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

• "The arc is homeomorphic to the circle." No, it isn't. An arc is disconnected by removing a point; a circle isn't. Jan 3, 2019 at 2:35
• @GerryMyerson yeah! My mistake. Thank you. Jan 3, 2019 at 2:37
• The strokes have width $0$ then? Otherwise $9$ and $0$ are homeomorphic. Jan 3, 2019 at 10:15
• 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. Jan 3, 2019 at 11:13
• 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 $b^*\in B^*$. Jan 3, 2019 at 11:24

$$0$$ 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.

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