3
$\begingroup$

I was looking through the definition of an induced subgraph, and it states that if $G'$ is an induced subgraph of $G$, then $V(G')\subseteq V(G)$ and $E(G')=E(G)\cap$$V(G')\choose2$.

My question is, if you have a big enough graph $G$, and choose two completely separate vertices such that there is no edge between them, then is the resulting graph $G'$ an induced subgraph (where $G'$ is just two separate nodes)?

It contains no edges, so it is just the two nodes, but I saw nothing of this mentioned in the book I'm reading so I had the doubt.

Similarly, if you chose one single vertex of $G$, is it also an induced subgraph?

$\endgroup$

1 Answer 1

5
$\begingroup$

Yes. Even when it looks strange at first, if the definition says it‘s true, then it‘s true indeed. You do not need to have any doubts here. Instead, proudly apply the definition! And moreover, that there are no edges between the vertices is no strange thing at all, I have to say, or is it?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .