Every Interval $I subset mathbb{R}$ is connected












0












$begingroup$


The proof I've been given is




If not, there is a non-constant continuous $f$ from $I$ to
discrete ${0, 1}$. Then $f: I rightarrow mathbb{R}$ is also continuous which contradicts the Intermediate Value Theorem.




Why must $f: I rightarrow mathbb{R}$ also be continuous?






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    $f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
    $endgroup$
    – fleablood
    May 20 '17 at 15:58






  • 1




    $begingroup$
    I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
    $endgroup$
    – fleablood
    May 20 '17 at 15:59






  • 1




    $begingroup$
    It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
    $endgroup$
    – Matematleta
    May 20 '17 at 16:02


















0












$begingroup$


The proof I've been given is




If not, there is a non-constant continuous $f$ from $I$ to
discrete ${0, 1}$. Then $f: I rightarrow mathbb{R}$ is also continuous which contradicts the Intermediate Value Theorem.




Why must $f: I rightarrow mathbb{R}$ also be continuous?






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    $f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
    $endgroup$
    – fleablood
    May 20 '17 at 15:58






  • 1




    $begingroup$
    I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
    $endgroup$
    – fleablood
    May 20 '17 at 15:59






  • 1




    $begingroup$
    It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
    $endgroup$
    – Matematleta
    May 20 '17 at 16:02
















0












0








0





$begingroup$


The proof I've been given is




If not, there is a non-constant continuous $f$ from $I$ to
discrete ${0, 1}$. Then $f: I rightarrow mathbb{R}$ is also continuous which contradicts the Intermediate Value Theorem.




Why must $f: I rightarrow mathbb{R}$ also be continuous?






general-topology







share|cite|improve this question











$endgroup$




The proof I've been given is




If not, there is a non-constant continuous $f$ from $I$ to
discrete ${0, 1}$. Then $f: I rightarrow mathbb{R}$ is also continuous which contradicts the Intermediate Value Theorem.




Why must $f: I rightarrow mathbb{R}$ also be continuous?






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 0:19









amWhy

1




1










asked May 20 '17 at 15:44









user440780user440780

143




143












  • $begingroup$
    $f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
    $endgroup$
    – fleablood
    May 20 '17 at 15:58






  • 1




    $begingroup$
    I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
    $endgroup$
    – fleablood
    May 20 '17 at 15:59






  • 1




    $begingroup$
    It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
    $endgroup$
    – Matematleta
    May 20 '17 at 16:02




















  • $begingroup$
    $f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
    $endgroup$
    – fleablood
    May 20 '17 at 15:58






  • 1




    $begingroup$
    I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
    $endgroup$
    – fleablood
    May 20 '17 at 15:59






  • 1




    $begingroup$
    It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
    $endgroup$
    – Matematleta
    May 20 '17 at 16:02


















$begingroup$
$f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
$endgroup$
– fleablood
May 20 '17 at 15:58




$begingroup$
$f: I rightarrow {0,1} subset mathbb R$. Viewing the space containing the codomain as a subset of a larger set does not affect whether a function is continuous or not. Continuity is determined by the behavior of the domain.
$endgroup$
– fleablood
May 20 '17 at 15:58




1




1




$begingroup$
I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
$endgroup$
– fleablood
May 20 '17 at 15:59




$begingroup$
I have to wonder though if that proof isn't circular and it is my impression that the Intermediate Value Theorem relies upon the connectedness of intervals in its proof.
$endgroup$
– fleablood
May 20 '17 at 15:59




1




1




$begingroup$
It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
$endgroup$
– Matematleta
May 20 '17 at 16:02






$begingroup$
It $is$ circular. I never liked this proof because just as you say, the various proofs of IVT use at some point, the fact that intervals are connected.
$endgroup$
– Matematleta
May 20 '17 at 16:02












2 Answers
2






active

oldest

votes


















1












$begingroup$

The inclusion ${0,1} to mathbb{R}$ is continuous because ${0,1}$ has the discrete topology.



Compositions of continuous functions are continuous.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If $I$ is not connected, then by definition there are $A$ and $B$ both open such that $A cup B=I$, if you consider $f$ such that $f(A)=0$ and $f(B)=1$ than this is a continuos function by definition, infact in $ {0,1} $ as a subset of $mathbb R$ there is the discrete topology and $ f^{-1}( {0,1})=I $ that is open, $ f^{-1}( {0})=A $ that is open, $ f^{-1}( {1})=B $ that is open and
    $ f^{-1}(emptyset)=emptyset $ that is open so you have a contraddiction with the IVT. However this demonstration is circular because you have to use the connectedness of $I$ for proving the IVT.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      The inclusion ${0,1} to mathbb{R}$ is continuous because ${0,1}$ has the discrete topology.



      Compositions of continuous functions are continuous.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The inclusion ${0,1} to mathbb{R}$ is continuous because ${0,1}$ has the discrete topology.



        Compositions of continuous functions are continuous.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The inclusion ${0,1} to mathbb{R}$ is continuous because ${0,1}$ has the discrete topology.



          Compositions of continuous functions are continuous.






          share|cite|improve this answer









          $endgroup$



          The inclusion ${0,1} to mathbb{R}$ is continuous because ${0,1}$ has the discrete topology.



          Compositions of continuous functions are continuous.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 20 '17 at 15:47









          user133281user133281

          13.6k22551




          13.6k22551























              0












              $begingroup$

              If $I$ is not connected, then by definition there are $A$ and $B$ both open such that $A cup B=I$, if you consider $f$ such that $f(A)=0$ and $f(B)=1$ than this is a continuos function by definition, infact in $ {0,1} $ as a subset of $mathbb R$ there is the discrete topology and $ f^{-1}( {0,1})=I $ that is open, $ f^{-1}( {0})=A $ that is open, $ f^{-1}( {1})=B $ that is open and
              $ f^{-1}(emptyset)=emptyset $ that is open so you have a contraddiction with the IVT. However this demonstration is circular because you have to use the connectedness of $I$ for proving the IVT.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                If $I$ is not connected, then by definition there are $A$ and $B$ both open such that $A cup B=I$, if you consider $f$ such that $f(A)=0$ and $f(B)=1$ than this is a continuos function by definition, infact in $ {0,1} $ as a subset of $mathbb R$ there is the discrete topology and $ f^{-1}( {0,1})=I $ that is open, $ f^{-1}( {0})=A $ that is open, $ f^{-1}( {1})=B $ that is open and
                $ f^{-1}(emptyset)=emptyset $ that is open so you have a contraddiction with the IVT. However this demonstration is circular because you have to use the connectedness of $I$ for proving the IVT.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If $I$ is not connected, then by definition there are $A$ and $B$ both open such that $A cup B=I$, if you consider $f$ such that $f(A)=0$ and $f(B)=1$ than this is a continuos function by definition, infact in $ {0,1} $ as a subset of $mathbb R$ there is the discrete topology and $ f^{-1}( {0,1})=I $ that is open, $ f^{-1}( {0})=A $ that is open, $ f^{-1}( {1})=B $ that is open and
                  $ f^{-1}(emptyset)=emptyset $ that is open so you have a contraddiction with the IVT. However this demonstration is circular because you have to use the connectedness of $I$ for proving the IVT.






                  share|cite|improve this answer











                  $endgroup$



                  If $I$ is not connected, then by definition there are $A$ and $B$ both open such that $A cup B=I$, if you consider $f$ such that $f(A)=0$ and $f(B)=1$ than this is a continuos function by definition, infact in $ {0,1} $ as a subset of $mathbb R$ there is the discrete topology and $ f^{-1}( {0,1})=I $ that is open, $ f^{-1}( {0})=A $ that is open, $ f^{-1}( {1})=B $ that is open and
                  $ f^{-1}(emptyset)=emptyset $ that is open so you have a contraddiction with the IVT. However this demonstration is circular because you have to use the connectedness of $I$ for proving the IVT.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited May 20 '17 at 16:36

























                  answered May 20 '17 at 15:56









                  StabiloBossStabiloBoss

                  916




                  916






























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