Show that this graph is $k$-connected











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Given a $k$-connected graph $G$, let $G'$ be a graph obtained from $G$ by adding a new vertex $x$ and connecting it to any $k$ vertices of $G$. From first principles, show that $G'$ is $k$-connected.



I understand visually how this is, just don't know how to "show it from first principles".






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  • I'm not sure what first principles mean either but, presumably, it means to prove it from the definition of connectedness; Much like proving limits from first principles means to prove it from $epsilon$-$delta$. So you ought to prove by removing up to $k-1$ vertices the graph $G'$ remains connected.
    – Sisyphus
    15 hours ago















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Given a $k$-connected graph $G$, let $G'$ be a graph obtained from $G$ by adding a new vertex $x$ and connecting it to any $k$ vertices of $G$. From first principles, show that $G'$ is $k$-connected.



I understand visually how this is, just don't know how to "show it from first principles".






graph-theory







share|cite|improve this question






















  • I'm not sure what first principles mean either but, presumably, it means to prove it from the definition of connectedness; Much like proving limits from first principles means to prove it from $epsilon$-$delta$. So you ought to prove by removing up to $k-1$ vertices the graph $G'$ remains connected.
    – Sisyphus
    15 hours ago













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Given a $k$-connected graph $G$, let $G'$ be a graph obtained from $G$ by adding a new vertex $x$ and connecting it to any $k$ vertices of $G$. From first principles, show that $G'$ is $k$-connected.



I understand visually how this is, just don't know how to "show it from first principles".






graph-theory







share|cite|improve this question













Given a $k$-connected graph $G$, let $G'$ be a graph obtained from $G$ by adding a new vertex $x$ and connecting it to any $k$ vertices of $G$. From first principles, show that $G'$ is $k$-connected.



I understand visually how this is, just don't know how to "show it from first principles".






graph-theory




graph-theory






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asked 15 hours ago









user499701

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947












  • I'm not sure what first principles mean either but, presumably, it means to prove it from the definition of connectedness; Much like proving limits from first principles means to prove it from $epsilon$-$delta$. So you ought to prove by removing up to $k-1$ vertices the graph $G'$ remains connected.
    – Sisyphus
    15 hours ago


















  • I'm not sure what first principles mean either but, presumably, it means to prove it from the definition of connectedness; Much like proving limits from first principles means to prove it from $epsilon$-$delta$. So you ought to prove by removing up to $k-1$ vertices the graph $G'$ remains connected.
    – Sisyphus
    15 hours ago
















I'm not sure what first principles mean either but, presumably, it means to prove it from the definition of connectedness; Much like proving limits from first principles means to prove it from $epsilon$-$delta$. So you ought to prove by removing up to $k-1$ vertices the graph $G'$ remains connected.
– Sisyphus
15 hours ago




I'm not sure what first principles mean either but, presumably, it means to prove it from the definition of connectedness; Much like proving limits from first principles means to prove it from $epsilon$-$delta$. So you ought to prove by removing up to $k-1$ vertices the graph $G'$ remains connected.
– Sisyphus
15 hours ago










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$G$ is a graph that is k-connected means that removing less than $k$ vertices from $G$ and $G$ is still connected. Removing any set of vertices that do not contain $v$ from $G'$ of size less than $K$, doesn't disconnect $v$ from $G'$ as $v$ is connected to $k$ vertices in $G$. It doesn't disconnect $G$ as we removed less than k vertices. Now if you remove a set of vertices from $G'$ containing $v$ of size less than $k$, you don't disconnect $G$ as you removed less than $k$ vertices in $G$. so $G'$ is $k$ connected by definition.






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    $G$ is a graph that is k-connected means that removing less than $k$ vertices from $G$ and $G$ is still connected. Removing any set of vertices that do not contain $v$ from $G'$ of size less than $K$, doesn't disconnect $v$ from $G'$ as $v$ is connected to $k$ vertices in $G$. It doesn't disconnect $G$ as we removed less than k vertices. Now if you remove a set of vertices from $G'$ containing $v$ of size less than $k$, you don't disconnect $G$ as you removed less than $k$ vertices in $G$. so $G'$ is $k$ connected by definition.






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      $G$ is a graph that is k-connected means that removing less than $k$ vertices from $G$ and $G$ is still connected. Removing any set of vertices that do not contain $v$ from $G'$ of size less than $K$, doesn't disconnect $v$ from $G'$ as $v$ is connected to $k$ vertices in $G$. It doesn't disconnect $G$ as we removed less than k vertices. Now if you remove a set of vertices from $G'$ containing $v$ of size less than $k$, you don't disconnect $G$ as you removed less than $k$ vertices in $G$. so $G'$ is $k$ connected by definition.






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        $G$ is a graph that is k-connected means that removing less than $k$ vertices from $G$ and $G$ is still connected. Removing any set of vertices that do not contain $v$ from $G'$ of size less than $K$, doesn't disconnect $v$ from $G'$ as $v$ is connected to $k$ vertices in $G$. It doesn't disconnect $G$ as we removed less than k vertices. Now if you remove a set of vertices from $G'$ containing $v$ of size less than $k$, you don't disconnect $G$ as you removed less than $k$ vertices in $G$. so $G'$ is $k$ connected by definition.






        share|cite|improve this answer












        $G$ is a graph that is k-connected means that removing less than $k$ vertices from $G$ and $G$ is still connected. Removing any set of vertices that do not contain $v$ from $G'$ of size less than $K$, doesn't disconnect $v$ from $G'$ as $v$ is connected to $k$ vertices in $G$. It doesn't disconnect $G$ as we removed less than k vertices. Now if you remove a set of vertices from $G'$ containing $v$ of size less than $k$, you don't disconnect $G$ as you removed less than $k$ vertices in $G$. so $G'$ is $k$ connected by definition.







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        answered 15 hours ago









        mathnoob

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