Inductively defined sequence of graph neighborhoods












5












$begingroup$


I'm trying to solve the following problem:




Let $v_0$ be a vertex in a graph $G$, and $D_0 := {v_0}$.




  1. For $n = 1, 2 dots$ inductively define
    $D_n := N(D_0 cup D_1 cup dots cup D_{n-1})$.


  2. Show that $D_n = {v | d(v_0, v) = n}$ and
    $D_{n+1} subseteq N(D_n) subseteq D_{n-1} cup D_{n+1}$ for all
    $n in mathbb{N}$.





Here, $N_G(V') = N(V')$ is the neighborhood of $V' subseteq V(G)$, and
$d_G(u, v) = d(u, v)$ is the distance in $G$ of two vertices $u$, $v$.



I'm unable to solve the task because I am not sure if it's correctly defined. Every type of graph seems to be a contradiction to the statements above. Consider for simplicity the following example:



Let $P$ be a path, and $v_0$ be the central vertex in $P$. Then



$D_0 = {v_0}$,



$D_1 = N(D_0) = {v | d(v_0, v) = 1}$,



$D_2 = N(D_0 cup D_1) = N({v_0} cup N({v_0})) =
{v | d(v_0, v) leq 2}$,



and for all $n geq 2$ we obtain $D_n = {v | d(v_0, v) leq n}$. In particular, it follows that $V(P) = D_r$, where $r$ is the radius of $P$.



This example contradicts both statements in the second part of the task. Can you, please, help me find out, is there anything I am missing or interpreting incorrectly?






graph-theory







share|cite|improve this question









$endgroup$

















    5












    $begingroup$


    I'm trying to solve the following problem:




    Let $v_0$ be a vertex in a graph $G$, and $D_0 := {v_0}$.




    1. For $n = 1, 2 dots$ inductively define
      $D_n := N(D_0 cup D_1 cup dots cup D_{n-1})$.


    2. Show that $D_n = {v | d(v_0, v) = n}$ and
      $D_{n+1} subseteq N(D_n) subseteq D_{n-1} cup D_{n+1}$ for all
      $n in mathbb{N}$.





    Here, $N_G(V') = N(V')$ is the neighborhood of $V' subseteq V(G)$, and
    $d_G(u, v) = d(u, v)$ is the distance in $G$ of two vertices $u$, $v$.



    I'm unable to solve the task because I am not sure if it's correctly defined. Every type of graph seems to be a contradiction to the statements above. Consider for simplicity the following example:



    Let $P$ be a path, and $v_0$ be the central vertex in $P$. Then



    $D_0 = {v_0}$,



    $D_1 = N(D_0) = {v | d(v_0, v) = 1}$,



    $D_2 = N(D_0 cup D_1) = N({v_0} cup N({v_0})) =
    {v | d(v_0, v) leq 2}$    ,



    and for all $n geq 2$ we obtain $D_n = {v | d(v_0, v) leq n}$. In particular, it follows that $V(P) = D_r$, where $r$ is the radius of $P$.



    This example contradicts both statements in the second part of the task. Can you, please, help me find out, is there anything I am missing or interpreting incorrectly?






    graph-theory







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      I'm trying to solve the following problem:




      Let $v_0$ be a vertex in a graph $G$, and $D_0 := {v_0}$.




      1. For $n = 1, 2 dots$ inductively define
        $D_n := N(D_0 cup D_1 cup dots cup D_{n-1})$.


      2. Show that $D_n = {v | d(v_0, v) = n}$ and
        $D_{n+1} subseteq N(D_n) subseteq D_{n-1} cup D_{n+1}$ for all
        $n in mathbb{N}$.





      Here, $N_G(V') = N(V')$ is the neighborhood of $V' subseteq V(G)$, and
      $d_G(u, v) = d(u, v)$ is the distance in $G$ of two vertices $u$, $v$.



      I'm unable to solve the task because I am not sure if it's correctly defined. Every type of graph seems to be a contradiction to the statements above. Consider for simplicity the following example:



      Let $P$ be a path, and $v_0$ be the central vertex in $P$. Then



      $D_0 = {v_0}$,



      $D_1 = N(D_0) = {v | d(v_0, v) = 1}$,



      $D_2 = N(D_0 cup D_1) = N({v_0} cup N({v_0})) =
      {v | d(v_0, v) leq 2}$      ,



      and for all $n geq 2$ we obtain $D_n = {v | d(v_0, v) leq n}$. In particular, it follows that $V(P) = D_r$, where $r$ is the radius of $P$.



      This example contradicts both statements in the second part of the task. Can you, please, help me find out, is there anything I am missing or interpreting incorrectly?






      graph-theory







      share|cite|improve this question









      $endgroup$




      I'm trying to solve the following problem:




      Let $v_0$ be a vertex in a graph $G$, and $D_0 := {v_0}$.




      1. For $n = 1, 2 dots$ inductively define
        $D_n := N(D_0 cup D_1 cup dots cup D_{n-1})$.


      2. Show that $D_n = {v | d(v_0, v) = n}$ and
        $D_{n+1} subseteq N(D_n) subseteq D_{n-1} cup D_{n+1}$ for all
        $n in mathbb{N}$.





      Here, $N_G(V') = N(V')$ is the neighborhood of $V' subseteq V(G)$, and
      $d_G(u, v) = d(u, v)$ is the distance in $G$ of two vertices $u$, $v$.



      I'm unable to solve the task because I am not sure if it's correctly defined. Every type of graph seems to be a contradiction to the statements above. Consider for simplicity the following example:



      Let $P$ be a path, and $v_0$ be the central vertex in $P$. Then



      $D_0 = {v_0}$,



      $D_1 = N(D_0) = {v | d(v_0, v) = 1}$,



      $D_2 = N(D_0 cup D_1) = N({v_0} cup N({v_0})) =
      {v | d(v_0, v) leq 2}$      ,



      and for all $n geq 2$ we obtain $D_n = {v | d(v_0, v) leq n}$. In particular, it follows that $V(P) = D_r$, where $r$ is the radius of $P$.



      This example contradicts both statements in the second part of the task. Can you, please, help me find out, is there anything I am missing or interpreting incorrectly?






      graph-theory




      graph-theory






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Feb 6 at 8:16









      Sanjar AdylovSanjar Adylov

      284




      284






















          2 Answers
          2






          active

          oldest

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          4












          $begingroup$

          You are mixing closed and open neighbourhood. I think when they are stating $D_n=N(D_0cupldotscup D_{n-1})$, they are implying the open neighbourhood



          The open neighbourhood $N(S)$ of some set of vertices $S$ does not include $S$. Therefore, using your exampla of the path $P$:



          $D_0={v_0}$



          $D_1=N(D_0)={v mid d(v_0,v)=1}$



          $D_2=N(D_0cup D_1)={v mid d(v_0,v)=2}$
          and so on.



          More precisely if you define $P$ as:
          $$ ldots v_{-n} ldots v_{-2} v_1 v_0 v_1 v_2 ldots v_n ldots$$
          Then $D_0={v_0}$, $D_1=N(D_0)={v_1,v_{-1}}$, and
          $$D_2=N(D_0cup D_1)= N({v_1,v_0,v_{-1}})={v_2,v_{-2}}$$



          Therefore the statements holds as
          begin{align*}
          D_{n+1}={v_{n+1},v_{-(n+1)}} &subset N(D_n)=N({v_{n},v_{-n}})={v_{n+1},v_{n-1},v_{-(n-1)},v_{-(n+1)}}\
          &subset D_{n-1} cup D_{n+1}
          end{align*}
          with an equality in your specific case of $P$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Yes, this. I've seen people use the notation $N(X)$ for open neighbourhood and $N[X]$ for closed neighbourhood.
            $endgroup$
            – Especially Lime
            Feb 6 at 9:17










          • $begingroup$
            Indeed $N(X)$ can mean a few different things, after all this time I am not really sure which is the standard. $N(X)$ could be $S_1 doteq {v: vx in E$ for some $x in X}$... $N(X)$ could mean $S_1 cup X$ or it could mean $S_1 setminus X$.
            $endgroup$
            – Mike
            Feb 6 at 21:10





















          3












          $begingroup$

          It depends on how you define the neighbourhood, but it seems that here it is defined so that $N(V')$ comprises the vertices adjacent to some vertex in $V'$, excluding vertices in $V'$.



          So in that case, $N(D_0cup D_1)neq {v|d(v_0,v)leq 2}$.






          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









            4












            $begingroup$

            You are mixing closed and open neighbourhood. I think when they are stating $D_n=N(D_0cupldotscup D_{n-1})$, they are implying the open neighbourhood



            The open neighbourhood $N(S)$ of some set of vertices $S$ does not include $S$. Therefore, using your exampla of the path $P$:



            $D_0={v_0}$



            $D_1=N(D_0)={v mid d(v_0,v)=1}$



            $D_2=N(D_0cup D_1)={v mid d(v_0,v)=2}$
            and so on.



            More precisely if you define $P$ as:
            $$ ldots v_{-n} ldots v_{-2} v_1 v_0 v_1 v_2 ldots v_n ldots$$
            Then $D_0={v_0}$, $D_1=N(D_0)={v_1,v_{-1}}$, and
            $$D_2=N(D_0cup D_1)= N({v_1,v_0,v_{-1}})={v_2,v_{-2}}$$



            Therefore the statements holds as
            begin{align*}
            D_{n+1}={v_{n+1},v_{-(n+1)}} &subset N(D_n)=N({v_{n},v_{-n}})={v_{n+1},v_{n-1},v_{-(n-1)},v_{-(n+1)}}\
            &subset D_{n-1} cup D_{n+1}
            end{align*}
            with an equality in your specific case of $P$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Yes, this. I've seen people use the notation $N(X)$ for open neighbourhood and $N[X]$ for closed neighbourhood.
              $endgroup$
              – Especially Lime
              Feb 6 at 9:17










            • $begingroup$
              Indeed $N(X)$ can mean a few different things, after all this time I am not really sure which is the standard. $N(X)$ could be $S_1 doteq {v: vx in E$ for some $x in X}$... $N(X)$ could mean $S_1 cup X$ or it could mean $S_1 setminus X$.
              $endgroup$
              – Mike
              Feb 6 at 21:10


















            4












            $begingroup$

            You are mixing closed and open neighbourhood. I think when they are stating $D_n=N(D_0cupldotscup D_{n-1})$, they are implying the open neighbourhood



            The open neighbourhood $N(S)$ of some set of vertices $S$ does not include $S$. Therefore, using your exampla of the path $P$:



            $D_0={v_0}$



            $D_1=N(D_0)={v mid d(v_0,v)=1}$



            $D_2=N(D_0cup D_1)={v mid d(v_0,v)=2}$
            and so on.



            More precisely if you define $P$ as:
            $$ ldots v_{-n} ldots v_{-2} v_1 v_0 v_1 v_2 ldots v_n ldots$$
            Then $D_0={v_0}$, $D_1=N(D_0)={v_1,v_{-1}}$, and
            $$D_2=N(D_0cup D_1)= N({v_1,v_0,v_{-1}})={v_2,v_{-2}}$$



            Therefore the statements holds as
            begin{align*}
            D_{n+1}={v_{n+1},v_{-(n+1)}} &subset N(D_n)=N({v_{n},v_{-n}})={v_{n+1},v_{n-1},v_{-(n-1)},v_{-(n+1)}}\
            &subset D_{n-1} cup D_{n+1}
            end{align*}
            with an equality in your specific case of $P$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Yes, this. I've seen people use the notation $N(X)$ for open neighbourhood and $N[X]$ for closed neighbourhood.
              $endgroup$
              – Especially Lime
              Feb 6 at 9:17










            • $begingroup$
              Indeed $N(X)$ can mean a few different things, after all this time I am not really sure which is the standard. $N(X)$ could be $S_1 doteq {v: vx in E$ for some $x in X}$... $N(X)$ could mean $S_1 cup X$ or it could mean $S_1 setminus X$.
              $endgroup$
              – Mike
              Feb 6 at 21:10
















            4












            4








            4





            $begingroup$

            You are mixing closed and open neighbourhood. I think when they are stating $D_n=N(D_0cupldotscup D_{n-1})$, they are implying the open neighbourhood



            The open neighbourhood $N(S)$ of some set of vertices $S$ does not include $S$. Therefore, using your exampla of the path $P$:



            $D_0={v_0}$



            $D_1=N(D_0)={v mid d(v_0,v)=1}$



            $D_2=N(D_0cup D_1)={v mid d(v_0,v)=2}$
            and so on.



            More precisely if you define $P$ as:
            $$ ldots v_{-n} ldots v_{-2} v_1 v_0 v_1 v_2 ldots v_n ldots$$
            Then $D_0={v_0}$, $D_1=N(D_0)={v_1,v_{-1}}$, and
            $$D_2=N(D_0cup D_1)= N({v_1,v_0,v_{-1}})={v_2,v_{-2}}$$



            Therefore the statements holds as
            begin{align*}
            D_{n+1}={v_{n+1},v_{-(n+1)}} &subset N(D_n)=N({v_{n},v_{-n}})={v_{n+1},v_{n-1},v_{-(n-1)},v_{-(n+1)}}\
            &subset D_{n-1} cup D_{n+1}
            end{align*}
            with an equality in your specific case of $P$






            share|cite|improve this answer









            $endgroup$



            You are mixing closed and open neighbourhood. I think when they are stating $D_n=N(D_0cupldotscup D_{n-1})$, they are implying the open neighbourhood



            The open neighbourhood $N(S)$ of some set of vertices $S$ does not include $S$. Therefore, using your exampla of the path $P$:



            $D_0={v_0}$



            $D_1=N(D_0)={v mid d(v_0,v)=1}$



            $D_2=N(D_0cup D_1)={v mid d(v_0,v)=2}$
            and so on.



            More precisely if you define $P$ as:
            $$ ldots v_{-n} ldots v_{-2} v_1 v_0 v_1 v_2 ldots v_n ldots$$
            Then $D_0={v_0}$, $D_1=N(D_0)={v_1,v_{-1}}$, and
            $$D_2=N(D_0cup D_1)= N({v_1,v_0,v_{-1}})={v_2,v_{-2}}$$



            Therefore the statements holds as
            begin{align*}
            D_{n+1}={v_{n+1},v_{-(n+1)}} &subset N(D_n)=N({v_{n},v_{-n}})={v_{n+1},v_{n-1},v_{-(n-1)},v_{-(n+1)}}\
            &subset D_{n-1} cup D_{n+1}
            end{align*}
            with an equality in your specific case of $P$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 6 at 9:16









            Thomas LesgourguesThomas Lesgourgues

            1,285220




            1,285220








            • 1




              $begingroup$
              Yes, this. I've seen people use the notation $N(X)$ for open neighbourhood and $N[X]$ for closed neighbourhood.
              $endgroup$
              – Especially Lime
              Feb 6 at 9:17










            • $begingroup$
              Indeed $N(X)$ can mean a few different things, after all this time I am not really sure which is the standard. $N(X)$ could be $S_1 doteq {v: vx in E$ for some $x in X}$... $N(X)$ could mean $S_1 cup X$ or it could mean $S_1 setminus X$.
              $endgroup$
              – Mike
              Feb 6 at 21:10
















            • 1




              $begingroup$
              Yes, this. I've seen people use the notation $N(X)$ for open neighbourhood and $N[X]$ for closed neighbourhood.
              $endgroup$
              – Especially Lime
              Feb 6 at 9:17










            • $begingroup$
              Indeed $N(X)$ can mean a few different things, after all this time I am not really sure which is the standard. $N(X)$ could be $S_1 doteq {v: vx in E$ for some $x in X}$... $N(X)$ could mean $S_1 cup X$ or it could mean $S_1 setminus X$.
              $endgroup$
              – Mike
              Feb 6 at 21:10










            1




            1




            $begingroup$
            Yes, this. I've seen people use the notation $N(X)$ for open neighbourhood and $N[X]$ for closed neighbourhood.
            $endgroup$
            – Especially Lime
            Feb 6 at 9:17




            $begingroup$
            Yes, this. I've seen people use the notation $N(X)$ for open neighbourhood and $N[X]$ for closed neighbourhood.
            $endgroup$
            – Especially Lime
            Feb 6 at 9:17












            $begingroup$
            Indeed $N(X)$ can mean a few different things, after all this time I am not really sure which is the standard. $N(X)$ could be $S_1 doteq {v: vx in E$ for some $x in X}$... $N(X)$ could mean $S_1 cup X$ or it could mean $S_1 setminus X$.
            $endgroup$
            – Mike
            Feb 6 at 21:10






            $begingroup$
            Indeed $N(X)$ can mean a few different things, after all this time I am not really sure which is the standard. $N(X)$ could be $S_1 doteq {v: vx in E$ for some $x in X}$... $N(X)$ could mean $S_1 cup X$ or it could mean $S_1 setminus X$.
            $endgroup$
            – Mike
            Feb 6 at 21:10













            3












            $begingroup$

            It depends on how you define the neighbourhood, but it seems that here it is defined so that $N(V')$ comprises the vertices adjacent to some vertex in $V'$, excluding vertices in $V'$.



            So in that case, $N(D_0cup D_1)neq {v|d(v_0,v)leq 2}$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              It depends on how you define the neighbourhood, but it seems that here it is defined so that $N(V')$ comprises the vertices adjacent to some vertex in $V'$, excluding vertices in $V'$.



              So in that case, $N(D_0cup D_1)neq {v|d(v_0,v)leq 2}$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                It depends on how you define the neighbourhood, but it seems that here it is defined so that $N(V')$ comprises the vertices adjacent to some vertex in $V'$, excluding vertices in $V'$.



                So in that case, $N(D_0cup D_1)neq {v|d(v_0,v)leq 2}$.






                share|cite|improve this answer









                $endgroup$



                It depends on how you define the neighbourhood, but it seems that here it is defined so that $N(V')$ comprises the vertices adjacent to some vertex in $V'$, excluding vertices in $V'$.



                So in that case, $N(D_0cup D_1)neq {v|d(v_0,v)leq 2}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 6 at 9:15









                broncoAbiertobroncoAbierto

                28819




                28819






























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                    張江高科駅

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                    張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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