homeomorphism

[Anonymous No.8761076]

anyone cares to explain why these two are not topologically equivalent? The second figure does not have the 'tip' included.
they both have the infinite number of 2-points (the vertical line) and no 3-points.

[Anonymous No.8761079]

It seems that if you shrink the vertical line down to a point, then the figure on the right will have a hole but the one on the left won't. I can't remember if this is homeomorphism or some homotopy condition.

[Anonymous No.8761188]

>>8761076
Is this the tip just one point? Like the left one is a circle with a half-open interval attachhed and the second one is with a closed interval attached?

[Anonymous No.8761196]

>>8761076
If the tip is like a representation of a closed interval like >>8761188 says then the one in the left isn't compact but the ritght one is

[Anonymous No.8761352]

>>8761196
yeah, that was what I was gonna say and it is the right answer and it is trivial and OP should have thought of that. No offense, it's just sometimes I think about using big fancy theorems while there's an easy solution right in front of me.

[Anonymous No.8761534]

The sets have the subset topology of [math]\mathbb{R}^2[/math]. If two set are homeomorphic and one is compact, then the other is compact as well. As already noted, one of the two sets is not compact, while the other is.
1. One is closed (prove this) and limited hence compact (Heine-Borel).
2. The other is not compact: use the same reasoning you use for [math]\{1/n\}[/math].

[Anonymous No.8761550]

>>8761534
what is the quickest way of proving that the left one is closed ?

[Anonymous No.8761566]

>>8761550
Prove that the circle is closed (easy, since preimage of a closed point under a continuous map) and that an interval embedded in R^2 is closed. Then the left one is a union of two closed sets and hence closed.

[Anonymous No.8761613]

>why is an open interval not homeomorphic to a closed interval

rly mak eu think

[Anonymous No.8761678]

>>8761550
>what is the quickest way of proving that the left one is closed ?
Because it's equal to it's closure (show this*), which is an intersection of closed sets, hence closed.
Hope I'm not mistaken.
* See the definition of closure and use circles as open sets.
In the same manner you can check whether the other set is open (using the interior instead).

[Anonymous No.8761680]

>>8761678
>it's
its.

[Anonymous No.8761714]

>>8761613
>>why is an open interval not homeomorphic to a closed interval
Let [math]X[/math] and [math]A,B\subset X[/math] with the subset topology.
[math]A[/math] is open and [math]B[/math] is closed.
[If they are bounded, [math]A[/math] cannot be compact (at least in this case for Heine-Borel) thus the sets are not homeomorphic.]
What is the general proof?

[Anonymous No.8761774]

>>8761714
dude what ?

[Anonymous No.8761783]

>>8761774
>dude what ?
Did I make a mistake or the post is not clear?

[Anonymous No.8761799]

>>8761783
>>8761714
>what is the general proof?
general proof of what please?

[Anonymous No.8761804]

>>8761799
>general proof of what please?
Is this
>>8761714
>Let X and A,B⊂X with the subset topology.
>A is open and B is closed.
>A and B cannot be homeomorphic.
true?

[Anonymous No.8761810]

>>8761804
consider A=B=X

[Anonymous No.8761817]

>>8761804
It's definitely not true: let A = B = {empty set}. I think it might be true if you insist that X is connected and A and B are proper nonempty subsets, but I can't think of a proof or counterexample at the moment.

[Anonymous No.8761828]

>>8761810
>>8761817
Thank you.
So, in this case, are there simpler alternatives to:
1. showing that one of the sets is open, one is closed and invoking Heine-Borel,
2. showing by hand that one is compact and the other is not?

[Anonymous No.8761840]

>>8761828
In this particular case, you may notice that removing any point from an open interval makes it disconnected, whereas in a closed interval, you can remove the boundary points and it will stay connected.

[Anonymous No.8761858]

no, but it's simple enough. it really boils down to showing that the left subset is not closed while the second one is. the spaces are so nicely embedded that this can be proven straight from the definition using euclidean or equivalent metric, you won't even get your hands dirty.

[Anonymous No.8761863]

>>8761840
Go it, but one last question.
When can subsets of [math]\mathbb{R}^{2}[/math] be called intervals?
Something like being homeomorphic to intervals in [math]\mathbb{R}[/math]?

[Anonymous No.8761865]

>>8761858
was reacting to >>8761828

[Anonymous No.8761875]

>>8761863
well either you consider two-dimensional intervals which are just rectangles but this is obviously not the case.

the correct term would be "embedded interval" which is an image of an embedding of some interval. this is of course equivalent to being homeomorphic to an interval in the subspace topology.

[Anonymous No.8761876]

>>8761863
I don't think it's really common terminology. That being said, you can define a segment in the plane as the convex hull of two points, ie. [math][A,B] = \{tA + (1-t)B, t \in [0,1]\}[/math]

[Anonymous No.8761886]

>>8761876
I'm asking because you said
>>8761840
>that removing any point from an open interval...
and I was wondering by what definition are the sets in OP's post considered intervals.
I otherwise completely understand you post referred to the real line.

[Anonymous No.8761938]

>>8761886
Oh, yeah I did mean in the real line. In OP's pic, I think the compactness argument is the best one (which is not to be phrased as "uhh one of them is closed and the other is not" because being "closed" is a relative property, whereas compactness is an intrinsic property)

[Anonymous No.8761943]

>>8761865
>>8761938
Thank you both and thanks to OP.

[Anonymous No.8761947]

compactness makes sense. what about these two? compactness clearly doesn't work in this case

[Anonymous No.8761960]

>>8761947
there should exist an argument using just connectedness but I don't see it right now. anyway, those spaces have different fundamental groups. do you know what that is?

[Anonymous No.8761963]

>>8761947
does the fact that one set of 2-points is closed and the other is not? i.e. we can't have a homeomorphism between those sets, and if we had a homeomorphism between those two figures and reduce it to 2-points, it'd still have to be homeomorphism

[Anonymous No.8761971]

>>8761963
ugh, I think I haven't explained myself well here.

My reasoning is as follows:
suppose we have a homeomorphism between the two figures. call it f. we know that if x is a 2-cut point in figure A, f(x) is a 2-cut point in figure B. so by reducing f to set of 2-cut points in figure A, we see that there can't exist a homeomorphism as one of the sets is closed and the other is not

[Anonymous No.8761976]

>>8761960
now I know, that works. thanks!

[Anonymous No.8761981]

got something. on the bottom one, there is a point with the following property: every connected neighborhood of the point remains connected after the removal of the point. the upper one doesn't have such point.

[Anonymous No.8762110]

Second figure is compact.

[Anonymous No.8762113]

>>8762110
Eh, I couldn't read your drawings. First one is compact.