrow mathbb{R}$ be continious. Then for every $y_0in[min(f(a),f(b)), max (f(a),f(b))] $ there exists at least one $x_0in[a,b]$ with $f(x_0)=y_0$.



Proof: Without loss of generality let $f(a)leq f(b)$.



We define the set $ M$



$M:={xin[a,b]:f(x)geq y_0}=f^{-1}([y_0,+infty))subseteq [a,b].$



Then $bin M$, thus $M neq emptyset$. Let $x_0 := inf M in [a,b]$. There exists a sequence $(x_n)$ in $M$ with $x_nrightarrow x_0$. Because $f$ is continious, we have $f(x_n)rightarrow f(x_0)$, and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$, in conclusion $x_0in M$, thus $x_0= min M$.



I have understood the rest of the proof but if somebody wants me to write it down I will do it.



For an infinum we need a lower bound what would be the lower bound in this case, and how do we know that the infinum sits in the intervall? Also I don't understand this part "...and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$..." Every element of the sequence is $geq y_0$, why must the same also apply to the Limit?






real-analysis







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    2












    $begingroup$


    Let$ f:[a,b]rightarrow mathbb{R}$ be continious. Then for every $y_0in[min(f(a),f(b)), max (f(a),f(b))] $ there exists at least one $x_0in[a,b]$ with $f(x_0)=y_0$.



    Proof: Without loss of generality let $f(a)leq f(b)$.



    We define the set $ M$



    $M:={xin[a,b]:f(x)geq y_0}=f^{-1}([y_0,+infty))subseteq [a,b].$



    Then $bin M$, thus $M neq emptyset$. Let $x_0 := inf M in [a,b]$. There exists a sequence $(x_n)$ in $M$ with $x_nrightarrow x_0$. Because $f$ is continious, we have $f(x_n)rightarrow f(x_0)$, and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$, in conclusion $x_0in M$, thus $x_0= min M$.



    I have understood the rest of the proof but if somebody wants me to write it down I will do it.



    For an infinum we need a lower bound what would be the lower bound in this case, and how do we know that the infinum sits in the intervall? Also I don't understand this part "...and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$..." Every element of the sequence is $geq y_0$, why must the same also apply to the Limit?






    real-analysis







    share|cite|improve this question











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      2








      2


      0



      $begingroup$


      Let$ f:[a,b]rightarrow mathbb{R}$ be continious. Then for every $y_0in[min(f(a),f(b)), max (f(a),f(b))] $ there exists at least one $x_0in[a,b]$ with $f(x_0)=y_0$.



      Proof: Without loss of generality let $f(a)leq f(b)$.



      We define the set $ M$



      $M:={xin[a,b]:f(x)geq y_0}=f^{-1}([y_0,+infty))subseteq [a,b].$



      Then $bin M$, thus $M neq emptyset$. Let $x_0 := inf M in [a,b]$. There exists a sequence $(x_n)$ in $M$ with $x_nrightarrow x_0$. Because $f$ is continious, we have $f(x_n)rightarrow f(x_0)$, and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$, in conclusion $x_0in M$, thus $x_0= min M$.



      I have understood the rest of the proof but if somebody wants me to write it down I will do it.



      For an infinum we need a lower bound what would be the lower bound in this case, and how do we know that the infinum sits in the intervall? Also I don't understand this part "...and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$..." Every element of the sequence is $geq y_0$, why must the same also apply to the Limit?






      real-analysis







      share|cite|improve this question











      $endgroup$




      Let$ f:[a,b]rightarrow mathbb{R}$ be continious. Then for every $y_0in[min(f(a),f(b)), max (f(a),f(b))] $ there exists at least one $x_0in[a,b]$ with $f(x_0)=y_0$.



      Proof: Without loss of generality let $f(a)leq f(b)$.



      We define the set $ M$



      $M:={xin[a,b]:f(x)geq y_0}=f^{-1}([y_0,+infty))subseteq [a,b].$



      Then $bin M$, thus $M neq emptyset$. Let $x_0 := inf M in [a,b]$. There exists a sequence $(x_n)$ in $M$ with $x_nrightarrow x_0$. Because $f$ is continious, we have $f(x_n)rightarrow f(x_0)$, and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$, in conclusion $x_0in M$, thus $x_0= min M$.



      I have understood the rest of the proof but if somebody wants me to write it down I will do it.



      For an infinum we need a lower bound what would be the lower bound in this case, and how do we know that the infinum sits in the intervall? Also I don't understand this part "...and because $f(x_n)geq y_0$ we also have $f(x_0)geq y_0$..." Every element of the sequence is $geq y_0$, why must the same also apply to the Limit?






      real-analysis




      real-analysis






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      edited Jan 11 at 13:20









      Song

      12.2k630




      12.2k630










      asked Jan 11 at 13:05









      RM777RM777

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          $begingroup$


          • Since $M subset [a,b]$ it follows that $a$ is a lower bound.

          • For the same reason you have $aleq inf M$.

          • In general it holds $c leq y_n stackrel{ntoinfty}{rightarrow} y_0 Rightarrow c leq y_0$. You can see this quickly by contradiction:


          Assume $y_0 < c$. Choose $epsilon > 0$ with $y_0 + epsilon < c stackrel{y_n stackrel{ntoinfty}{rightarrow} y_0}{Longrightarrow} c leq y_n < y_0 + epsilon < c$ for sufficient large $n$. Contradiction!






          share|cite|improve this answer









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            $begingroup$

            The infimum of a set lies in the closure of the set (see here). So $inf M$ lies in $overline M subseteq overline {[a, b]} = [a, b]$. Basically, the infimum is in $[a, b]$ because it is a closed interval.



            For your other question, yes: if all elements of a converging sequence are $geq y_0$, then the limit is also $geq y_0$. The same thing holds for "$leq$". You can prove it by contradiction: assume the limit is $lneqq y$. So you have $d := y_0-f(x_0) gneqq 0$. By definition of convergence, there exists $N$ sufficiently big such that $f(x_n)-f(x_0) leq frac d2$, therefore $y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) gneqq d - frac d2 gneqq 0$ for $n > N$ and this is a contradiction.






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              $begingroup$


              • Since $M subset [a,b]$ it follows that $a$ is a lower bound.

              • For the same reason you have $aleq inf M$.

              • In general it holds $c leq y_n stackrel{ntoinfty}{rightarrow} y_0 Rightarrow c leq y_0$. You can see this quickly by contradiction:


              Assume $y_0 < c$. Choose $epsilon > 0$ with $y_0 + epsilon < c stackrel{y_n stackrel{ntoinfty}{rightarrow} y_0}{Longrightarrow} c leq y_n < y_0 + epsilon < c$ for sufficient large $n$. Contradiction!






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$


                • Since $M subset [a,b]$ it follows that $a$ is a lower bound.

                • For the same reason you have $aleq inf M$.

                • In general it holds $c leq y_n stackrel{ntoinfty}{rightarrow} y_0 Rightarrow c leq y_0$. You can see this quickly by contradiction:


                Assume $y_0 < c$. Choose $epsilon > 0$ with $y_0 + epsilon < c stackrel{y_n stackrel{ntoinfty}{rightarrow} y_0}{Longrightarrow} c leq y_n < y_0 + epsilon < c$ for sufficient large $n$. Contradiction!






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$


                  • Since $M subset [a,b]$ it follows that $a$ is a lower bound.

                  • For the same reason you have $aleq inf M$.

                  • In general it holds $c leq y_n stackrel{ntoinfty}{rightarrow} y_0 Rightarrow c leq y_0$. You can see this quickly by contradiction:


                  Assume $y_0 < c$. Choose $epsilon > 0$ with $y_0 + epsilon < c stackrel{y_n stackrel{ntoinfty}{rightarrow} y_0}{Longrightarrow} c leq y_n < y_0 + epsilon < c$ for sufficient large $n$. Contradiction!






                  share|cite|improve this answer









                  $endgroup$




                  • Since $M subset [a,b]$ it follows that $a$ is a lower bound.

                  • For the same reason you have $aleq inf M$.

                  • In general it holds $c leq y_n stackrel{ntoinfty}{rightarrow} y_0 Rightarrow c leq y_0$. You can see this quickly by contradiction:


                  Assume $y_0 < c$. Choose $epsilon > 0$ with $y_0 + epsilon < c stackrel{y_n stackrel{ntoinfty}{rightarrow} y_0}{Longrightarrow} c leq y_n < y_0 + epsilon < c$ for sufficient large $n$. Contradiction!







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 11 at 13:19









                  trancelocationtrancelocation

                  11k1723




                  11k1723























                      2












                      $begingroup$

                      The infimum of a set lies in the closure of the set (see here). So $inf M$ lies in $overline M subseteq overline {[a, b]} = [a, b]$. Basically, the infimum is in $[a, b]$ because it is a closed interval.



                      For your other question, yes: if all elements of a converging sequence are $geq y_0$, then the limit is also $geq y_0$. The same thing holds for "$leq$". You can prove it by contradiction: assume the limit is $lneqq y$. So you have $d := y_0-f(x_0) gneqq 0$. By definition of convergence, there exists $N$ sufficiently big such that $f(x_n)-f(x_0) leq frac d2$, therefore $y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) gneqq d - frac d2 gneqq 0$ for $n > N$ and this is a contradiction.






                      share|cite|improve this answer











                      $endgroup$


















                        2












                        $begingroup$

                        The infimum of a set lies in the closure of the set (see here). So $inf M$ lies in $overline M subseteq overline {[a, b]} = [a, b]$. Basically, the infimum is in $[a, b]$ because it is a closed interval.



                        For your other question, yes: if all elements of a converging sequence are $geq y_0$, then the limit is also $geq y_0$. The same thing holds for "$leq$". You can prove it by contradiction: assume the limit is $lneqq y$. So you have $d := y_0-f(x_0) gneqq 0$. By definition of convergence, there exists $N$ sufficiently big such that $f(x_n)-f(x_0) leq frac d2$, therefore $y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) gneqq d - frac d2 gneqq 0$ for $n > N$ and this is a contradiction.






                        share|cite|improve this answer











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          The infimum of a set lies in the closure of the set (see here). So $inf M$ lies in $overline M subseteq overline {[a, b]} = [a, b]$. Basically, the infimum is in $[a, b]$ because it is a closed interval.



                          For your other question, yes: if all elements of a converging sequence are $geq y_0$, then the limit is also $geq y_0$. The same thing holds for "$leq$". You can prove it by contradiction: assume the limit is $lneqq y$. So you have $d := y_0-f(x_0) gneqq 0$. By definition of convergence, there exists $N$ sufficiently big such that $f(x_n)-f(x_0) leq frac d2$, therefore $y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) gneqq d - frac d2 gneqq 0$ for $n > N$ and this is a contradiction.






                          share|cite|improve this answer











                          $endgroup$



                          The infimum of a set lies in the closure of the set (see here). So $inf M$ lies in $overline M subseteq overline {[a, b]} = [a, b]$. Basically, the infimum is in $[a, b]$ because it is a closed interval.



                          For your other question, yes: if all elements of a converging sequence are $geq y_0$, then the limit is also $geq y_0$. The same thing holds for "$leq$". You can prove it by contradiction: assume the limit is $lneqq y$. So you have $d := y_0-f(x_0) gneqq 0$. By definition of convergence, there exists $N$ sufficiently big such that $f(x_n)-f(x_0) leq frac d2$, therefore $y_0 - f(x_n) = (y_0 - f(x_0)) + (f(x_0)-f(x_n)) gneqq d - frac d2 gneqq 0$ for $n > N$ and this is a contradiction.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 11 at 13:47

























                          answered Jan 11 at 13:19









                          BermudesBermudes

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                          18713






























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