1. cut-point

How would I show that a circle and a circle with a line bisecting it(with the end-points of the line touching the circumference) aren't homeomorphic?

I know I would have to use the 'cut-point principle' which states that homeomorphic sets have the same number of n-points for each n, but I don't know how to find the n-points....

2. For any circle (a simple closed curve) removing any two points leaves exactly two connected subsets.
We know that a simple closed curve is the union of two arcs the intersection of which is the set of their endpoints.

Is that true of the second set in this problem? What about the endpoints of the chord?

3. In a circle, any single point will disconnect the set right... Why specify 2?

I don't understand how to find the n-points?

For example, I know $\displaystyle s^1$, the unit circle, every point is a 1-point and [0,1)
is not homeomorphic to $\displaystyle s^1$, since [0,1) has some 2-points whereas $\displaystyle S^1$ doesn't....can anyone explain why!?

4. [QUOTE=bigdoggy;396352]In a circle, any single point will disconnect the set right... Why specify 2?[QUOTE]
How in the world could a single point's removal disconnect a circle?
How much do you know about simple closed curves and about arcs?

5. How in the world could a single point's removal disconnect a circle?
I was told that every point of a circle is a 1-point which I believed implied by removing a single point would disconnect it...?

I'm trying to understand how to find the n-points in general so I can then decide if two sets are homeomorphic....

6. Either you are being totally false and incompetent instruction or you do not understand.
The whole area of ‘cut-points’ as applied to arcs and simple closed curves is such a rich area of study.
It is a pity that you seem to be so confused by this topic.

7. Originally Posted by Plato
Either you are being totally false and incompetent instruction or you do not understand.
The whole area of ‘cut-points’ as applied to arcs and simple closed curves is such a rich area of study.
It is a pity that you seem to be so confused by this topic.
I don't see the value in that comment Plato.
Surely it is apparent that I'm failing to 'connect the dots' here, hence I was hoping for some pointers!!

8. I think from your definition of cut points a "1-point" means you are left with 1 connected path component. So of course, on the unit circle all points are 1-points.
To show 2 sets X and Y aren't homeomorphic with cut points, either show that X (or Y) contains an n-point that isn't in Y (or X), or show that X contains more n-points than Y (for a specified n)

9. Originally Posted by bigdoggy
I was told that every point of a circle is a 1-point which I believed implied by removing a single point would disconnect it...?

I'm trying to understand how to find the n-points in general so I can then decide if two sets are homeomorphic....
removing one point from a circle allows you to "open it up" to be a line segment but that is still connected. Removing another point from that line segment makes it unconnected.

What, exactly, is the definition of "n-points"?

10. Originally Posted by HallsofIvy
removing one point from a circle allows you to "open it up" to be a line segment but that is still connected. Removing another point from that line segment makes it unconnected.

What, exactly, is the definition of "n-points"?
Say the space X is path-connected, a point $\displaystyle x \ \varepsilon X$ is called an n-point if $\displaystyle X-\{x\}$ has n path-compnents...

11. Originally Posted by bigdoggy
Say the space X is path-connected, a point $\displaystyle x \ \varepsilon X$ is called an n-point if $\displaystyle X-\{x\}$ has n path-components...
That is a completely understandable definition.
No one could argue with its clarity.

BUT, under that definition a 1-point is not a cut-point.
Authors are not free to co-op definitions.
That is the basis of the arguments in this thread.

The idea of a ‘cut-point’ goes back as far as a 1909 paper by R.L. Moore, the founder of point set topology.
Again a “1-point is not a cut-point.”