Can I use mathematical induction here?

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Given any digraph $D=(V,E)$ such that for every finite subset $Ssubseteq V$ there exists a vertex $v_Sin V$ capable of reaching any vertex in $S$ via a directed path.



Can one deduce there is a vertex $uin V$ capable of reaching any vertex in $V$ via a directed path?



Do I have to assume $V$ itself is finite for this to work? Or can I make this deduction irregardless of the cardinality of $V$? I know if I have two subsets $A,Bsubseteq V$, where every vertex in $A$ is reachable by $v_A$ and every vertex in $B$ is reachble by $v_B$ then, since $v_A,v_Bsubseteq V$ there must exist a vertex $u$ capable of reaching both $v_A$ and $v_B$ which means $u$ is capable of reaching every vertex in $Acup B$.










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    Given any digraph $D=(V,E)$ such that for every finite subset $Ssubseteq V$ there exists a vertex $v_Sin V$ capable of reaching any vertex in $S$ via a directed path.



    Can one deduce there is a vertex $uin V$ capable of reaching any vertex in $V$ via a directed path?



    Do I have to assume $V$ itself is finite for this to work? Or can I make this deduction irregardless of the cardinality of $V$? I know if I have two subsets $A,Bsubseteq V$, where every vertex in $A$ is reachable by $v_A$ and every vertex in $B$ is reachble by $v_B$ then, since $v_A,v_Bsubseteq V$ there must exist a vertex $u$ capable of reaching both $v_A$ and $v_B$ which means $u$ is capable of reaching every vertex in $Acup B$.










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      up vote
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      up vote
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      down vote

      favorite











      Given any digraph $D=(V,E)$ such that for every finite subset $Ssubseteq V$ there exists a vertex $v_Sin V$ capable of reaching any vertex in $S$ via a directed path.



      Can one deduce there is a vertex $uin V$ capable of reaching any vertex in $V$ via a directed path?



      Do I have to assume $V$ itself is finite for this to work? Or can I make this deduction irregardless of the cardinality of $V$? I know if I have two subsets $A,Bsubseteq V$, where every vertex in $A$ is reachable by $v_A$ and every vertex in $B$ is reachble by $v_B$ then, since $v_A,v_Bsubseteq V$ there must exist a vertex $u$ capable of reaching both $v_A$ and $v_B$ which means $u$ is capable of reaching every vertex in $Acup B$.










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      Given any digraph $D=(V,E)$ such that for every finite subset $Ssubseteq V$ there exists a vertex $v_Sin V$ capable of reaching any vertex in $S$ via a directed path.



      Can one deduce there is a vertex $uin V$ capable of reaching any vertex in $V$ via a directed path?



      Do I have to assume $V$ itself is finite for this to work? Or can I make this deduction irregardless of the cardinality of $V$? I know if I have two subsets $A,Bsubseteq V$, where every vertex in $A$ is reachable by $v_A$ and every vertex in $B$ is reachble by $v_B$ then, since $v_A,v_Bsubseteq V$ there must exist a vertex $u$ capable of reaching both $v_A$ and $v_B$ which means $u$ is capable of reaching every vertex in $Acup B$.







      graph-theory induction transfinite-induction






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      asked Aug 30 at 0:09









      davey5554

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          No, you can't deduce that such a vertex $u$ exists.



          Consider the graph with vertex set $mathbb Z$, and an edge from $n$ to $n+1$ for every $n in mathbb Z$.



          Then for every finite set $S$, the vertex $min S$ can reach any vertex in $S$. However, for every vertex $n$, there are vertices (such as $n-1$) that can't be reached from $n$.



          (On the other hand, if $V$ is finite, then the statement is - trivially - true. In that case, $V$ itself is a finite subset of $V$.)






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            No, you can't deduce that such a vertex $u$ exists.



            Consider the graph with vertex set $mathbb Z$, and an edge from $n$ to $n+1$ for every $n in mathbb Z$.



            Then for every finite set $S$, the vertex $min S$ can reach any vertex in $S$. However, for every vertex $n$, there are vertices (such as $n-1$) that can't be reached from $n$.



            (On the other hand, if $V$ is finite, then the statement is - trivially - true. In that case, $V$ itself is a finite subset of $V$.)






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              up vote
              1
              down vote



              accepted










              No, you can't deduce that such a vertex $u$ exists.



              Consider the graph with vertex set $mathbb Z$, and an edge from $n$ to $n+1$ for every $n in mathbb Z$.



              Then for every finite set $S$, the vertex $min S$ can reach any vertex in $S$. However, for every vertex $n$, there are vertices (such as $n-1$) that can't be reached from $n$.



              (On the other hand, if $V$ is finite, then the statement is - trivially - true. In that case, $V$ itself is a finite subset of $V$.)






              share|cite|improve this answer






















                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                No, you can't deduce that such a vertex $u$ exists.



                Consider the graph with vertex set $mathbb Z$, and an edge from $n$ to $n+1$ for every $n in mathbb Z$.



                Then for every finite set $S$, the vertex $min S$ can reach any vertex in $S$. However, for every vertex $n$, there are vertices (such as $n-1$) that can't be reached from $n$.



                (On the other hand, if $V$ is finite, then the statement is - trivially - true. In that case, $V$ itself is a finite subset of $V$.)






                share|cite|improve this answer












                No, you can't deduce that such a vertex $u$ exists.



                Consider the graph with vertex set $mathbb Z$, and an edge from $n$ to $n+1$ for every $n in mathbb Z$.



                Then for every finite set $S$, the vertex $min S$ can reach any vertex in $S$. However, for every vertex $n$, there are vertices (such as $n-1$) that can't be reached from $n$.



                (On the other hand, if $V$ is finite, then the statement is - trivially - true. In that case, $V$ itself is a finite subset of $V$.)







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                answered Aug 30 at 0:42









                Misha Lavrov

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