Proving that two real functions are equal [duplicate]












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  • Continuous function and dense set

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Let $f : mathbb{R} to mathbb{R}$ and $g : mathbb{R} to mathbb{R}$ be continuous functions. Suppose that $D ⊆ mathbb{R}$,
and that $D$ is dense in $mathbb{R}$. Suppose that $f(x) = g(x)$ for every $x ∈ D$. Prove that
$f(x) = g(x)$ for every $x ∈ mathbb{R}$.




Any tips for where I can start here? I'm pulling up blanks for this.



EDIT: there is a very similar question asked, but I have no idea what a metric space is so the whole thing didn't mean a ton to me unfortunately.






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 Nov 21 '18 at 7:24


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    0















    This question already has an answer here:




    • Continuous function and dense set

      2 answers





    Let $f : mathbb{R} to mathbb{R}$ and $g : mathbb{R} to mathbb{R}$ be continuous functions. Suppose that $D ⊆ mathbb{R}$,
    and that $D$ is dense in $mathbb{R}$. Suppose that $f(x) = g(x)$ for every $x ∈ D$. Prove that
    $f(x) = g(x)$ for every $x ∈ mathbb{R}$.




    Any tips for where I can start here? I'm pulling up blanks for this.



    EDIT: there is a very similar question asked, but I have no idea what a metric space is so the whole thing didn't mean a ton to me unfortunately.






    real-analysis







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    marked as duplicate by José Carlos Santos real-analysis
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     Nov 21 '18 at 7:24


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      This question already has an answer here:




      • Continuous function and dense set

        2 answers





      Let $f : mathbb{R} to mathbb{R}$ and $g : mathbb{R} to mathbb{R}$ be continuous functions. Suppose that $D ⊆ mathbb{R}$,
      and that $D$ is dense in $mathbb{R}$. Suppose that $f(x) = g(x)$ for every $x ∈ D$. Prove that
      $f(x) = g(x)$ for every $x ∈ mathbb{R}$.




      Any tips for where I can start here? I'm pulling up blanks for this.



      EDIT: there is a very similar question asked, but I have no idea what a metric space is so the whole thing didn't mean a ton to me unfortunately.






      real-analysis







      share|cite|improve this question
















      This question already has an answer here:




      • Continuous function and dense set

        2 answers





      Let $f : mathbb{R} to mathbb{R}$ and $g : mathbb{R} to mathbb{R}$ be continuous functions. Suppose that $D ⊆ mathbb{R}$,
      and that $D$ is dense in $mathbb{R}$. Suppose that $f(x) = g(x)$ for every $x ∈ D$. Prove that
      $f(x) = g(x)$ for every $x ∈ mathbb{R}$.




      Any tips for where I can start here? I'm pulling up blanks for this.



      EDIT: there is a very similar question asked, but I have no idea what a metric space is so the whole thing didn't mean a ton to me unfortunately.





      This question already has an answer here:




      • Continuous function and dense set

        2 answers







      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 Nov 21 '18 at 7:33

























      asked Nov 21 '18 at 7:07









      ktuggle

      184




      184




      marked as duplicate by José Carlos Santos real-analysis
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       Nov 21 '18 at 7:24


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by José Carlos Santos real-analysis
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          4 Answers
          4






          active

          oldest

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          2














          Hint. Let $x in mathbb{R}setminus D$ then, since $D$ is dense in $mathbb{R}$, there exists a sequence ${d_n}_n$ in $D$ such that $d_nto x$. By your assumption $f(d_n)=g(d_n)$. Now use the continuity of $f$ and $g$ at $x$.






          share|cite|improve this answer























          • I don't understand the purpose of saying $xnot in mathbb{R}setminus D$, isn't the just a really roundabout way of saying x is in D?
            – ktuggle
            Nov 21 '18 at 8:10










          • @ktuggle Sorry, it is $x in mathbb{R}setminus D$, that is $xnotin D$.
            – Robert Z
            Nov 21 '18 at 8:50





















          1














          Hint: Consider $h(x) = f(x)-g(x)$. This should simplify your problem.






          share|cite|improve this answer





























            0














            The set $U:={,f(x)ne g(x)mid xinBbb R}$ is open. We are given that $Ucap D=emptyset$, but as $D$ is dense, it has non-empty intersection with every non-empty open set. We conclude that $U$ is empty.






            share|cite|improve this answer





























              0














              Hint: Prove the set $ A= {x in mathbb{R} vert f (x)=g (x)}$ is closed in $mathbb {R}$. Clearly, $D subseteq A $. Taking closure on both sides, we get $mathbb {R} subseteq A implies mathbb {R}=A $.






              share|cite|improve this answer




























                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                2














                Hint. Let $x in mathbb{R}setminus D$ then, since $D$ is dense in $mathbb{R}$, there exists a sequence ${d_n}_n$ in $D$ such that $d_nto x$. By your assumption $f(d_n)=g(d_n)$. Now use the continuity of $f$ and $g$ at $x$.






                share|cite|improve this answer























                • I don't understand the purpose of saying $xnot in mathbb{R}setminus D$, isn't the just a really roundabout way of saying x is in D?
                  – ktuggle
                  Nov 21 '18 at 8:10










                • @ktuggle Sorry, it is $x in mathbb{R}setminus D$, that is $xnotin D$.
                  – Robert Z
                  Nov 21 '18 at 8:50


















                2














                Hint. Let $x in mathbb{R}setminus D$ then, since $D$ is dense in $mathbb{R}$, there exists a sequence ${d_n}_n$ in $D$ such that $d_nto x$. By your assumption $f(d_n)=g(d_n)$. Now use the continuity of $f$ and $g$ at $x$.






                share|cite|improve this answer























                • I don't understand the purpose of saying $xnot in mathbb{R}setminus D$, isn't the just a really roundabout way of saying x is in D?
                  – ktuggle
                  Nov 21 '18 at 8:10










                • @ktuggle Sorry, it is $x in mathbb{R}setminus D$, that is $xnotin D$.
                  – Robert Z
                  Nov 21 '18 at 8:50
















                2












                2








                2






                Hint. Let $x in mathbb{R}setminus D$ then, since $D$ is dense in $mathbb{R}$, there exists a sequence ${d_n}_n$ in $D$ such that $d_nto x$. By your assumption $f(d_n)=g(d_n)$. Now use the continuity of $f$ and $g$ at $x$.






                share|cite|improve this answer














                Hint. Let $x in mathbb{R}setminus D$ then, since $D$ is dense in $mathbb{R}$, there exists a sequence ${d_n}_n$ in $D$ such that $d_nto x$. By your assumption $f(d_n)=g(d_n)$. Now use the continuity of $f$ and $g$ at $x$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 21 '18 at 8:50

























                answered Nov 21 '18 at 7:13









                Robert Z

                93.3k1061132




                93.3k1061132












                • I don't understand the purpose of saying $xnot in mathbb{R}setminus D$, isn't the just a really roundabout way of saying x is in D?
                  – ktuggle
                  Nov 21 '18 at 8:10










                • @ktuggle Sorry, it is $x in mathbb{R}setminus D$, that is $xnotin D$.
                  – Robert Z
                  Nov 21 '18 at 8:50




















                • I don't understand the purpose of saying $xnot in mathbb{R}setminus D$, isn't the just a really roundabout way of saying x is in D?
                  – ktuggle
                  Nov 21 '18 at 8:10










                • @ktuggle Sorry, it is $x in mathbb{R}setminus D$, that is $xnotin D$.
                  – Robert Z
                  Nov 21 '18 at 8:50


















                I don't understand the purpose of saying $xnot in mathbb{R}setminus D$, isn't the just a really roundabout way of saying x is in D?
                – ktuggle
                Nov 21 '18 at 8:10




                I don't understand the purpose of saying $xnot in mathbb{R}setminus D$, isn't the just a really roundabout way of saying x is in D?
                – ktuggle
                Nov 21 '18 at 8:10












                @ktuggle Sorry, it is $x in mathbb{R}setminus D$, that is $xnotin D$.
                – Robert Z
                Nov 21 '18 at 8:50






                @ktuggle Sorry, it is $x in mathbb{R}setminus D$, that is $xnotin D$.
                – Robert Z
                Nov 21 '18 at 8:50













                1














                Hint: Consider $h(x) = f(x)-g(x)$. This should simplify your problem.






                share|cite|improve this answer


























                  1














                  Hint: Consider $h(x) = f(x)-g(x)$. This should simplify your problem.






                  share|cite|improve this answer
























                    1












                    1








                    1






                    Hint: Consider $h(x) = f(x)-g(x)$. This should simplify your problem.






                    share|cite|improve this answer












                    Hint: Consider $h(x) = f(x)-g(x)$. This should simplify your problem.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 21 '18 at 7:09









                    Jacky Chong

                    17.7k21128




                    17.7k21128























                        0














                        The set $U:={,f(x)ne g(x)mid xinBbb R}$ is open. We are given that $Ucap D=emptyset$, but as $D$ is dense, it has non-empty intersection with every non-empty open set. We conclude that $U$ is empty.






                        share|cite|improve this answer


























                          0














                          The set $U:={,f(x)ne g(x)mid xinBbb R}$ is open. We are given that $Ucap D=emptyset$, but as $D$ is dense, it has non-empty intersection with every non-empty open set. We conclude that $U$ is empty.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            The set $U:={,f(x)ne g(x)mid xinBbb R}$ is open. We are given that $Ucap D=emptyset$, but as $D$ is dense, it has non-empty intersection with every non-empty open set. We conclude that $U$ is empty.






                            share|cite|improve this answer












                            The set $U:={,f(x)ne g(x)mid xinBbb R}$ is open. We are given that $Ucap D=emptyset$, but as $D$ is dense, it has non-empty intersection with every non-empty open set. We conclude that $U$ is empty.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 21 '18 at 7:20









                            Hagen von Eitzen

                            276k21269496




                            276k21269496























                                0














                                Hint: Prove the set $ A= {x in mathbb{R} vert f (x)=g (x)}$ is closed in $mathbb {R}$. Clearly, $D subseteq A $. Taking closure on both sides, we get $mathbb {R} subseteq A implies mathbb {R}=A $.






                                share|cite|improve this answer


























                                  0














                                  Hint: Prove the set $ A= {x in mathbb{R} vert f (x)=g (x)}$ is closed in $mathbb {R}$. Clearly, $D subseteq A $. Taking closure on both sides, we get $mathbb {R} subseteq A implies mathbb {R}=A $.






                                  share|cite|improve this answer
























                                    0












                                    0








                                    0






                                    Hint: Prove the set $ A= {x in mathbb{R} vert f (x)=g (x)}$ is closed in $mathbb {R}$. Clearly, $D subseteq A $. Taking closure on both sides, we get $mathbb {R} subseteq A implies mathbb {R}=A $.






                                    share|cite|improve this answer












                                    Hint: Prove the set $ A= {x in mathbb{R} vert f (x)=g (x)}$ is closed in $mathbb {R}$. Clearly, $D subseteq A $. Taking closure on both sides, we get $mathbb {R} subseteq A implies mathbb {R}=A $.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 21 '18 at 7:28









                                    Thomas Shelby

                                    1,489216




                                    1,489216















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