How can I prove that $d(A,B)=d(A,bar{B})=d(bar{A},bar{B})$












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$begingroup$


How can I prove that $d(A,B)=d(A,bar{B})=d(bar{A},bar{B})$



$bar{A}$ is the set of all accumulation points $A$



If $d(A,B)=inf{ d(a,b)$ with $ain A , bin B}=alpha$



$d(A,bar{B})=inf{ d(a,b)$ with $ain A , bin bar{B}}=beta$



$d(bar{A},bar{B})=inf{ d(a,b)$ with $ain bar{A} , bin bar{B}}=gamma$



I proved that $alpha geq beta geq gamma$






general-topology analysis







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  • 2




    $begingroup$
    I think you mean $geq$, not $>$. Otherwise you would have disproved your claim.
    $endgroup$
    – parsiad
    Jan 9 at 1:56










  • $begingroup$
    $inf$ has the special command inf
    $endgroup$
    – Chase Ryan Taylor
    Jan 9 at 2:30
















1












$begingroup$


How can I prove that $d(A,B)=d(A,bar{B})=d(bar{A},bar{B})$



$bar{A}$ is the set of all accumulation points $A$



If $d(A,B)=inf{ d(a,b)$ with $ain A , bin B}=alpha$



$d(A,bar{B})=inf{ d(a,b)$ with $ain A , bin bar{B}}=beta$



$d(bar{A},bar{B})=inf{ d(a,b)$ with $ain bar{A} , bin bar{B}}=gamma$



I proved that $alpha geq beta geq gamma$






general-topology analysis







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I think you mean $geq$, not $>$. Otherwise you would have disproved your claim.
    $endgroup$
    – parsiad
    Jan 9 at 1:56










  • $begingroup$
    $inf$ has the special command inf
    $endgroup$
    – Chase Ryan Taylor
    Jan 9 at 2:30














1












1








1


1



$begingroup$


How can I prove that $d(A,B)=d(A,bar{B})=d(bar{A},bar{B})$



$bar{A}$ is the set of all accumulation points $A$



If $d(A,B)=inf{ d(a,b)$ with $ain A , bin B}=alpha$



$d(A,bar{B})=inf{ d(a,b)$ with $ain A , bin bar{B}}=beta$



$d(bar{A},bar{B})=inf{ d(a,b)$ with $ain bar{A} , bin bar{B}}=gamma$



I proved that $alpha geq beta geq gamma$






general-topology analysis







share|cite|improve this question











$endgroup$




How can I prove that $d(A,B)=d(A,bar{B})=d(bar{A},bar{B})$



$bar{A}$ is the set of all accumulation points $A$



If $d(A,B)=inf{ d(a,b)$ with $ain A , bin B}=alpha$



$d(A,bar{B})=inf{ d(a,b)$ with $ain A , bin bar{B}}=beta$



$d(bar{A},bar{B})=inf{ d(a,b)$ with $ain bar{A} , bin bar{B}}=gamma$



I proved that $alpha geq beta geq gamma$






general-topology analysis




general-topology analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 5:22









Henno Brandsma

108k347114




108k347114










asked Jan 9 at 1:52









Ivan S. GuerraIvan S. Guerra

349113




349113








  • 2




    $begingroup$
    I think you mean $geq$, not $>$. Otherwise you would have disproved your claim.
    $endgroup$
    – parsiad
    Jan 9 at 1:56










  • $begingroup$
    $inf$ has the special command inf
    $endgroup$
    – Chase Ryan Taylor
    Jan 9 at 2:30














  • 2




    $begingroup$
    I think you mean $geq$, not $>$. Otherwise you would have disproved your claim.
    $endgroup$
    – parsiad
    Jan 9 at 1:56










  • $begingroup$
    $inf$ has the special command inf
    $endgroup$
    – Chase Ryan Taylor
    Jan 9 at 2:30








2




2




$begingroup$
I think you mean $geq$, not $>$. Otherwise you would have disproved your claim.
$endgroup$
– parsiad
Jan 9 at 1:56




$begingroup$
I think you mean $geq$, not $>$. Otherwise you would have disproved your claim.
$endgroup$
– parsiad
Jan 9 at 1:56












$begingroup$
$inf$ has the special command inf
$endgroup$
– Chase Ryan Taylor
Jan 9 at 2:30




$begingroup$
$inf$ has the special command inf
$endgroup$
– Chase Ryan Taylor
Jan 9 at 2:30










3 Answers
3






active

oldest

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1












$begingroup$

For every $c>0$, there exists $x in A, yin bar B$ such that $d(x,y)leq d(A,bar B)+c/2$. There exists $y'in B$ such that $d(y',y)<c/2$, $d(A,B)leq d(x,y')leq d(x,y)+d(y,y')leq d(A,bar B)+c/2+c/2$. This implies that $d(A,B)leq d(A,bar B)$ and $d(A,B)=d(A,bar B)$ from the equality that you have proven.



$d(A,B)=d(A,bar B)$ implies that $d(B,bar A)=d(B,A)$ and $d(bar A,B)=d(bar A,bar B)$, the result follows from the fact that $d(A,B)=d(B,A)$.






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    1












    $begingroup$

    For any $epsilon >0$ there exist $xin bar A, yin bar B$ with $d(x,y)<gamma+epsilon $ and there exist $x'in X, y'in Y$ with $d(x',x)<epsilon$ and $d(y,y')<epsilon.$ Then $$alpha le d(x',y')le d(x',x)+d(x,y)+d(y,y')<gamma+3epsilon.$$ Hence $$forall epsilon >0;(alpha<gamma+3epsilon)$$ which is not possible unless $$alphale gamma.$$ Since you already have $gamma le beta le alpha$, therefore $$alphale gammale beta le alpha;.$$






    share|cite|improve this answer











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      $begingroup$

      Clearly, $betaleqalpha$ since $Bsubsetbar{B}$.



      Suppose, in order to obtain a contradiction, that $beta<alpha$.
      Then, we can find $a$ in $A$ and $b$ in $bar{B}$ such that $d(a,b)leqgamma$ where $gammaequiv(beta+alpha)/2<alpha$.
      We can also find a sequence of points $(b_{n})_{n}$ in $B$ such that $b_{n}rightarrow b$.
      By continuity, $d(a,b_{n})rightarrow d(a, b)$, which in turn implies that we can find some $n$ sufficiently large with $d(a,b_{n})<alpha$, contradicting the minimality of $alpha$.



      We have established $alpha = beta$.



      To get $beta = gamma$, we can apply the same argument, reversing the roles of $A$ and $B$.






      share|cite|improve this answer











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        3 Answers
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        3 Answers
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        1












        $begingroup$

        For every $c>0$, there exists $x in A, yin bar B$ such that $d(x,y)leq d(A,bar B)+c/2$. There exists $y'in B$ such that $d(y',y)<c/2$, $d(A,B)leq d(x,y')leq d(x,y)+d(y,y')leq d(A,bar B)+c/2+c/2$. This implies that $d(A,B)leq d(A,bar B)$ and $d(A,B)=d(A,bar B)$ from the equality that you have proven.



        $d(A,B)=d(A,bar B)$ implies that $d(B,bar A)=d(B,A)$ and $d(bar A,B)=d(bar A,bar B)$, the result follows from the fact that $d(A,B)=d(B,A)$.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          For every $c>0$, there exists $x in A, yin bar B$ such that $d(x,y)leq d(A,bar B)+c/2$. There exists $y'in B$ such that $d(y',y)<c/2$, $d(A,B)leq d(x,y')leq d(x,y)+d(y,y')leq d(A,bar B)+c/2+c/2$. This implies that $d(A,B)leq d(A,bar B)$ and $d(A,B)=d(A,bar B)$ from the equality that you have proven.



          $d(A,B)=d(A,bar B)$ implies that $d(B,bar A)=d(B,A)$ and $d(bar A,B)=d(bar A,bar B)$, the result follows from the fact that $d(A,B)=d(B,A)$.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            For every $c>0$, there exists $x in A, yin bar B$ such that $d(x,y)leq d(A,bar B)+c/2$. There exists $y'in B$ such that $d(y',y)<c/2$, $d(A,B)leq d(x,y')leq d(x,y)+d(y,y')leq d(A,bar B)+c/2+c/2$. This implies that $d(A,B)leq d(A,bar B)$ and $d(A,B)=d(A,bar B)$ from the equality that you have proven.



            $d(A,B)=d(A,bar B)$ implies that $d(B,bar A)=d(B,A)$ and $d(bar A,B)=d(bar A,bar B)$, the result follows from the fact that $d(A,B)=d(B,A)$.






            share|cite|improve this answer









            $endgroup$



            For every $c>0$, there exists $x in A, yin bar B$ such that $d(x,y)leq d(A,bar B)+c/2$. There exists $y'in B$ such that $d(y',y)<c/2$, $d(A,B)leq d(x,y')leq d(x,y)+d(y,y')leq d(A,bar B)+c/2+c/2$. This implies that $d(A,B)leq d(A,bar B)$ and $d(A,B)=d(A,bar B)$ from the equality that you have proven.



            $d(A,B)=d(A,bar B)$ implies that $d(B,bar A)=d(B,A)$ and $d(bar A,B)=d(bar A,bar B)$, the result follows from the fact that $d(A,B)=d(B,A)$.







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            answered Jan 9 at 2:03









            Tsemo AristideTsemo Aristide

            57.6k11444




            57.6k11444























                1












                $begingroup$

                For any $epsilon >0$ there exist $xin bar A, yin bar B$ with $d(x,y)<gamma+epsilon $ and there exist $x'in X, y'in Y$ with $d(x',x)<epsilon$ and $d(y,y')<epsilon.$ Then $$alpha le d(x',y')le d(x',x)+d(x,y)+d(y,y')<gamma+3epsilon.$$ Hence $$forall epsilon >0;(alpha<gamma+3epsilon)$$ which is not possible unless $$alphale gamma.$$ Since you already have $gamma le beta le alpha$, therefore $$alphale gammale beta le alpha;.$$






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  For any $epsilon >0$ there exist $xin bar A, yin bar B$ with $d(x,y)<gamma+epsilon $ and there exist $x'in X, y'in Y$ with $d(x',x)<epsilon$ and $d(y,y')<epsilon.$ Then $$alpha le d(x',y')le d(x',x)+d(x,y)+d(y,y')<gamma+3epsilon.$$ Hence $$forall epsilon >0;(alpha<gamma+3epsilon)$$ which is not possible unless $$alphale gamma.$$ Since you already have $gamma le beta le alpha$, therefore $$alphale gammale beta le alpha;.$$






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    For any $epsilon >0$ there exist $xin bar A, yin bar B$ with $d(x,y)<gamma+epsilon $ and there exist $x'in X, y'in Y$ with $d(x',x)<epsilon$ and $d(y,y')<epsilon.$ Then $$alpha le d(x',y')le d(x',x)+d(x,y)+d(y,y')<gamma+3epsilon.$$ Hence $$forall epsilon >0;(alpha<gamma+3epsilon)$$ which is not possible unless $$alphale gamma.$$ Since you already have $gamma le beta le alpha$, therefore $$alphale gammale beta le alpha;.$$






                    share|cite|improve this answer











                    $endgroup$



                    For any $epsilon >0$ there exist $xin bar A, yin bar B$ with $d(x,y)<gamma+epsilon $ and there exist $x'in X, y'in Y$ with $d(x',x)<epsilon$ and $d(y,y')<epsilon.$ Then $$alpha le d(x',y')le d(x',x)+d(x,y)+d(y,y')<gamma+3epsilon.$$ Hence $$forall epsilon >0;(alpha<gamma+3epsilon)$$ which is not possible unless $$alphale gamma.$$ Since you already have $gamma le beta le alpha$, therefore $$alphale gammale beta le alpha;.$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 9 at 8:19

























                    answered Jan 9 at 8:14









                    DanielWainfleetDanielWainfleet

                    34.9k31648




                    34.9k31648























                        0












                        $begingroup$

                        Clearly, $betaleqalpha$ since $Bsubsetbar{B}$.



                        Suppose, in order to obtain a contradiction, that $beta<alpha$.
                        Then, we can find $a$ in $A$ and $b$ in $bar{B}$ such that $d(a,b)leqgamma$ where $gammaequiv(beta+alpha)/2<alpha$.
                        We can also find a sequence of points $(b_{n})_{n}$ in $B$ such that $b_{n}rightarrow b$.
                        By continuity, $d(a,b_{n})rightarrow d(a, b)$, which in turn implies that we can find some $n$ sufficiently large with $d(a,b_{n})<alpha$, contradicting the minimality of $alpha$.



                        We have established $alpha = beta$.



                        To get $beta = gamma$, we can apply the same argument, reversing the roles of $A$ and $B$.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          Clearly, $betaleqalpha$ since $Bsubsetbar{B}$.



                          Suppose, in order to obtain a contradiction, that $beta<alpha$.
                          Then, we can find $a$ in $A$ and $b$ in $bar{B}$ such that $d(a,b)leqgamma$ where $gammaequiv(beta+alpha)/2<alpha$.
                          We can also find a sequence of points $(b_{n})_{n}$ in $B$ such that $b_{n}rightarrow b$.
                          By continuity, $d(a,b_{n})rightarrow d(a, b)$, which in turn implies that we can find some $n$ sufficiently large with $d(a,b_{n})<alpha$, contradicting the minimality of $alpha$.



                          We have established $alpha = beta$.



                          To get $beta = gamma$, we can apply the same argument, reversing the roles of $A$ and $B$.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Clearly, $betaleqalpha$ since $Bsubsetbar{B}$.



                            Suppose, in order to obtain a contradiction, that $beta<alpha$.
                            Then, we can find $a$ in $A$ and $b$ in $bar{B}$ such that $d(a,b)leqgamma$ where $gammaequiv(beta+alpha)/2<alpha$.
                            We can also find a sequence of points $(b_{n})_{n}$ in $B$ such that $b_{n}rightarrow b$.
                            By continuity, $d(a,b_{n})rightarrow d(a, b)$, which in turn implies that we can find some $n$ sufficiently large with $d(a,b_{n})<alpha$, contradicting the minimality of $alpha$.



                            We have established $alpha = beta$.



                            To get $beta = gamma$, we can apply the same argument, reversing the roles of $A$ and $B$.






                            share|cite|improve this answer











                            $endgroup$



                            Clearly, $betaleqalpha$ since $Bsubsetbar{B}$.



                            Suppose, in order to obtain a contradiction, that $beta<alpha$.
                            Then, we can find $a$ in $A$ and $b$ in $bar{B}$ such that $d(a,b)leqgamma$ where $gammaequiv(beta+alpha)/2<alpha$.
                            We can also find a sequence of points $(b_{n})_{n}$ in $B$ such that $b_{n}rightarrow b$.
                            By continuity, $d(a,b_{n})rightarrow d(a, b)$, which in turn implies that we can find some $n$ sufficiently large with $d(a,b_{n})<alpha$, contradicting the minimality of $alpha$.



                            We have established $alpha = beta$.



                            To get $beta = gamma$, we can apply the same argument, reversing the roles of $A$ and $B$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 9 at 2:18

























                            answered Jan 9 at 2:11









                            parsiadparsiad

                            17.1k32353




                            17.1k32353






























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