Are 0,8 and 9 homeomorphic topological space?
Consider the topological spaces "0","8" and "9" in $mathbb{R}^{2}$. Are they homeomorphic?
I have an approach that doesnt look very rigorous to me. I wanted to know how to formalize this if its correct.
0 and 8 are not homeomorphic since excluding one point of 0 the space is still connected, but excluding the "tangent point" of 8, we have a disconnected space.
Same idea for 8 and 9.
The space 9 is union of one circle and one arc. The arc is homeomorphic to the circle, so we can view 9 as a union of two circles, then 8 and 9 are homeomorphic
PS: the topology of the spaces is induced by topology of $mathbb{R}^{2}$.
general-topology metric-spaces
add a comment |
Consider the topological spaces "0","8" and "9" in $mathbb{R}^{2}$. Are they homeomorphic?
I have an approach that doesnt look very rigorous to me. I wanted to know how to formalize this if its correct.
0 and 8 are not homeomorphic since excluding one point of 0 the space is still connected, but excluding the "tangent point" of 8, we have a disconnected space.
Same idea for 8 and 9.
The space 9 is union of one circle and one arc. The arc is homeomorphic to the circle, so we can view 9 as a union of two circles, then 8 and 9 are homeomorphic
PS: the topology of the spaces is induced by topology of $mathbb{R}^{2}$.
general-topology metric-spaces
2
"The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
– Gerry Myerson
46 mins ago
1
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
44 mins ago
add a comment |
Consider the topological spaces "0","8" and "9" in $mathbb{R}^{2}$. Are they homeomorphic?
I have an approach that doesnt look very rigorous to me. I wanted to know how to formalize this if its correct.
0 and 8 are not homeomorphic since excluding one point of 0 the space is still connected, but excluding the "tangent point" of 8, we have a disconnected space.
Same idea for 8 and 9.
The space 9 is union of one circle and one arc. The arc is homeomorphic to the circle, so we can view 9 as a union of two circles, then 8 and 9 are homeomorphic
PS: the topology of the spaces is induced by topology of $mathbb{R}^{2}$.
general-topology metric-spaces
Consider the topological spaces "0","8" and "9" in $mathbb{R}^{2}$. Are they homeomorphic?
I have an approach that doesnt look very rigorous to me. I wanted to know how to formalize this if its correct.
0 and 8 are not homeomorphic since excluding one point of 0 the space is still connected, but excluding the "tangent point" of 8, we have a disconnected space.
Same idea for 8 and 9.
The space 9 is union of one circle and one arc. The arc is homeomorphic to the circle, so we can view 9 as a union of two circles, then 8 and 9 are homeomorphic
PS: the topology of the spaces is induced by topology of $mathbb{R}^{2}$.
general-topology metric-spaces
general-topology metric-spaces
asked 1 hour ago
Lucas Corrêa
1,4661321
1,4661321
2
"The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
– Gerry Myerson
46 mins ago
1
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
44 mins ago
add a comment |
2
"The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
– Gerry Myerson
46 mins ago
1
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
44 mins ago
2
2
"The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
– Gerry Myerson
46 mins ago
"The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
– Gerry Myerson
46 mins ago
1
1
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
44 mins ago
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
44 mins ago
add a comment |
1 Answer
1
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0 has no cut points.
8 has exactly one cut point.
9 has infinitely many cutpoints.
To show there are no homeomorphisms
among 0,8,9 use the exercise.
Exercise. Prove if f:X -> Y is homeomorphism and
p cutpoint of X, then f(p) is cutpoint of Y.
Also show an arc is not homeomorphic to a circle.
Nice! Thanks for the hint!
– Lucas Corrêa
43 mins ago
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
0 has no cut points.
8 has exactly one cut point.
9 has infinitely many cutpoints.
To show there are no homeomorphisms
among 0,8,9 use the exercise.
Exercise. Prove if f:X -> Y is homeomorphism and
p cutpoint of X, then f(p) is cutpoint of Y.
Also show an arc is not homeomorphic to a circle.
Nice! Thanks for the hint!
– Lucas Corrêa
43 mins ago
add a comment |
0 has no cut points.
8 has exactly one cut point.
9 has infinitely many cutpoints.
To show there are no homeomorphisms
among 0,8,9 use the exercise.
Exercise. Prove if f:X -> Y is homeomorphism and
p cutpoint of X, then f(p) is cutpoint of Y.
Also show an arc is not homeomorphic to a circle.
Nice! Thanks for the hint!
– Lucas Corrêa
43 mins ago
add a comment |
0 has no cut points.
8 has exactly one cut point.
9 has infinitely many cutpoints.
To show there are no homeomorphisms
among 0,8,9 use the exercise.
Exercise. Prove if f:X -> Y is homeomorphism and
p cutpoint of X, then f(p) is cutpoint of Y.
Also show an arc is not homeomorphic to a circle.
0 has no cut points.
8 has exactly one cut point.
9 has infinitely many cutpoints.
To show there are no homeomorphisms
among 0,8,9 use the exercise.
Exercise. Prove if f:X -> Y is homeomorphism and
p cutpoint of X, then f(p) is cutpoint of Y.
Also show an arc is not homeomorphic to a circle.
answered 46 mins ago
William Elliot
7,2082519
7,2082519
Nice! Thanks for the hint!
– Lucas Corrêa
43 mins ago
add a comment |
Nice! Thanks for the hint!
– Lucas Corrêa
43 mins ago
Nice! Thanks for the hint!
– Lucas Corrêa
43 mins ago
Nice! Thanks for the hint!
– Lucas Corrêa
43 mins ago
add a comment |
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2
"The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
– Gerry Myerson
46 mins ago
1
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
44 mins ago