The subgraph induced by the neighbourhood of a vertex

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I have the connected graph as in picture 1 attached.

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

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.

asked 12 hours ago

Buddhini Angelika

1047

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up vote
0
down vote

favorite

I have the connected graph as in picture 1 attached.

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

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.

asked 12 hours ago

Buddhini Angelika

1047

add a comment |

up vote
0
down vote

favorite

I have the connected graph as in picture 1 attached.

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

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.

asked 12 hours ago

Buddhini Angelika

1047

I have the connected graph as in picture 1 attached.

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

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

asked 12 hours ago

Buddhini Angelika

1047

asked 12 hours ago

Buddhini Angelika

1047

asked 12 hours ago

Buddhini Angelika

1047

asked 12 hours ago

Buddhini Angelika

1047

asked 12 hours ago

Buddhini Angelika

1047

add a comment |

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 $a$).

answered 12 hours ago

Michal Adamaszek

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

|
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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 $a$).

answered 12 hours ago

Michal Adamaszek

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

|
show 3 more comments

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 $a$).

answered 12 hours ago

Michal Adamaszek

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

|
show 3 more comments

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 $a$).

answered 12 hours ago

Michal Adamaszek

1,99248

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 $a$).

answered 12 hours ago

Michal Adamaszek

1,99248

answered 12 hours ago

Michal Adamaszek

1,99248

answered 12 hours ago

Michal Adamaszek

1,99248

answered 12 hours ago

Michal Adamaszek

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

|
show 3 more comments

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

|
show 3 more comments

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