A topological space is connected if and only if the associated graph is connected
Similar to this question A topological space is path connected if and only if the associated graph is connected, but I can't seem to make the connection.
My problem is:
$G$ is a graph. Show that $G$ is connected as a topological space if and only if $G$ is connected as a graph.
I can show that $G$ is connected as a topological space if it is connected as a graph, but how do I show, that if $G$ is connected as a topological space, it is connected as a graph?
general-topology graph-theory connectedness
asked Nov 26 at 11:49
Rasmus Søgaard Christensen
597
add a comment |
Similar to this question A topological space is path connected if and only if the associated graph is connected, but I can't seem to make the connection.
My problem is:
$G$ is a graph. Show that $G$ is connected as a topological space if and only if $G$ is connected as a graph.
I can show that $G$ is connected as a topological space if it is connected as a graph, but how do I show, that if $G$ is connected as a topological space, it is connected as a graph?
general-topology graph-theory connectedness
asked Nov 26 at 11:49
Rasmus Søgaard Christensen
597
How is $G$ a topological space, presumably it's a $1$-simplex, right? It will follow from a connected graph as space being path-connected, I suppose.
– Henno Brandsma
Nov 26 at 12:16
1
If it is not connected as a graph, then there are two disconnected components, yes? I'd bet that those components are each open (topologically). If this is true, then this is a topological separation of your space.
– Prototank
Nov 26 at 16:04
If you are talking about geometric realization of a graph (i.e. each edge corresponds to $1$-simplex) then it is enough to show that two vertices can be connected by a path in topological sense if and only if they can be connected by a path in graph sense. This is easy because each edge is an image of $[0,1]$ on one hand and it is a path (in graph sense) of length $1$.
– freakish
Nov 26 at 16:10
Some duplicates: math.stackexchange.com/q/3021280 and math.stackexchange.com/q/3017661.
– Paul Frost
Dec 1 at 13:11
add a comment |
Similar to this question A topological space is path connected if and only if the associated graph is connected, but I can't seem to make the connection.
My problem is:
$G$ is a graph. Show that $G$ is connected as a topological space if and only if $G$ is connected as a graph.
I can show that $G$ is connected as a topological space if it is connected as a graph, but how do I show, that if $G$ is connected as a topological space, it is connected as a graph?
general-topology graph-theory connectedness
asked Nov 26 at 11:49
Rasmus Søgaard Christensen
597
Similar to this question A topological space is path connected if and only if the associated graph is connected, but I can't seem to make the connection.
My problem is:
$G$ is a graph. Show that $G$ is connected as a topological space if and only if $G$ is connected as a graph.
I can show that $G$ is connected as a topological space if it is connected as a graph, but how do I show, that if $G$ is connected as a topological space, it is connected as a graph?
general-topology graph-theory connectedness
general-topology graph-theory connectedness
asked Nov 26 at 11:49
Rasmus Søgaard Christensen
597
asked Nov 26 at 11:49
Rasmus Søgaard Christensen
597
asked Nov 26 at 11:49
Rasmus Søgaard Christensen
597
asked Nov 26 at 11:49
Rasmus Søgaard Christensen
597
asked Nov 26 at 11:49
Rasmus Søgaard Christensen
597
597
How is $G$ a topological space, presumably it's a $1$-simplex, right? It will follow from a connected graph as space being path-connected, I suppose.
– Henno Brandsma
Nov 26 at 12:16
1
If it is not connected as a graph, then there are two disconnected components, yes? I'd bet that those components are each open (topologically). If this is true, then this is a topological separation of your space.
– Prototank
Nov 26 at 16:04
If you are talking about geometric realization of a graph (i.e. each edge corresponds to $1$-simplex) then it is enough to show that two vertices can be connected by a path in topological sense if and only if they can be connected by a path in graph sense. This is easy because each edge is an image of $[0,1]$ on one hand and it is a path (in graph sense) of length $1$.
– freakish
Nov 26 at 16:10
Some duplicates: math.stackexchange.com/q/3021280 and math.stackexchange.com/q/3017661.
– Paul Frost
Dec 1 at 13:11
add a comment |
How is $G$ a topological space, presumably it's a $1$-simplex, right? It will follow from a connected graph as space being path-connected, I suppose.
– Henno Brandsma
Nov 26 at 12:16
1
If it is not connected as a graph, then there are two disconnected components, yes? I'd bet that those components are each open (topologically). If this is true, then this is a topological separation of your space.
– Prototank
Nov 26 at 16:04
If you are talking about geometric realization of a graph (i.e. each edge corresponds to $1$-simplex) then it is enough to show that two vertices can be connected by a path in topological sense if and only if they can be connected by a path in graph sense. This is easy because each edge is an image of $[0,1]$ on one hand and it is a path (in graph sense) of length $1$.
– freakish
Nov 26 at 16:10
Some duplicates: math.stackexchange.com/q/3021280 and math.stackexchange.com/q/3017661.
– Paul Frost
Dec 1 at 13:11
How is $G$ a topological space, presumably it's a $1$-simplex, right? It will follow from a connected graph as space being path-connected, I suppose.
– Henno Brandsma
Nov 26 at 12:16
How is $G$ a topological space, presumably it's a $1$-simplex, right? It will follow from a connected graph as space being path-connected, I suppose.
– Henno Brandsma
Nov 26 at 12:16
1
1
If it is not connected as a graph, then there are two disconnected components, yes? I'd bet that those components are each open (topologically). If this is true, then this is a topological separation of your space.
– Prototank
Nov 26 at 16:04
If it is not connected as a graph, then there are two disconnected components, yes? I'd bet that those components are each open (topologically). If this is true, then this is a topological separation of your space.
– Prototank
Nov 26 at 16:04
If you are talking about geometric realization of a graph (i.e. each edge corresponds to $1$-simplex) then it is enough to show that two vertices can be connected by a path in topological sense if and only if they can be connected by a path in graph sense. This is easy because each edge is an image of $[0,1]$ on one hand and it is a path (in graph sense) of length $1$.
– freakish
Nov 26 at 16:10
If you are talking about geometric realization of a graph (i.e. each edge corresponds to $1$-simplex) then it is enough to show that two vertices can be connected by a path in topological sense if and only if they can be connected by a path in graph sense. This is easy because each edge is an image of $[0,1]$ on one hand and it is a path (in graph sense) of length $1$.
– freakish
Nov 26 at 16:10
Some duplicates: math.stackexchange.com/q/3021280 and math.stackexchange.com/q/3017661.
– Paul Frost
Dec 1 at 13:11
Some duplicates: math.stackexchange.com/q/3021280 and math.stackexchange.com/q/3017661.
– Paul Frost
Dec 1 at 13:11
add a comment |
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How is $G$ a topological space, presumably it's a $1$-simplex, right? It will follow from a connected graph as space being path-connected, I suppose.
– Henno Brandsma
Nov 26 at 12:16
1
If it is not connected as a graph, then there are two disconnected components, yes? I'd bet that those components are each open (topologically). If this is true, then this is a topological separation of your space.
– Prototank
Nov 26 at 16:04
If you are talking about geometric realization of a graph (i.e. each edge corresponds to $1$-simplex) then it is enough to show that two vertices can be connected by a path in topological sense if and only if they can be connected by a path in graph sense. This is easy because each edge is an image of $[0,1]$ on one hand and it is a path (in graph sense) of length $1$.
– freakish
Nov 26 at 16:10
Some duplicates: math.stackexchange.com/q/3021280 and math.stackexchange.com/q/3017661.
– Paul Frost
Dec 1 at 13:11