Are spaces shaped like the digits 0, 8 and 9 homeomorphic topological spaces?












7












$begingroup$



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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    "The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
    $endgroup$
    – Gerry Myerson
    Jan 3 at 2:35






  • 1




    $begingroup$
    @GerryMyerson yeah! My mistake. Thank you.
    $endgroup$
    – Lucas Corrêa
    Jan 3 at 2:37








  • 3




    $begingroup$
    The strokes have width $0$ then? Otherwise $9$ and $0$ are homeomorphic.
    $endgroup$
    – Carsten S
    Jan 3 at 10:15






  • 1




    $begingroup$
    Remove the point in $9$ where the circle meets the arc and it becomes disconnected But $8$ and $0$ are still connected if you remove any one point from either of them... Remove the center-point and any other point of $8$ and it is still connected, but if you remove any 2 points from $0$ it is disconnected..... This $is$ a valid approach. E.g. suppose $f:8to 0$ was a homeomorphism. Let $c$ be the center-point of $8$ and let $d$ be another point of $8$. Then the image $f(8$ ${c,d})=0$ ${f(c),f(d)}$ is connected, which is absurd.
    $endgroup$
    – DanielWainfleet
    Jan 3 at 11:13








  • 1




    $begingroup$
    Another approach is to look at boundaries of members of bases for the spaces. If $B$ is a base (basis) for $8$ and $c$ is the center-point of $8$ then there exists $bin B$ with $cin b$, such that $partial b$ has at least 4 members. But $0$ has a base $B^*$ such that $partial b^*$ has exactly 2 members for each $b^*in B^*$.
    $endgroup$
    – DanielWainfleet
    Jan 3 at 11:24


















7












$begingroup$



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







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    "The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
    $endgroup$
    – Gerry Myerson
    Jan 3 at 2:35






  • 1




    $begingroup$
    @GerryMyerson yeah! My mistake. Thank you.
    $endgroup$
    – Lucas Corrêa
    Jan 3 at 2:37








  • 3




    $begingroup$
    The strokes have width $0$ then? Otherwise $9$ and $0$ are homeomorphic.
    $endgroup$
    – Carsten S
    Jan 3 at 10:15






  • 1




    $begingroup$
    Remove the point in $9$ where the circle meets the arc and it becomes disconnected But $8$ and $0$ are still connected if you remove any one point from either of them... Remove the center-point and any other point of $8$ and it is still connected, but if you remove any 2 points from $0$ it is disconnected..... This $is$ a valid approach. E.g. suppose $f:8to 0$ was a homeomorphism. Let $c$ be the center-point of $8$ and let $d$ be another point of $8$. Then the image $f(8$ ${c,d})=0$ ${f(c),f(d)}$ is connected, which is absurd.
    $endgroup$
    – DanielWainfleet
    Jan 3 at 11:13








  • 1




    $begingroup$
    Another approach is to look at boundaries of members of bases for the spaces. If $B$ is a base (basis) for $8$ and $c$ is the center-point of $8$ then there exists $bin B$ with $cin b$, such that $partial b$ has at least 4 members. But $0$ has a base $B^*$ such that $partial b^*$ has exactly 2 members for each $b^*in B^*$.
    $endgroup$
    – DanielWainfleet
    Jan 3 at 11:24
















7












7








7





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 4:31









Tanner Swett

4,2581639




4,2581639










asked Jan 3 at 1:39









Lucas CorrêaLucas Corrêa

1,6321321




1,6321321








  • 3




    $begingroup$
    "The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
    $endgroup$
    – Gerry Myerson
    Jan 3 at 2:35






  • 1




    $begingroup$
    @GerryMyerson yeah! My mistake. Thank you.
    $endgroup$
    – Lucas Corrêa
    Jan 3 at 2:37








  • 3




    $begingroup$
    The strokes have width $0$ then? Otherwise $9$ and $0$ are homeomorphic.
    $endgroup$
    – Carsten S
    Jan 3 at 10:15






  • 1




    $begingroup$
    Remove the point in $9$ where the circle meets the arc and it becomes disconnected But $8$ and $0$ are still connected if you remove any one point from either of them... Remove the center-point and any other point of $8$ and it is still connected, but if you remove any 2 points from $0$ it is disconnected..... This $is$ a valid approach. E.g. suppose $f:8to 0$ was a homeomorphism. Let $c$ be the center-point of $8$ and let $d$ be another point of $8$. Then the image $f(8$ ${c,d})=0$ ${f(c),f(d)}$ is connected, which is absurd.
    $endgroup$
    – DanielWainfleet
    Jan 3 at 11:13








  • 1




    $begingroup$
    Another approach is to look at boundaries of members of bases for the spaces. If $B$ is a base (basis) for $8$ and $c$ is the center-point of $8$ then there exists $bin B$ with $cin b$, such that $partial b$ has at least 4 members. But $0$ has a base $B^*$ such that $partial b^*$ has exactly 2 members for each $b^*in B^*$.
    $endgroup$
    – DanielWainfleet
    Jan 3 at 11:24
















  • 3




    $begingroup$
    "The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
    $endgroup$
    – Gerry Myerson
    Jan 3 at 2:35






  • 1




    $begingroup$
    @GerryMyerson yeah! My mistake. Thank you.
    $endgroup$
    – Lucas Corrêa
    Jan 3 at 2:37








  • 3




    $begingroup$
    The strokes have width $0$ then? Otherwise $9$ and $0$ are homeomorphic.
    $endgroup$
    – Carsten S
    Jan 3 at 10:15






  • 1




    $begingroup$
    Remove the point in $9$ where the circle meets the arc and it becomes disconnected But $8$ and $0$ are still connected if you remove any one point from either of them... Remove the center-point and any other point of $8$ and it is still connected, but if you remove any 2 points from $0$ it is disconnected..... This $is$ a valid approach. E.g. suppose $f:8to 0$ was a homeomorphism. Let $c$ be the center-point of $8$ and let $d$ be another point of $8$. Then the image $f(8$ ${c,d})=0$ ${f(c),f(d)}$ is connected, which is absurd.
    $endgroup$
    – DanielWainfleet
    Jan 3 at 11:13








  • 1




    $begingroup$
    Another approach is to look at boundaries of members of bases for the spaces. If $B$ is a base (basis) for $8$ and $c$ is the center-point of $8$ then there exists $bin B$ with $cin b$, such that $partial b$ has at least 4 members. But $0$ has a base $B^*$ such that $partial b^*$ has exactly 2 members for each $b^*in B^*$.
    $endgroup$
    – DanielWainfleet
    Jan 3 at 11:24










3




3




$begingroup$
"The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
$endgroup$
– Gerry Myerson
Jan 3 at 2:35




$begingroup$
"The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't.
$endgroup$
– Gerry Myerson
Jan 3 at 2:35




1




1




$begingroup$
@GerryMyerson yeah! My mistake. Thank you.
$endgroup$
– Lucas Corrêa
Jan 3 at 2:37






$begingroup$
@GerryMyerson yeah! My mistake. Thank you.
$endgroup$
– Lucas Corrêa
Jan 3 at 2:37






3




3




$begingroup$
The strokes have width $0$ then? Otherwise $9$ and $0$ are homeomorphic.
$endgroup$
– Carsten S
Jan 3 at 10:15




$begingroup$
The strokes have width $0$ then? Otherwise $9$ and $0$ are homeomorphic.
$endgroup$
– Carsten S
Jan 3 at 10:15




1




1




$begingroup$
Remove the point in $9$ where the circle meets the arc and it becomes disconnected But $8$ and $0$ are still connected if you remove any one point from either of them... Remove the center-point and any other point of $8$ and it is still connected, but if you remove any 2 points from $0$ it is disconnected..... This $is$ a valid approach. E.g. suppose $f:8to 0$ was a homeomorphism. Let $c$ be the center-point of $8$ and let $d$ be another point of $8$. Then the image $f(8$ ${c,d})=0$ ${f(c),f(d)}$ is connected, which is absurd.
$endgroup$
– DanielWainfleet
Jan 3 at 11:13






$begingroup$
Remove the point in $9$ where the circle meets the arc and it becomes disconnected But $8$ and $0$ are still connected if you remove any one point from either of them... Remove the center-point and any other point of $8$ and it is still connected, but if you remove any 2 points from $0$ it is disconnected..... This $is$ a valid approach. E.g. suppose $f:8to 0$ was a homeomorphism. Let $c$ be the center-point of $8$ and let $d$ be another point of $8$. Then the image $f(8$ ${c,d})=0$ ${f(c),f(d)}$ is connected, which is absurd.
$endgroup$
– DanielWainfleet
Jan 3 at 11:13






1




1




$begingroup$
Another approach is to look at boundaries of members of bases for the spaces. If $B$ is a base (basis) for $8$ and $c$ is the center-point of $8$ then there exists $bin B$ with $cin b$, such that $partial b$ has at least 4 members. But $0$ has a base $B^*$ such that $partial b^*$ has exactly 2 members for each $b^*in B^*$.
$endgroup$
– DanielWainfleet
Jan 3 at 11:24






$begingroup$
Another approach is to look at boundaries of members of bases for the spaces. If $B$ is a base (basis) for $8$ and $c$ is the center-point of $8$ then there exists $bin B$ with $cin b$, such that $partial b$ has at least 4 members. But $0$ has a base $B^*$ such that $partial b^*$ has exactly 2 members for each $b^*in B^*$.
$endgroup$
– DanielWainfleet
Jan 3 at 11:24












1 Answer
1






active

oldest

votes


















15












$begingroup$

$0$ 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.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice! Thanks for the hint!
    $endgroup$
    – Lucas Corrêa
    Jan 3 at 2:37










  • $begingroup$
    I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
    $endgroup$
    – user 170039
    Jan 3 at 4:30






  • 1




    $begingroup$
    @user170039, Pointless.
    $endgroup$
    – William Elliot
    Jan 3 at 6:20











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









15












$begingroup$

$0$ 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.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice! Thanks for the hint!
    $endgroup$
    – Lucas Corrêa
    Jan 3 at 2:37










  • $begingroup$
    I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
    $endgroup$
    – user 170039
    Jan 3 at 4:30






  • 1




    $begingroup$
    @user170039, Pointless.
    $endgroup$
    – William Elliot
    Jan 3 at 6:20
















15












$begingroup$

$0$ 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.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice! Thanks for the hint!
    $endgroup$
    – Lucas Corrêa
    Jan 3 at 2:37










  • $begingroup$
    I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
    $endgroup$
    – user 170039
    Jan 3 at 4:30






  • 1




    $begingroup$
    @user170039, Pointless.
    $endgroup$
    – William Elliot
    Jan 3 at 6:20














15












15








15





$begingroup$

$0$ 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.







share|cite|improve this answer











$endgroup$



$0$ 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.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 4:30









user 170039

10.5k42466




10.5k42466










answered Jan 3 at 2:34









William ElliotWilliam Elliot

8,5922720




8,5922720












  • $begingroup$
    Nice! Thanks for the hint!
    $endgroup$
    – Lucas Corrêa
    Jan 3 at 2:37










  • $begingroup$
    I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
    $endgroup$
    – user 170039
    Jan 3 at 4:30






  • 1




    $begingroup$
    @user170039, Pointless.
    $endgroup$
    – William Elliot
    Jan 3 at 6:20


















  • $begingroup$
    Nice! Thanks for the hint!
    $endgroup$
    – Lucas Corrêa
    Jan 3 at 2:37










  • $begingroup$
    I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
    $endgroup$
    – user 170039
    Jan 3 at 4:30






  • 1




    $begingroup$
    @user170039, Pointless.
    $endgroup$
    – William Elliot
    Jan 3 at 6:20
















$begingroup$
Nice! Thanks for the hint!
$endgroup$
– Lucas Corrêa
Jan 3 at 2:37




$begingroup$
Nice! Thanks for the hint!
$endgroup$
– Lucas Corrêa
Jan 3 at 2:37












$begingroup$
I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
$endgroup$
– user 170039
Jan 3 at 4:30




$begingroup$
I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit.
$endgroup$
– user 170039
Jan 3 at 4:30




1




1




$begingroup$
@user170039, Pointless.
$endgroup$
– William Elliot
Jan 3 at 6:20




$begingroup$
@user170039, Pointless.
$endgroup$
– William Elliot
Jan 3 at 6:20


















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