Can a set be connected without being polygonally-connected?

[Bachelier]

I'm having a hard time using the definition.

To show that a set S is connected, does it suffice to find two open disjoint sets in the metric space that contain S. (Basically one will contain S while the other wouldn't)?

Thanks

 

[deluks917]

If you can find two disjoint open sets that contain S then S is dis-connected. If S is connected you CAN'T find the disjoint open sets.

 

[Bachelier]

If you can find two disjoint open sets that contain S then S is dis-connected. If S is connected you CAN'T find the disjoint open sets.

In other words you are saying that to prove connectedness, you can only do it by contradiction?

 

[Robert1986]

In other words you are saying that to prove connectedness, you can only do it by contradiction?

One good way is to suppose that there is a disconnection and then show that this implies that the unit interval is disconnected. There is also a theorem that says that a set, G, is connected if and only if any two points, x and y, in G can be connected by a polygonal curve (that is, a curve made of line segments) such that this curve lies entirely in G.

 

[HallsofIvy]

I won't say that you can only prove connectedness by contradiction but since "connected" is defined in a negative way- "A set X is connected if and only if it is NOT the union of two separated sets"- that is the most natural way.

Just proving that, if X is contained in the union of two open sets, U and V, then it is contained in either U or V, is not sufficient because you have not said anything about the connectedness of U and V.

 

[Bachelier]

Thank you.

 

[spamiam]

There is also a theorem that says that a set, G, is connected if and only if any two points, x and y, in G can be connected by a polygonal curve (that is, a curve made of line segments) such that this curve lies entirely in G.

I don't think this is true. This is the standard counterexample for "connected if and only if path-connected," (and I think it works for polygonally connected as well): http://en.wikipedia.org/wiki/Topologist's_sine_curve.

That said, I think path-connectedness does imply connectedness.

 

[Robert1986]

I don't think this is true. This is the standard counterexample for "connected if and only if path-connected," (and I think it works for polygonally connected as well): http://en.wikipedia.org/wiki/Topologist's_sine_curve.

That said, I think path-connectedness does imply connectedness.

It is true. Path connected and polygonally connected are not the same. Something can be path connected without being polygonally connected. The intuition behind the proof of what I said is that if G is polygonally connected, and if G_1 and G_2 form a disconnection, then there is a line segment from a point g_1 in G_1 to a point g_2 in G_2 that is contained within G. Since this line segment must pass the "boundary" between G_1 and G_2, it can be shown that this implies that the unit interval is disconnected, which isn't true.


The proof breaks down when you start having to use non-line segments to connect two points. That is why it doesn't work for general path-connected sets.

 

[spamiam]

The intuition behind the proof of what I said is that if G is polygonally connected, and if G_1 and G_2 form a disconnection, then there is a line segment from a point g_1 in G_1 to a point g_2 in G_2 that is contained within G. Since this line segment must pass the "boundary" between G_1 and G_2, it can be shown that this implies that the unit interval is disconnected, which isn't true.

It's the converse that doesn't work: connected doesn't imply polygonnaly-connected. For instance, the curve y=x² is connected, but not polygonally-connected. So polygonally-connected implies connected, but not vice-versa.