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

  • 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. $\endgroup$ Commented Jan 3, 2019 at 2:35
  • 1
    $\begingroup$ @GerryMyerson yeah! My mistake. Thank you. $\endgroup$
    – Lucas
    Commented Jan 3, 2019 at 2:37
  • 3
    $\begingroup$ The strokes have width $0$ then? Otherwise $9$ and $0$ are homeomorphic. $\endgroup$
    – Carsten S
    Commented 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. $\endgroup$ Commented 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 $b^*\in B^*$. $\endgroup$ Commented Jan 3, 2019 at 11:24

1 Answer 1


$0$ 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.

  • $\begingroup$ Nice! Thanks for the hint! $\endgroup$
    – Lucas
    Commented 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
    Commented Jan 3, 2019 at 4:30
  • 1
    $\begingroup$ @user170039, Pointless. $\endgroup$ Commented Jan 3, 2019 at 6:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .