Are spaces shaped like the digits 0, 8 and 9 homeomorphic topological spaces?
Consider the topological spaces shaped like the numerals "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 shaped like the numerals "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
3 hours ago
1
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
3 hours ago
add a comment |
Consider the topological spaces shaped like the numerals "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 shaped like the numerals "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
edited 1 hour ago
Tanner Swett
3,9941638
3,9941638
asked 4 hours 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
3 hours ago
1
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
3 hours 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
3 hours ago
1
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
3 hours 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
3 hours 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
3 hours ago
1
1
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
3 hours ago
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
3 hours ago
add a comment |
1 Answer
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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 $f:Xto Y$ 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
3 hours ago
I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
– user 170039
1 hour ago
add a comment |
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1 Answer
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1 Answer
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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 $f:Xto Y$ 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
3 hours ago
I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
– user 170039
1 hour 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 $f:Xto Y$ 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
3 hours ago
I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
– user 170039
1 hour 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 $f:Xto Y$ 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 $f:Xto Y$ is homeomorphism and $p$ cutpoint of $X$, then $f(p)$ is cutpoint of $Y$. Also show an arc is not homeomorphic to a circle.
edited 1 hour ago
user 170039
10.4k42465
10.4k42465
answered 3 hours ago
William Elliot
7,2382519
7,2382519
Nice! Thanks for the hint!
– Lucas Corrêa
3 hours ago
I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
– user 170039
1 hour ago
add a comment |
Nice! Thanks for the hint!
– Lucas Corrêa
3 hours ago
I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
– user 170039
1 hour ago
Nice! Thanks for the hint!
– Lucas Corrêa
3 hours ago
Nice! Thanks for the hint!
– Lucas Corrêa
3 hours ago
I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
– user 170039
1 hour ago
I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
– user 170039
1 hour 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
3 hours ago
1
@GerryMyerson yeah! My mistake. Thank you.
– Lucas Corrêa
3 hours ago