How do i prove that these 2 path connected subsets are not homeomorphic?

I have the following 2 path connected subets:

(i) O-

(ii) O-O

(Imagine that the line intervals are connected to the circles in each case, i just wasnt able to draw them touching but they are)

and for part (i) the end point of the lin interval IS included in the subset.

How can i prove that subset (i) is NOT homeomorphic to subset (ii) ?

I thought first that this involves studying the cut points and showing that they have different numbers of cut points of each type.

However:
. (i) and (ii) both have infinite cut points of type 1
. they both have infinite cut points of type 2
. they both have zero cut points of type 3,4,...

So they have the same number of cut points of each type which doesnt help me.

Then i thought i must have to study cut pairs.

But:
. (i) and (ii) have infinite cut pairs of type 1
. they both have infinite cut pairs of type 2
. they both have infinite cut pairs of type 3
. then i don't think its possible to find any cut pairs of type 4 in either (i) or (ii).

So they have same number of cut pairs of each type...no use.

Can anyone help show me how they are not homeomorphic?

Thanks

Quote:

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Originally Posted by Siknature

I have the following 2 path connected subets:

(i) O-

(ii) O-O

(Imagine that the line intervals are connected to the circles in each case, i just wasnt able to draw them touching but they are)

and for part (i) the end point of the lin interval IS included in the subset.

How can i prove that subset (i) is NOT homeomorphic to subset (ii) ?

I thought first that this involves studying the cut points and showing that they have different numbers of cut points of each type.

However:
. (i) and (ii) both have infinite cut points of type 1
. they both have infinite cut points of type 2
. they both have zero cut points of type 3,4,...

So they have the same number of cut points of each type which doesnt help me.

Then i thought i must have to study cut pairs.

But:
. (i) and (ii) have infinite cut pairs of type 1
. they both have infinite cut pairs of type 2
. they both have infinite cut pairs of type 3
. then i don't think its possible to find any cut pairs of type 4 in either (i) or (ii).

So they have same number of cut pairs of each type...no use.

Can anyone help show me how they are not homeomorphic?

Thanks

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You can take away two points from set (ii) and the remaining set is path-connected, whereas any two points you remove from set (i) will give a set that is not path-connected: can you see this?

Tonio

Quote:

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Originally Posted by tonio

You can take away two points from set (ii) and the remaining set is path-connected, whereas any two points you remove from set (i) will give a set that is not path-connected: can you see this?

Tonio

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Thanks for your help however I think that there is an example that would make what you have said incorrect

I stated that the end point of the line interval for (i) IS included in the subset.

Therefore surely it is possible to remove 2 points in (i), namely the end point of the line interval and some point on the circle that it is connected to and it will still remain path connected.

Is that not true?

Thanks

Quote:

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Originally Posted by Siknature

Thanks for your help however I think that there is an example that would make what you have said incorrect

I stated that the end point of the line interval for (i) IS included in the subset.

Therefore surely it is possible to remove 2 points in (i), namely the end point of the line interval and some point on the circle that it is connected to and it will still remain path connected.

Is that not true?

Thanks

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Indeed. I didn't see that end-point included thing. Let us try the following:
suppose f: (i) --> (ii) is a homeom. between the two given sets, and let I denote the line in (i) with both end-points included (the right one and the left one where it makes contact with the circle) ==> As f is a closed map it must be f(I) is a closed set and, in fact, homeom. to a closed bounded interval (we may think of [0,1]).
The question is: what is f(I) in (ii)? It can be one of the following options:

a) A closed arc in some of the two circles

b) The closed line joining both circles (included both end-points)

c) Part of a closed arc in one of the circles, or in both, and part (or all) the line segment joining both circles


Now, as (i) - I has two CONTRACTIBLE (path-)components, it must be that (ii) - f(I) also has two contractible components, but :

(a) cannot be since we're left with a loop

(b) cannot be since we're left with three components

(c) cannot be because we're left with

A fourth option would be one whole circle and part of the line segment (and or part of a closed arc of the second circle), but this can't be since I is contractible and something containing a whole circle isn't.

Check the above, I think it's a fair chance to work this time.

Tonio

Quote:

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Originally Posted by Siknature

I have the following 2 path connected subets:

(i) O-

(ii) O-O

(Imagine that the line intervals are connected to the circles in each case, i just wasnt able to draw them touching but they are)

and for part (i) the end point of the lin interval IS included in the subset.

How can i prove that subset (i) is NOT homeomorphic to subset (ii) ?

I thought first that this involves studying the cut points and showing that they have different numbers of cut points of each type.

However:
. (i) and (ii) both have infinite cut points of type 1
. they both have infinite cut points of type 2
. they both have zero cut points of type 3,4,...

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Does i) not have 1 cut-point of type 2? Since if the intersection point of the line and circle is removed, we have 2 path components...Whereas if any point is removed from the circle, there is only 1 path component?

Quote:

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Originally Posted by bigdoggy

Does i) not have 1 cut-point of type 2? Since if the intersection point of the line and circle is removed, we have 2 path components...Whereas if any point is removed from the circle, there is only 1 path component?

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But what about anywhere else along the straight line in (i).

You could remove any point along this straight line and it will create 2 path components.

So there are infinite cut points of type 2.

Does shape ii) not only have a finite number of cut points of type 1?

namely the points above and below where the line touches each circle.

edit: actually using your arguement about cutting anywhere along the line of i) leaves 2 path components, I can see why it is difficult to find the difference.

Quote:

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Originally Posted by disenchanted

Does shape ii) not only have a finite number of cut points of type 1?

namely the points above and below where the line touches each circle.

edit: actually using your arguement about cutting anywhere along the line of i) leaves 2 path components, I can see why it is difficult to find the difference.

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The closed interval [a,b] has exactly 2 cut points of type 1....can't this be used as an analogy for the line between the circles...?

Quote:

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Originally Posted by bigdoggy

The closed interval [a,b] has exactly 2 cut points of type 1....can't this be used as an analogy for the line between the circles...?

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The cut points of type one for a closed interval you have described is the end points.
However, using the end points of the line for shape ii) disconnects either one or both circles, hence making it a type 2 or 3.

Quote:

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Originally Posted by Siknature

Thanks for your help however I think that there is an example that would make what you have said incorrect

I stated that the end point of the line interval for (i) IS included in the subset.

Therefore surely it is possible to remove 2 points in (i), namely the end point of the line interval and some point on the circle that it is connected to and it will still remain path connected.

Is that not true?

Thanks

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Thinking about what you have mentioned here, is it possible to say that you can not take any 3 points out of ii) without disconnecting it, whilst for shape i) you can take 3 of the end points of the line, one after the other and it is still path connected?

edit: actually you could keep taking end points for shape ii) ... posted too early.

Quote:

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Originally Posted by disenchanted

Thinking about what you have mentioned here, is it possible to say that you can not take any 3 points out of ii) without disconnecting it, whilst for shape i) you can take 3 of the end points of the line, one after the other and it is still path connected?

edit: actually you could keep taking end points for shape ii) ... posted too early.

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Thanks for your suggestions anyway. You are giving me new ideas of how to go about answering the question.

Quote:

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Originally Posted by Siknature

Thanks for your suggestions anyway. You are giving me new ideas of how to go about answering the question.

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If you take any point from the circle from the shape O-O, it would be left with something like O-C , are you sure that is still 1 path components and not 2 path components?

To prove that these 2 sets aren't homeomorphic use the fact that homeomorphic sets must have homeomorphic cut point sets. (ie. the set of 1-points must be homeomorphic to the other set of 1-points etc.)

Quote:

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Originally Posted by skamoni

To prove that these 2 sets aren't homeomorphic use the fact that homeomorphic sets must have homeomorphic cut point sets. (ie. the set of 1-points must be homeomorphic to the other set of 1-points etc.)

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Here, do both spaces not have infinitely many 1-points?
i.e anywhere on the circles...

Would anyone like to elaborate on this?
I understand that both i and ii have infinitely many 1-point and 2-point cutpoints. Would you have to look at cut-pairs in this example? Or is there another way of going about it?