Difference between continuous over a set and continue in every point of that set












0












$begingroup$


In one of the first chapters in my class about mathematical analysis we learned about continuity, but when we talked about continuity in sets i found something that is in my opinion kinda weird.



Suppose that $f$ is a function with domain $mathcal{D}$, then $f$ is continuous in (or over) a set $A subseteq mathcal{D}$ if the restriction $f/A$ is continuous in every point:



begin{equation}
limlimits_{tto x,tin A} f(t) = f(x), hspace{1cm} forall xin A
end{equation}
or also if:



begin{equation}
(forall x in A)(forall varepsilon > 0)(existsdelta >0)(forall t in A)(|t-x|<delta Longrightarrow |f(t)-f(x)|< varepsilon)
end{equation}



And this is all well and good but under the definition in a note they say:
This formula doesn't express that $f$ is continuous in every point of $A$ because the formula for that is:



begin{equation}
(forall x in A)(forall varepsilon > 0)(existsdelta >0)(forall t in mathcal{D})(|t-x|<delta Longrightarrow |f(t)-f(x)|< varepsilon)
end{equation}



And also analogous, '$f$ is continuous in $A$' isn't equal to:



begin{equation}
limlimits_{tto x} f(t) = f(x), hspace{1cm} forall xin A
end{equation}



The difference between both formula is that in the $4^{th}$ pair of brackets the set $A$ is replaced with the domain $mathcal{D}$. But i can't quite wrap my head around that. I don't quite get what the difference actually is and how this difference expresses itself.



Would someone be able to explain it, maybe even with an example?






real-analysis continuity epsilon-delta







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    In one of the first chapters in my class about mathematical analysis we learned about continuity, but when we talked about continuity in sets i found something that is in my opinion kinda weird.



    Suppose that $f$ is a function with domain $mathcal{D}$, then $f$ is continuous in (or over) a set $A subseteq mathcal{D}$ if the restriction $f/A$ is continuous in every point:



    begin{equation}
    limlimits_{tto x,tin A} f(t) = f(x), hspace{1cm} forall xin A
    end{equation}
    or also if:



    begin{equation}
    (forall x in A)(forall varepsilon > 0)(existsdelta >0)(forall t in A)(|t-x|<delta Longrightarrow |f(t)-f(x)|< varepsilon)
    end{equation}



    And this is all well and good but under the definition in a note they say:
    This formula doesn't express that $f$ is continuous in every point of $A$ because the formula for that is:



    begin{equation}
    (forall x in A)(forall varepsilon > 0)(existsdelta >0)(forall t in mathcal{D})(|t-x|<delta Longrightarrow |f(t)-f(x)|< varepsilon)
    end{equation}



    And also analogous, '$f$ is continuous in $A$' isn't equal to:



    begin{equation}
    limlimits_{tto x} f(t) = f(x), hspace{1cm} forall xin A
    end{equation}



    The difference between both formula is that in the $4^{th}$ pair of brackets the set $A$ is replaced with the domain $mathcal{D}$. But i can't quite wrap my head around that. I don't quite get what the difference actually is and how this difference expresses itself.



    Would someone be able to explain it, maybe even with an example?






    real-analysis continuity epsilon-delta







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      In one of the first chapters in my class about mathematical analysis we learned about continuity, but when we talked about continuity in sets i found something that is in my opinion kinda weird.



      Suppose that $f$ is a function with domain $mathcal{D}$, then $f$ is continuous in (or over) a set $A subseteq mathcal{D}$ if the restriction $f/A$ is continuous in every point:



      begin{equation}
      limlimits_{tto x,tin A} f(t) = f(x), hspace{1cm} forall xin A
      end{equation}
      or also if:



      begin{equation}
      (forall x in A)(forall varepsilon > 0)(existsdelta >0)(forall t in A)(|t-x|<delta Longrightarrow |f(t)-f(x)|< varepsilon)
      end{equation}



      And this is all well and good but under the definition in a note they say:
      This formula doesn't express that $f$ is continuous in every point of $A$ because the formula for that is:



      begin{equation}
      (forall x in A)(forall varepsilon > 0)(existsdelta >0)(forall t in mathcal{D})(|t-x|<delta Longrightarrow |f(t)-f(x)|< varepsilon)
      end{equation}



      And also analogous, '$f$ is continuous in $A$' isn't equal to:



      begin{equation}
      limlimits_{tto x} f(t) = f(x), hspace{1cm} forall xin A
      end{equation}



      The difference between both formula is that in the $4^{th}$ pair of brackets the set $A$ is replaced with the domain $mathcal{D}$. But i can't quite wrap my head around that. I don't quite get what the difference actually is and how this difference expresses itself.



      Would someone be able to explain it, maybe even with an example?






      real-analysis continuity epsilon-delta







      share|cite|improve this question











      $endgroup$




      In one of the first chapters in my class about mathematical analysis we learned about continuity, but when we talked about continuity in sets i found something that is in my opinion kinda weird.



      Suppose that $f$ is a function with domain $mathcal{D}$, then $f$ is continuous in (or over) a set $A subseteq mathcal{D}$ if the restriction $f/A$ is continuous in every point:



      begin{equation}
      limlimits_{tto x,tin A} f(t) = f(x), hspace{1cm} forall xin A
      end{equation}
      or also if:



      begin{equation}
      (forall x in A)(forall varepsilon > 0)(existsdelta >0)(forall t in A)(|t-x|<delta Longrightarrow |f(t)-f(x)|< varepsilon)
      end{equation}



      And this is all well and good but under the definition in a note they say:
      This formula doesn't express that $f$ is continuous in every point of $A$ because the formula for that is:



      begin{equation}
      (forall x in A)(forall varepsilon > 0)(existsdelta >0)(forall t in mathcal{D})(|t-x|<delta Longrightarrow |f(t)-f(x)|< varepsilon)
      end{equation}



      And also analogous, '$f$ is continuous in $A$' isn't equal to:



      begin{equation}
      limlimits_{tto x} f(t) = f(x), hspace{1cm} forall xin A
      end{equation}



      The difference between both formula is that in the $4^{th}$ pair of brackets the set $A$ is replaced with the domain $mathcal{D}$. But i can't quite wrap my head around that. I don't quite get what the difference actually is and how this difference expresses itself.



      Would someone be able to explain it, maybe even with an example?






      real-analysis continuity epsilon-delta




      real-analysis continuity epsilon-delta






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 25 '18 at 19:21









      José Carlos Santos

      167k22132235




      167k22132235










      asked Dec 25 '18 at 19:12









      ViktorViktor

      1389




      1389






















          1 Answer
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          3












          $begingroup$

          Here's an example: consider the map$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}1&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}end{array}$$Then $f$ is discontinuous at every point of its domain. However, $f|_{mathbb Q}$ is continuous.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for answering, I think I understand now, but just a quick question. Wouldn't $f$ be continuous between 2 consecutive points in $mathbb{Q}$ since there are infinitely many points in $mathbb{R}$ between those points in $mathbb{Q}$, or am i missing something here?
            $endgroup$
            – Viktor
            Dec 25 '18 at 20:07








          • 1




            $begingroup$
            You are missing the fact that there is no such thing as “consecutive points in $mathbb Q$”, since for each two distinct rational numbers there are infinitely many other rational numbers between them.
            $endgroup$
            – José Carlos Santos
            Dec 25 '18 at 20:09










          • $begingroup$
            To the proposer: Constant functions are continuous. The function $f|_{Bbb Q}$ is constant on its domain, $Bbb Q.$
            $endgroup$
            – DanielWainfleet
            Dec 26 '18 at 1:16











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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

          votes









          3












          $begingroup$

          Here's an example: consider the map$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}1&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}end{array}$$Then $f$ is discontinuous at every point of its domain. However, $f|_{mathbb Q}$ is continuous.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for answering, I think I understand now, but just a quick question. Wouldn't $f$ be continuous between 2 consecutive points in $mathbb{Q}$ since there are infinitely many points in $mathbb{R}$ between those points in $mathbb{Q}$, or am i missing something here?
            $endgroup$
            – Viktor
            Dec 25 '18 at 20:07








          • 1




            $begingroup$
            You are missing the fact that there is no such thing as “consecutive points in $mathbb Q$”, since for each two distinct rational numbers there are infinitely many other rational numbers between them.
            $endgroup$
            – José Carlos Santos
            Dec 25 '18 at 20:09










          • $begingroup$
            To the proposer: Constant functions are continuous. The function $f|_{Bbb Q}$ is constant on its domain, $Bbb Q.$
            $endgroup$
            – DanielWainfleet
            Dec 26 '18 at 1:16
















          3












          $begingroup$

          Here's an example: consider the map$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}1&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}end{array}$$Then $f$ is discontinuous at every point of its domain. However, $f|_{mathbb Q}$ is continuous.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for answering, I think I understand now, but just a quick question. Wouldn't $f$ be continuous between 2 consecutive points in $mathbb{Q}$ since there are infinitely many points in $mathbb{R}$ between those points in $mathbb{Q}$, or am i missing something here?
            $endgroup$
            – Viktor
            Dec 25 '18 at 20:07








          • 1




            $begingroup$
            You are missing the fact that there is no such thing as “consecutive points in $mathbb Q$”, since for each two distinct rational numbers there are infinitely many other rational numbers between them.
            $endgroup$
            – José Carlos Santos
            Dec 25 '18 at 20:09










          • $begingroup$
            To the proposer: Constant functions are continuous. The function $f|_{Bbb Q}$ is constant on its domain, $Bbb Q.$
            $endgroup$
            – DanielWainfleet
            Dec 26 '18 at 1:16














          3












          3








          3





          $begingroup$

          Here's an example: consider the map$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}1&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}end{array}$$Then $f$ is discontinuous at every point of its domain. However, $f|_{mathbb Q}$ is continuous.






          share|cite|improve this answer









          $endgroup$



          Here's an example: consider the map$$begin{array}{rccc}fcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begin{cases}1&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}end{array}$$Then $f$ is discontinuous at every point of its domain. However, $f|_{mathbb Q}$ is continuous.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 19:18









          José Carlos SantosJosé Carlos Santos

          167k22132235




          167k22132235












          • $begingroup$
            Thanks for answering, I think I understand now, but just a quick question. Wouldn't $f$ be continuous between 2 consecutive points in $mathbb{Q}$ since there are infinitely many points in $mathbb{R}$ between those points in $mathbb{Q}$, or am i missing something here?
            $endgroup$
            – Viktor
            Dec 25 '18 at 20:07








          • 1




            $begingroup$
            You are missing the fact that there is no such thing as “consecutive points in $mathbb Q$”, since for each two distinct rational numbers there are infinitely many other rational numbers between them.
            $endgroup$
            – José Carlos Santos
            Dec 25 '18 at 20:09










          • $begingroup$
            To the proposer: Constant functions are continuous. The function $f|_{Bbb Q}$ is constant on its domain, $Bbb Q.$
            $endgroup$
            – DanielWainfleet
            Dec 26 '18 at 1:16


















          • $begingroup$
            Thanks for answering, I think I understand now, but just a quick question. Wouldn't $f$ be continuous between 2 consecutive points in $mathbb{Q}$ since there are infinitely many points in $mathbb{R}$ between those points in $mathbb{Q}$, or am i missing something here?
            $endgroup$
            – Viktor
            Dec 25 '18 at 20:07








          • 1




            $begingroup$
            You are missing the fact that there is no such thing as “consecutive points in $mathbb Q$”, since for each two distinct rational numbers there are infinitely many other rational numbers between them.
            $endgroup$
            – José Carlos Santos
            Dec 25 '18 at 20:09










          • $begingroup$
            To the proposer: Constant functions are continuous. The function $f|_{Bbb Q}$ is constant on its domain, $Bbb Q.$
            $endgroup$
            – DanielWainfleet
            Dec 26 '18 at 1:16
















          $begingroup$
          Thanks for answering, I think I understand now, but just a quick question. Wouldn't $f$ be continuous between 2 consecutive points in $mathbb{Q}$ since there are infinitely many points in $mathbb{R}$ between those points in $mathbb{Q}$, or am i missing something here?
          $endgroup$
          – Viktor
          Dec 25 '18 at 20:07






          $begingroup$
          Thanks for answering, I think I understand now, but just a quick question. Wouldn't $f$ be continuous between 2 consecutive points in $mathbb{Q}$ since there are infinitely many points in $mathbb{R}$ between those points in $mathbb{Q}$, or am i missing something here?
          $endgroup$
          – Viktor
          Dec 25 '18 at 20:07






          1




          1




          $begingroup$
          You are missing the fact that there is no such thing as “consecutive points in $mathbb Q$”, since for each two distinct rational numbers there are infinitely many other rational numbers between them.
          $endgroup$
          – José Carlos Santos
          Dec 25 '18 at 20:09




          $begingroup$
          You are missing the fact that there is no such thing as “consecutive points in $mathbb Q$”, since for each two distinct rational numbers there are infinitely many other rational numbers between them.
          $endgroup$
          – José Carlos Santos
          Dec 25 '18 at 20:09












          $begingroup$
          To the proposer: Constant functions are continuous. The function $f|_{Bbb Q}$ is constant on its domain, $Bbb Q.$
          $endgroup$
          – DanielWainfleet
          Dec 26 '18 at 1:16




          $begingroup$
          To the proposer: Constant functions are continuous. The function $f|_{Bbb Q}$ is constant on its domain, $Bbb Q.$
          $endgroup$
          – DanielWainfleet
          Dec 26 '18 at 1:16


















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