8
$\begingroup$

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 $\mathbb{R}^{2}$.

$\endgroup$
6
  • 4
    $\begingroup$ "The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't. $\endgroup$ Jan 3, 2019 at 2:35
  • 1
    $\begingroup$ @GerryMyerson yeah! My mistake. Thank you. $\endgroup$
    – Lucas
    Jan 3, 2019 at 2:37
  • 3
    $\begingroup$ The strokes have width $0$ then? Otherwise $9$ and $0$ are homeomorphic. $\endgroup$
    – Carsten S
    Jan 3, 2019 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 $f:8\to 0$ 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. $\endgroup$ Jan 3, 2019 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 $b\in B$ with $c\in b$, 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 $b^*\in B^*$. $\endgroup$ Jan 3, 2019 at 11:24

1 Answer 1

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 $f:X\to Y$ is homeomorphism and $p$ cutpoint of $X$, then $f(p)$ is cutpoint of $Y$. Also show an arc is not homeomorphic to a circle.

$\endgroup$
3
  • $\begingroup$ Nice! Thanks for the hint! $\endgroup$
    – Lucas
    Jan 3, 2019 at 2:37
  • 2
    $\begingroup$ I have made an edit to your post. Feel free to revert to its original version if you disagree with my edit. $\endgroup$
    – user170039
    Jan 3, 2019 at 4:30
  • 1
    $\begingroup$ @user170039, Pointless. $\endgroup$ Jan 3, 2019 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.