The definition of closure.












1












$begingroup$


Recently, i have seen the definition of closure as follows,
begin{equation}
text{cl}~C = bigcap_{varepsilon>0}(C + varepsilon B),
end{equation}
where $B$ is Euclidean unit ball: $B={x mid |x| leq 1}$.
I think it is right but i feel a little confused, because
begin{equation}
bigcap_{varepsilon>0} varepsilon B = lim_{varepsilon downarrow 0}varepsilon B = {boldsymbol{0}}.
end{equation}
Therefore, could i say
begin{equation}
begin{aligned}
text{cl}~C &= bigcap_{varepsilon > 0}(C + varepsilon B) \
&= lim_{varepsilon downarrow 0}(C + varepsilon B) \
&= C + lim_{varepsilon downarrow 0} varepsilon B \
&= C + {boldsymbol{0}}.
end{aligned}
end{equation}
I think it is incorrect but i don't know why, thanks in advance.






real-analysis linear-algebra convex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the definition of $B$?
    $endgroup$
    – José Carlos Santos
    Dec 18 '18 at 15:15










  • $begingroup$
    $A+B$ means ${a+bmid ain A, bin B}.
    $endgroup$
    – Jiu
    Dec 18 '18 at 15:16










  • $begingroup$
    That doesn't answer my question.
    $endgroup$
    – José Carlos Santos
    Dec 18 '18 at 15:17










  • $begingroup$
    I am sorry for my carelessness and i have added the definition of $B$.
    $endgroup$
    – Ze-Nan Li
    Dec 18 '18 at 15:54










  • $begingroup$
    @JoséCarlosSantos Acctually we better ask what the hell is the definition $lim_{varepsilon downarrow 0}(varepsilon B) $ ?
    $endgroup$
    – Red shoes
    Dec 22 '18 at 0:29


















1












$begingroup$


Recently, i have seen the definition of closure as follows,
begin{equation}
text{cl}~C = bigcap_{varepsilon>0}(C + varepsilon B),
end{equation}
where $B$ is Euclidean unit ball: $B={x mid |x| leq 1}$.
I think it is right but i feel a little confused, because
begin{equation}
bigcap_{varepsilon>0} varepsilon B = lim_{varepsilon downarrow 0}varepsilon B = {boldsymbol{0}}.
end{equation}
Therefore, could i say
begin{equation}
begin{aligned}
text{cl}~C &= bigcap_{varepsilon > 0}(C + varepsilon B) \
&= lim_{varepsilon downarrow 0}(C + varepsilon B) \
&= C + lim_{varepsilon downarrow 0} varepsilon B \
&= C + {boldsymbol{0}}.
end{aligned}
end{equation}
I think it is incorrect but i don't know why, thanks in advance.






real-analysis linear-algebra convex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the definition of $B$?
    $endgroup$
    – José Carlos Santos
    Dec 18 '18 at 15:15










  • $begingroup$
    $A+B$ means ${a+bmid ain A, bin B}.
    $endgroup$
    – Jiu
    Dec 18 '18 at 15:16










  • $begingroup$
    That doesn't answer my question.
    $endgroup$
    – José Carlos Santos
    Dec 18 '18 at 15:17










  • $begingroup$
    I am sorry for my carelessness and i have added the definition of $B$.
    $endgroup$
    – Ze-Nan Li
    Dec 18 '18 at 15:54










  • $begingroup$
    @JoséCarlosSantos Acctually we better ask what the hell is the definition $lim_{varepsilon downarrow 0}(varepsilon B) $ ?
    $endgroup$
    – Red shoes
    Dec 22 '18 at 0:29
















1












1








1





$begingroup$


Recently, i have seen the definition of closure as follows,
begin{equation}
text{cl}~C = bigcap_{varepsilon>0}(C + varepsilon B),
end{equation}
where $B$ is Euclidean unit ball: $B={x mid |x| leq 1}$.
I think it is right but i feel a little confused, because
begin{equation}
bigcap_{varepsilon>0} varepsilon B = lim_{varepsilon downarrow 0}varepsilon B = {boldsymbol{0}}.
end{equation}
Therefore, could i say
begin{equation}
begin{aligned}
text{cl}~C &= bigcap_{varepsilon > 0}(C + varepsilon B) \
&= lim_{varepsilon downarrow 0}(C + varepsilon B) \
&= C + lim_{varepsilon downarrow 0} varepsilon B \
&= C + {boldsymbol{0}}.
end{aligned}
end{equation}
I think it is incorrect but i don't know why, thanks in advance.






real-analysis linear-algebra convex-analysis







share|cite|improve this question











$endgroup$




Recently, i have seen the definition of closure as follows,
begin{equation}
text{cl}~C = bigcap_{varepsilon>0}(C + varepsilon B),
end{equation}
where $B$ is Euclidean unit ball: $B={x mid |x| leq 1}$.
I think it is right but i feel a little confused, because
begin{equation}
bigcap_{varepsilon>0} varepsilon B = lim_{varepsilon downarrow 0}varepsilon B = {boldsymbol{0}}.
end{equation}
Therefore, could i say
begin{equation}
begin{aligned}
text{cl}~C &= bigcap_{varepsilon > 0}(C + varepsilon B) \
&= lim_{varepsilon downarrow 0}(C + varepsilon B) \
&= C + lim_{varepsilon downarrow 0} varepsilon B \
&= C + {boldsymbol{0}}.
end{aligned}
end{equation}
I think it is incorrect but i don't know why, thanks in advance.






real-analysis linear-algebra convex-analysis




real-analysis linear-algebra convex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 15:51







Ze-Nan Li

















asked Dec 18 '18 at 15:09









Ze-Nan LiZe-Nan Li

286




286












  • $begingroup$
    What is the definition of $B$?
    $endgroup$
    – José Carlos Santos
    Dec 18 '18 at 15:15










  • $begingroup$
    $A+B$ means ${a+bmid ain A, bin B}.
    $endgroup$
    – Jiu
    Dec 18 '18 at 15:16










  • $begingroup$
    That doesn't answer my question.
    $endgroup$
    – José Carlos Santos
    Dec 18 '18 at 15:17










  • $begingroup$
    I am sorry for my carelessness and i have added the definition of $B$.
    $endgroup$
    – Ze-Nan Li
    Dec 18 '18 at 15:54










  • $begingroup$
    @JoséCarlosSantos Acctually we better ask what the hell is the definition $lim_{varepsilon downarrow 0}(varepsilon B) $ ?
    $endgroup$
    – Red shoes
    Dec 22 '18 at 0:29




















  • $begingroup$
    What is the definition of $B$?
    $endgroup$
    – José Carlos Santos
    Dec 18 '18 at 15:15










  • $begingroup$
    $A+B$ means ${a+bmid ain A, bin B}.
    $endgroup$
    – Jiu
    Dec 18 '18 at 15:16










  • $begingroup$
    That doesn't answer my question.
    $endgroup$
    – José Carlos Santos
    Dec 18 '18 at 15:17










  • $begingroup$
    I am sorry for my carelessness and i have added the definition of $B$.
    $endgroup$
    – Ze-Nan Li
    Dec 18 '18 at 15:54










  • $begingroup$
    @JoséCarlosSantos Acctually we better ask what the hell is the definition $lim_{varepsilon downarrow 0}(varepsilon B) $ ?
    $endgroup$
    – Red shoes
    Dec 22 '18 at 0:29


















$begingroup$
What is the definition of $B$?
$endgroup$
– José Carlos Santos
Dec 18 '18 at 15:15




$begingroup$
What is the definition of $B$?
$endgroup$
– José Carlos Santos
Dec 18 '18 at 15:15












$begingroup$
$A+B$ means ${a+bmid ain A, bin B}.
$endgroup$
– Jiu
Dec 18 '18 at 15:16




$begingroup$
$A+B$ means ${a+bmid ain A, bin B}.
$endgroup$
– Jiu
Dec 18 '18 at 15:16












$begingroup$
That doesn't answer my question.
$endgroup$
– José Carlos Santos
Dec 18 '18 at 15:17




$begingroup$
That doesn't answer my question.
$endgroup$
– José Carlos Santos
Dec 18 '18 at 15:17












$begingroup$
I am sorry for my carelessness and i have added the definition of $B$.
$endgroup$
– Ze-Nan Li
Dec 18 '18 at 15:54




$begingroup$
I am sorry for my carelessness and i have added the definition of $B$.
$endgroup$
– Ze-Nan Li
Dec 18 '18 at 15:54












$begingroup$
@JoséCarlosSantos Acctually we better ask what the hell is the definition $lim_{varepsilon downarrow 0}(varepsilon B) $ ?
$endgroup$
– Red shoes
Dec 22 '18 at 0:29






$begingroup$
@JoséCarlosSantos Acctually we better ask what the hell is the definition $lim_{varepsilon downarrow 0}(varepsilon B) $ ?
$endgroup$
– Red shoes
Dec 22 '18 at 0:29












3 Answers
3






active

oldest

votes


















1












$begingroup$

The easiest way to analyze such a proof is by using a small example. Take $C=(-infty,0)$, so the closure is $(-infty,0]$. You say $lim_{varepsilon downarrow 0}((-infty,0) + varepsilon B) = (-infty,0) + lim_{varepsilon downarrow 0}(varepsilon B)$, which does not seem right.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    This is the part where you made a mistake:
    $$
    lim_{varepsilon downarrow 0}(C + varepsilon B) = C + lim_{varepsilon downarrow 0} varepsilon B
    $$
    The operation of taking limit here is not distributive with respect to Minkowski sum.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I got it, thanks. By the way, i have never seen a closure could be written in such a limit ($lim(C+varepsilon B)$). Can I write like this?
      $endgroup$
      – Ze-Nan Li
      Dec 18 '18 at 15:59






    • 1




      $begingroup$
      Yes, you can do that for a metric space. It essentially captures the fact that a closed set contains all of its limit points.
      $endgroup$
      – BigbearZzz
      Dec 18 '18 at 16:06



















    1












    $begingroup$

    Writing an intersection $bigcap_{varepsilon>0}^infty A_varepsilon$ of sets satisfying $A_x subset A_y$ for $x<y$ as a limit is misleading to say the least. In this case it lead you to the false believe that this intersection is compatible with (Minkowski) sums.



    That is, in general:
    $$
    bigcap_{varepsilon>0} (A_varepsilon+B_varepsilon) color{red}neq bigcap_{varepsilon>0} A_varepsilon + bigcap_{varepsilon>0} B_varepsilon.
    $$




    Consider the example where $C = (0,1)subset Bbb R$ and $B=[-1,1]$ is the closed unit ball. Note that
    $$
    bigcap_{varepsilon > 0} (C+varepsilon B) = bigcap_{varepsilon > 0} (-varepsilon, 1+varepsilon) = [0,1] = operatorname{cl},(0,1)
    $$
    while
    $$
    C + bigcap_{varepsilon > 0} varepsilon B = C + {0} = C = (0,1) color{red}neq operatorname{cl},(0,1).
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks, i find that the $limit$ brings some misleading.
      $endgroup$
      – Ze-Nan Li
      Dec 18 '18 at 16:07






    • 1




      $begingroup$
      Well, you can describe it as a kind of limit and this actually can be made more precise, but it behaves very different from the kind of point-wise limits you know from calculus. – Be careful when importing intuitions from other subjects that have related notations!
      $endgroup$
      – Christoph
      Dec 18 '18 at 16:09











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The easiest way to analyze such a proof is by using a small example. Take $C=(-infty,0)$, so the closure is $(-infty,0]$. You say $lim_{varepsilon downarrow 0}((-infty,0) + varepsilon B) = (-infty,0) + lim_{varepsilon downarrow 0}(varepsilon B)$, which does not seem right.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The easiest way to analyze such a proof is by using a small example. Take $C=(-infty,0)$, so the closure is $(-infty,0]$. You say $lim_{varepsilon downarrow 0}((-infty,0) + varepsilon B) = (-infty,0) + lim_{varepsilon downarrow 0}(varepsilon B)$, which does not seem right.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The easiest way to analyze such a proof is by using a small example. Take $C=(-infty,0)$, so the closure is $(-infty,0]$. You say $lim_{varepsilon downarrow 0}((-infty,0) + varepsilon B) = (-infty,0) + lim_{varepsilon downarrow 0}(varepsilon B)$, which does not seem right.






        share|cite|improve this answer









        $endgroup$



        The easiest way to analyze such a proof is by using a small example. Take $C=(-infty,0)$, so the closure is $(-infty,0]$. You say $lim_{varepsilon downarrow 0}((-infty,0) + varepsilon B) = (-infty,0) + lim_{varepsilon downarrow 0}(varepsilon B)$, which does not seem right.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 15:16









        LinAlgLinAlg

        9,8791521




        9,8791521























            1












            $begingroup$

            This is the part where you made a mistake:
            $$
            lim_{varepsilon downarrow 0}(C + varepsilon B) = C + lim_{varepsilon downarrow 0} varepsilon B
            $$
            The operation of taking limit here is not distributive with respect to Minkowski sum.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I got it, thanks. By the way, i have never seen a closure could be written in such a limit ($lim(C+varepsilon B)$). Can I write like this?
              $endgroup$
              – Ze-Nan Li
              Dec 18 '18 at 15:59






            • 1




              $begingroup$
              Yes, you can do that for a metric space. It essentially captures the fact that a closed set contains all of its limit points.
              $endgroup$
              – BigbearZzz
              Dec 18 '18 at 16:06
















            1












            $begingroup$

            This is the part where you made a mistake:
            $$
            lim_{varepsilon downarrow 0}(C + varepsilon B) = C + lim_{varepsilon downarrow 0} varepsilon B
            $$
            The operation of taking limit here is not distributive with respect to Minkowski sum.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I got it, thanks. By the way, i have never seen a closure could be written in such a limit ($lim(C+varepsilon B)$). Can I write like this?
              $endgroup$
              – Ze-Nan Li
              Dec 18 '18 at 15:59






            • 1




              $begingroup$
              Yes, you can do that for a metric space. It essentially captures the fact that a closed set contains all of its limit points.
              $endgroup$
              – BigbearZzz
              Dec 18 '18 at 16:06














            1












            1








            1





            $begingroup$

            This is the part where you made a mistake:
            $$
            lim_{varepsilon downarrow 0}(C + varepsilon B) = C + lim_{varepsilon downarrow 0} varepsilon B
            $$
            The operation of taking limit here is not distributive with respect to Minkowski sum.






            share|cite|improve this answer









            $endgroup$



            This is the part where you made a mistake:
            $$
            lim_{varepsilon downarrow 0}(C + varepsilon B) = C + lim_{varepsilon downarrow 0} varepsilon B
            $$
            The operation of taking limit here is not distributive with respect to Minkowski sum.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 18 '18 at 15:28









            BigbearZzzBigbearZzz

            8,80921652




            8,80921652












            • $begingroup$
              I got it, thanks. By the way, i have never seen a closure could be written in such a limit ($lim(C+varepsilon B)$). Can I write like this?
              $endgroup$
              – Ze-Nan Li
              Dec 18 '18 at 15:59






            • 1




              $begingroup$
              Yes, you can do that for a metric space. It essentially captures the fact that a closed set contains all of its limit points.
              $endgroup$
              – BigbearZzz
              Dec 18 '18 at 16:06


















            • $begingroup$
              I got it, thanks. By the way, i have never seen a closure could be written in such a limit ($lim(C+varepsilon B)$). Can I write like this?
              $endgroup$
              – Ze-Nan Li
              Dec 18 '18 at 15:59






            • 1




              $begingroup$
              Yes, you can do that for a metric space. It essentially captures the fact that a closed set contains all of its limit points.
              $endgroup$
              – BigbearZzz
              Dec 18 '18 at 16:06
















            $begingroup$
            I got it, thanks. By the way, i have never seen a closure could be written in such a limit ($lim(C+varepsilon B)$). Can I write like this?
            $endgroup$
            – Ze-Nan Li
            Dec 18 '18 at 15:59




            $begingroup$
            I got it, thanks. By the way, i have never seen a closure could be written in such a limit ($lim(C+varepsilon B)$). Can I write like this?
            $endgroup$
            – Ze-Nan Li
            Dec 18 '18 at 15:59




            1




            1




            $begingroup$
            Yes, you can do that for a metric space. It essentially captures the fact that a closed set contains all of its limit points.
            $endgroup$
            – BigbearZzz
            Dec 18 '18 at 16:06




            $begingroup$
            Yes, you can do that for a metric space. It essentially captures the fact that a closed set contains all of its limit points.
            $endgroup$
            – BigbearZzz
            Dec 18 '18 at 16:06











            1












            $begingroup$

            Writing an intersection $bigcap_{varepsilon>0}^infty A_varepsilon$ of sets satisfying $A_x subset A_y$ for $x<y$ as a limit is misleading to say the least. In this case it lead you to the false believe that this intersection is compatible with (Minkowski) sums.



            That is, in general:
            $$
            bigcap_{varepsilon>0} (A_varepsilon+B_varepsilon) color{red}neq bigcap_{varepsilon>0} A_varepsilon + bigcap_{varepsilon>0} B_varepsilon.
            $$




            Consider the example where $C = (0,1)subset Bbb R$ and $B=[-1,1]$ is the closed unit ball. Note that
            $$
            bigcap_{varepsilon > 0} (C+varepsilon B) = bigcap_{varepsilon > 0} (-varepsilon, 1+varepsilon) = [0,1] = operatorname{cl},(0,1)
            $$
            while
            $$
            C + bigcap_{varepsilon > 0} varepsilon B = C + {0} = C = (0,1) color{red}neq operatorname{cl},(0,1).
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks, i find that the $limit$ brings some misleading.
              $endgroup$
              – Ze-Nan Li
              Dec 18 '18 at 16:07






            • 1




              $begingroup$
              Well, you can describe it as a kind of limit and this actually can be made more precise, but it behaves very different from the kind of point-wise limits you know from calculus. – Be careful when importing intuitions from other subjects that have related notations!
              $endgroup$
              – Christoph
              Dec 18 '18 at 16:09
















            1












            $begingroup$

            Writing an intersection $bigcap_{varepsilon>0}^infty A_varepsilon$ of sets satisfying $A_x subset A_y$ for $x<y$ as a limit is misleading to say the least. In this case it lead you to the false believe that this intersection is compatible with (Minkowski) sums.



            That is, in general:
            $$
            bigcap_{varepsilon>0} (A_varepsilon+B_varepsilon) color{red}neq bigcap_{varepsilon>0} A_varepsilon + bigcap_{varepsilon>0} B_varepsilon.
            $$




            Consider the example where $C = (0,1)subset Bbb R$ and $B=[-1,1]$ is the closed unit ball. Note that
            $$
            bigcap_{varepsilon > 0} (C+varepsilon B) = bigcap_{varepsilon > 0} (-varepsilon, 1+varepsilon) = [0,1] = operatorname{cl},(0,1)
            $$
            while
            $$
            C + bigcap_{varepsilon > 0} varepsilon B = C + {0} = C = (0,1) color{red}neq operatorname{cl},(0,1).
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks, i find that the $limit$ brings some misleading.
              $endgroup$
              – Ze-Nan Li
              Dec 18 '18 at 16:07






            • 1




              $begingroup$
              Well, you can describe it as a kind of limit and this actually can be made more precise, but it behaves very different from the kind of point-wise limits you know from calculus. – Be careful when importing intuitions from other subjects that have related notations!
              $endgroup$
              – Christoph
              Dec 18 '18 at 16:09














            1












            1








            1





            $begingroup$

            Writing an intersection $bigcap_{varepsilon>0}^infty A_varepsilon$ of sets satisfying $A_x subset A_y$ for $x<y$ as a limit is misleading to say the least. In this case it lead you to the false believe that this intersection is compatible with (Minkowski) sums.



            That is, in general:
            $$
            bigcap_{varepsilon>0} (A_varepsilon+B_varepsilon) color{red}neq bigcap_{varepsilon>0} A_varepsilon + bigcap_{varepsilon>0} B_varepsilon.
            $$




            Consider the example where $C = (0,1)subset Bbb R$ and $B=[-1,1]$ is the closed unit ball. Note that
            $$
            bigcap_{varepsilon > 0} (C+varepsilon B) = bigcap_{varepsilon > 0} (-varepsilon, 1+varepsilon) = [0,1] = operatorname{cl},(0,1)
            $$
            while
            $$
            C + bigcap_{varepsilon > 0} varepsilon B = C + {0} = C = (0,1) color{red}neq operatorname{cl},(0,1).
            $$






            share|cite|improve this answer











            $endgroup$



            Writing an intersection $bigcap_{varepsilon>0}^infty A_varepsilon$ of sets satisfying $A_x subset A_y$ for $x<y$ as a limit is misleading to say the least. In this case it lead you to the false believe that this intersection is compatible with (Minkowski) sums.



            That is, in general:
            $$
            bigcap_{varepsilon>0} (A_varepsilon+B_varepsilon) color{red}neq bigcap_{varepsilon>0} A_varepsilon + bigcap_{varepsilon>0} B_varepsilon.
            $$




            Consider the example where $C = (0,1)subset Bbb R$ and $B=[-1,1]$ is the closed unit ball. Note that
            $$
            bigcap_{varepsilon > 0} (C+varepsilon B) = bigcap_{varepsilon > 0} (-varepsilon, 1+varepsilon) = [0,1] = operatorname{cl},(0,1)
            $$
            while
            $$
            C + bigcap_{varepsilon > 0} varepsilon B = C + {0} = C = (0,1) color{red}neq operatorname{cl},(0,1).
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 18 '18 at 16:05

























            answered Dec 18 '18 at 15:59









            ChristophChristoph

            12.4k1642




            12.4k1642












            • $begingroup$
              Thanks, i find that the $limit$ brings some misleading.
              $endgroup$
              – Ze-Nan Li
              Dec 18 '18 at 16:07






            • 1




              $begingroup$
              Well, you can describe it as a kind of limit and this actually can be made more precise, but it behaves very different from the kind of point-wise limits you know from calculus. – Be careful when importing intuitions from other subjects that have related notations!
              $endgroup$
              – Christoph
              Dec 18 '18 at 16:09


















            • $begingroup$
              Thanks, i find that the $limit$ brings some misleading.
              $endgroup$
              – Ze-Nan Li
              Dec 18 '18 at 16:07






            • 1




              $begingroup$
              Well, you can describe it as a kind of limit and this actually can be made more precise, but it behaves very different from the kind of point-wise limits you know from calculus. – Be careful when importing intuitions from other subjects that have related notations!
              $endgroup$
              – Christoph
              Dec 18 '18 at 16:09
















            $begingroup$
            Thanks, i find that the $limit$ brings some misleading.
            $endgroup$
            – Ze-Nan Li
            Dec 18 '18 at 16:07




            $begingroup$
            Thanks, i find that the $limit$ brings some misleading.
            $endgroup$
            – Ze-Nan Li
            Dec 18 '18 at 16:07




            1




            1




            $begingroup$
            Well, you can describe it as a kind of limit and this actually can be made more precise, but it behaves very different from the kind of point-wise limits you know from calculus. – Be careful when importing intuitions from other subjects that have related notations!
            $endgroup$
            – Christoph
            Dec 18 '18 at 16:09




            $begingroup$
            Well, you can describe it as a kind of limit and this actually can be made more precise, but it behaves very different from the kind of point-wise limits you know from calculus. – Be careful when importing intuitions from other subjects that have related notations!
            $endgroup$
            – Christoph
            Dec 18 '18 at 16:09


















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