The subgraph induced by the neighbourhood of a vertex











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I have the connected graph as in picture 1 attached.



Picture 1



Is it correct if I take the subgraph induced by the neighbourhood of the vertex "a" as in the picture 2 below?



Picture 2



The neighbouring vertices of "a" are the vertices circled in red, right?



And the induced subgraph in Picture 2 is a connected graph right?



Can someone please help me with this question.



Thanks a lot in advance.






graph-theory connectedness graph-connectivity locally-connected







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    up vote
    0
    down vote

    favorite












    I have the connected graph as in picture 1 attached.



    Picture 1



    Is it correct if I take the subgraph induced by the neighbourhood of the vertex "a" as in the picture 2 below?



    Picture 2



    The neighbouring vertices of "a" are the vertices circled in red, right?



    And the induced subgraph in Picture 2 is a connected graph right?



    Can someone please help me with this question.



    Thanks a lot in advance.






    graph-theory connectedness graph-connectivity locally-connected







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have the connected graph as in picture 1 attached.



      Picture 1



      Is it correct if I take the subgraph induced by the neighbourhood of the vertex "a" as in the picture 2 below?



      Picture 2



      The neighbouring vertices of "a" are the vertices circled in red, right?



      And the induced subgraph in Picture 2 is a connected graph right?



      Can someone please help me with this question.



      Thanks a lot in advance.






      graph-theory connectedness graph-connectivity locally-connected







      share|cite|improve this question













      I have the connected graph as in picture 1 attached.



      Picture 1



      Is it correct if I take the subgraph induced by the neighbourhood of the vertex "a" as in the picture 2 below?



      Picture 2



      The neighbouring vertices of "a" are the vertices circled in red, right?



      And the induced subgraph in Picture 2 is a connected graph right?



      Can someone please help me with this question.



      Thanks a lot in advance.






      graph-theory connectedness graph-connectivity locally-connected




      graph-theory connectedness graph-connectivity locally-connected






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 12 hours ago









      Buddhini Angelika

      1047




      1047






















          1 Answer
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          Normally the subgraph induced by a set of vertices means those vertices and all edges between them that were in the original graph.



          So in your case it would be the four red vertices and the one edge between the topmost two of them.



          There is also sometimes a distinction between "open neibourhood" (all vertices adjacent to $a$) and "closed neibourhood" (all those and also $a$).






          share|cite|improve this answer





















          • Thanks a lot. Then in an open neighbourhood only the vertices adjacent to "a" and the edges between them which were in the original graph will be present? Without "a"?
            – Buddhini Angelika
            11 hours ago










          • @BuddhiniAngelika That is the most standard way to understand it, yes.
            – Michal Adamaszek
            11 hours ago










          • And in the closed neighbourhood when we have "a" then do we take the edges between the adjacent vertices and "a" as well to the induced subgraph?
            – Buddhini Angelika
            11 hours ago










          • @BuddhiniAngelika Once you fix a vertex set then "induced" means take all edges that were between them, so yes. In particular the subgraph induced by the closed neibhood of a is always connected.
            – Michal Adamaszek
            11 hours ago










          • Thanks a lot @MichalAdamaszek. There is a thorem, particularly theorem 1 in this document.core.ac.uk/download/pdf/81191533.pdf . In that in the paragraph above theorem 1 they have defined about locally connected graphs by thinking about the induced subgraphs of the neighbourhood.
            – Buddhini Angelika
            11 hours ago











          Your Answer





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          1 Answer
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          active

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          active

          oldest

          votes








          up vote
          0
          down vote













          Normally the subgraph induced by a set of vertices means those vertices and all edges between them that were in the original graph.



          So in your case it would be the four red vertices and the one edge between the topmost two of them.



          There is also sometimes a distinction between "open neibourhood" (all vertices adjacent to $a$) and "closed neibourhood" (all those and also $a$).






          share|cite|improve this answer





















          • Thanks a lot. Then in an open neighbourhood only the vertices adjacent to "a" and the edges between them which were in the original graph will be present? Without "a"?
            – Buddhini Angelika
            11 hours ago










          • @BuddhiniAngelika That is the most standard way to understand it, yes.
            – Michal Adamaszek
            11 hours ago










          • And in the closed neighbourhood when we have "a" then do we take the edges between the adjacent vertices and "a" as well to the induced subgraph?
            – Buddhini Angelika
            11 hours ago










          • @BuddhiniAngelika Once you fix a vertex set then "induced" means take all edges that were between them, so yes. In particular the subgraph induced by the closed neibhood of a is always connected.
            – Michal Adamaszek
            11 hours ago










          • Thanks a lot @MichalAdamaszek. There is a thorem, particularly theorem 1 in this document.core.ac.uk/download/pdf/81191533.pdf . In that in the paragraph above theorem 1 they have defined about locally connected graphs by thinking about the induced subgraphs of the neighbourhood.
            – Buddhini Angelika
            11 hours ago















          up vote
          0
          down vote













          Normally the subgraph induced by a set of vertices means those vertices and all edges between them that were in the original graph.



          So in your case it would be the four red vertices and the one edge between the topmost two of them.



          There is also sometimes a distinction between "open neibourhood" (all vertices adjacent to $a$) and "closed neibourhood" (all those and also $a$).






          share|cite|improve this answer





















          • Thanks a lot. Then in an open neighbourhood only the vertices adjacent to "a" and the edges between them which were in the original graph will be present? Without "a"?
            – Buddhini Angelika
            11 hours ago










          • @BuddhiniAngelika That is the most standard way to understand it, yes.
            – Michal Adamaszek
            11 hours ago










          • And in the closed neighbourhood when we have "a" then do we take the edges between the adjacent vertices and "a" as well to the induced subgraph?
            – Buddhini Angelika
            11 hours ago










          • @BuddhiniAngelika Once you fix a vertex set then "induced" means take all edges that were between them, so yes. In particular the subgraph induced by the closed neibhood of a is always connected.
            – Michal Adamaszek
            11 hours ago










          • Thanks a lot @MichalAdamaszek. There is a thorem, particularly theorem 1 in this document.core.ac.uk/download/pdf/81191533.pdf . In that in the paragraph above theorem 1 they have defined about locally connected graphs by thinking about the induced subgraphs of the neighbourhood.
            – Buddhini Angelika
            11 hours ago













          up vote
          0
          down vote










          up vote
          0
          down vote









          Normally the subgraph induced by a set of vertices means those vertices and all edges between them that were in the original graph.



          So in your case it would be the four red vertices and the one edge between the topmost two of them.



          There is also sometimes a distinction between "open neibourhood" (all vertices adjacent to $a$) and "closed neibourhood" (all those and also $a$).






          share|cite|improve this answer












          Normally the subgraph induced by a set of vertices means those vertices and all edges between them that were in the original graph.



          So in your case it would be the four red vertices and the one edge between the topmost two of them.



          There is also sometimes a distinction between "open neibourhood" (all vertices adjacent to $a$) and "closed neibourhood" (all those and also $a$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 12 hours ago









          Michal Adamaszek

          1,99248




          1,99248












          • Thanks a lot. Then in an open neighbourhood only the vertices adjacent to "a" and the edges between them which were in the original graph will be present? Without "a"?
            – Buddhini Angelika
            11 hours ago










          • @BuddhiniAngelika That is the most standard way to understand it, yes.
            – Michal Adamaszek
            11 hours ago










          • And in the closed neighbourhood when we have "a" then do we take the edges between the adjacent vertices and "a" as well to the induced subgraph?
            – Buddhini Angelika
            11 hours ago










          • @BuddhiniAngelika Once you fix a vertex set then "induced" means take all edges that were between them, so yes. In particular the subgraph induced by the closed neibhood of a is always connected.
            – Michal Adamaszek
            11 hours ago










          • Thanks a lot @MichalAdamaszek. There is a thorem, particularly theorem 1 in this document.core.ac.uk/download/pdf/81191533.pdf . In that in the paragraph above theorem 1 they have defined about locally connected graphs by thinking about the induced subgraphs of the neighbourhood.
            – Buddhini Angelika
            11 hours ago


















          • Thanks a lot. Then in an open neighbourhood only the vertices adjacent to "a" and the edges between them which were in the original graph will be present? Without "a"?
            – Buddhini Angelika
            11 hours ago










          • @BuddhiniAngelika That is the most standard way to understand it, yes.
            – Michal Adamaszek
            11 hours ago










          • And in the closed neighbourhood when we have "a" then do we take the edges between the adjacent vertices and "a" as well to the induced subgraph?
            – Buddhini Angelika
            11 hours ago










          • @BuddhiniAngelika Once you fix a vertex set then "induced" means take all edges that were between them, so yes. In particular the subgraph induced by the closed neibhood of a is always connected.
            – Michal Adamaszek
            11 hours ago










          • Thanks a lot @MichalAdamaszek. There is a thorem, particularly theorem 1 in this document.core.ac.uk/download/pdf/81191533.pdf . In that in the paragraph above theorem 1 they have defined about locally connected graphs by thinking about the induced subgraphs of the neighbourhood.
            – Buddhini Angelika
            11 hours ago
















          Thanks a lot. Then in an open neighbourhood only the vertices adjacent to "a" and the edges between them which were in the original graph will be present? Without "a"?
          – Buddhini Angelika
          11 hours ago




          Thanks a lot. Then in an open neighbourhood only the vertices adjacent to "a" and the edges between them which were in the original graph will be present? Without "a"?
          – Buddhini Angelika
          11 hours ago












          @BuddhiniAngelika That is the most standard way to understand it, yes.
          – Michal Adamaszek
          11 hours ago




          @BuddhiniAngelika That is the most standard way to understand it, yes.
          – Michal Adamaszek
          11 hours ago












          And in the closed neighbourhood when we have "a" then do we take the edges between the adjacent vertices and "a" as well to the induced subgraph?
          – Buddhini Angelika
          11 hours ago




          And in the closed neighbourhood when we have "a" then do we take the edges between the adjacent vertices and "a" as well to the induced subgraph?
          – Buddhini Angelika
          11 hours ago












          @BuddhiniAngelika Once you fix a vertex set then "induced" means take all edges that were between them, so yes. In particular the subgraph induced by the closed neibhood of a is always connected.
          – Michal Adamaszek
          11 hours ago




          @BuddhiniAngelika Once you fix a vertex set then "induced" means take all edges that were between them, so yes. In particular the subgraph induced by the closed neibhood of a is always connected.
          – Michal Adamaszek
          11 hours ago












          Thanks a lot @MichalAdamaszek. There is a thorem, particularly theorem 1 in this document.core.ac.uk/download/pdf/81191533.pdf . In that in the paragraph above theorem 1 they have defined about locally connected graphs by thinking about the induced subgraphs of the neighbourhood.
          – Buddhini Angelika
          11 hours ago




          Thanks a lot @MichalAdamaszek. There is a thorem, particularly theorem 1 in this document.core.ac.uk/download/pdf/81191533.pdf . In that in the paragraph above theorem 1 they have defined about locally connected graphs by thinking about the induced subgraphs of the neighbourhood.
          – Buddhini Angelika
          11 hours ago


















           

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