# Finding an example of nonhomeomorphic closed connected sets

Question: Find two closed, connected subsets in $\mathbb{R}^2$, $A$ and $B$, such that $A$ is not homeomorphic to $B$, but there is a continuous bijection $f:A \rightarrow B$ and a continuous bijection $g:B \rightarrow A$.

This is a homework question, so please only very small hints. I realize that both $A$ and $B$ must not be compact. Since they both must be closed, then they must be unbounded. However, I am having a hard time getting started on this. It is very easy to find two closed, unbounded, connected subsets of the plane that are not homeomorphic to each other, but it is hard to find the continuous bijections required.

I know the classic example of a continuous bijection with a discontinuous inverse is a map $f: [0,2\pi) \rightarrow \mathbb{S}^1$ given by $f(x) = (\cos x, \sin x)$. I am trying to use this map as a template to come up with my sets but I am having no success.

• This question has been asked before very recently. Must be the same assignment. – Matt Samuel Jan 24 '16 at 2:06
• Could you please link me to it? I tried searching, but I couldn't find it. – Jonathan Gafar Jan 24 '16 at 2:15
• math.stackexchange.com/questions/1623080/… – Silvia Ghinassi Jan 24 '16 at 3:45
• Thank you. I have been thinking about the problem for the last couple of hours. I think I am close. I will post here once I am finished. – Jonathan Gafar Jan 24 '16 at 4:24
• @JonathanGafar How about you talk with eachother and solve it together? That's how math works. – Paul K Jul 30 '16 at 19:06

## 2 Answers

For $T\subseteq \Bbb Z$ let $$C(T)=\bigl(\Bbb R\times\{0\}\bigr)\cup\bigcup_{n\in T}\bigl(\tfrac13S^1+(n,\tfrac13)\bigr) \cup \bigcup_{n\in\Bbb Z\setminus T}\bigl(\{n\}\times [0,\infty)\bigr),$$ i.e., $C(T)$ is the real line with either a small copy of $S^1$ or a ray attached to each integer point, depending on whether that integer is in $T$ or not. Note that $C(T)$ is path-connected and closed.

Lemma 1. We have $C(T)\approx C(T')$ iff $T'=\pm T+k$ for some $k\in\Bbb Z$.

Proof. Clearly, if $T'=\pm T+k$ for some $k\in\Bbb Z$, then $(x,y)\mapsto (\pm x+k,y)$ is a homeomorphism $C(T)\to C(T')$.

On the other hand, for a point $p\in C(T)$, we can detect what "type" of point it is:

• "glue-points": $C(T)\setminus\{p\}$ has three connected components iff $p\in\Bbb Z\times\{0\}$
• "circle-points": $C(T)\setminus\{p\}$ is connected iff $p$ is on one of the attached circles (but not the glueing point)
• "ray-poins": $C(T)\setminus \{p\}$ has two components and one of them contains no glue-points (or equivalently: is homeomorphic to $\Bbb R$) iff $p$ is on one of the attached rays (but not the glueing point)
• "backbone-points": $C(T)\setminus \{p\}$ has two components and both contain (infinitely many) glue-points iff $p\in(\Bbb R\setminus\Bbb Z)\times\{0\}$

The case of glue-points can be split up further:

• "$T$-points": $p$ is a glue-point in the closure of the set of circle-points
• "$T^c$-points": $p$ is a glue-point in the closure of the set of ray-points

Also, we can recover "betweenness" among glue-points: If $p_1,p_2,p_3$ are glue-points, then $p_2$ is between $p_1$ and $p_3$ iff $p_1$ and $p_3$ are in different components of $C(T)\setminus\{p_2\}$.

So assume $f\colon C(T)\to C(T')$ is a homeomorphism. By the above classification of points, this induces a bijection of $\Bbb Z\times\{0\}$ with itself - and via identification: of $\Bbb Z$ with itself. As this bijection respects betweenness, it must be of the form $n\mapsto \pm n+k$ with $k\in\Bbb Z$. Additionally, this bijection must respect "$T$-ness", i.e., it must induce a bijection $T\to T'$. This makes $T'=\pm T+k$. $\square$

Lemma 2. Assume $T\subseteq T'\subseteq \Bbb Z$. Then there exists a continuous bijection $C(T)\to C(T')$.

Proof. We can define $f\colon C(T)\to C(T')$ by letting it be the identity, except on rays $\{n\}\times [0,\infty)$ with $n\in T'\setminus T$, where we "wrap" the ray per $$(n,y)\mapsto \bigl(n+\tfrac 13 \sin h(y),\tfrac13-\tfrac13\cos h(y)\bigr),$$ where $h$ is a homeomorphism $[0,\infty)\to [0,2\pi)$, for example $h(t)=\frac{2\pi t}{1+t}$. $\square$

Now we can solve the original problem: Let $$A = C(\Bbb N_0), \quad B=C(\Bbb N_0\cup\{-2\}), \quad B'=C(\Bbb N_0\setminus\{1\}).$$ By lemma 2, we have continuous bijections $A\to B$ and $B'\to A$. By lemma 1, $B\approx B'$ and $A\not\approx B$. • Beautiful solution! – Cheerful Parsnip Aug 5 '16 at 17:39

This is not a complete answer, but I would like to share my picture here. Here I'm trying to draw two sets which are connected and there exists bijective continuous mapping from each other, but not homeomorphic. Here I'm drawing these two sets $A$ and $B$ ( the red line here means I'm excluding the $y-axis$ except the origin).

My first claim is that they are non-homeomorphic.

If they are homeomorphic , then image of a circle should be a circle. So the increasing sequence (of diameters) circles in $B$ should maps to a circles in the Hawiian ring of $A$, but then we can take an inifite descrete set out of those increasing circles whose image should have to be discrete, but which is not possible, so contradiction.

Now we can define a continuous bijective map from $A$ to $B$ in following sense... map the Hawiian earring to itself by a homeomorphism, and then map alternative lines bijectively onto a copy of circles and rest of the lines maps bijectively onto lines. So this will give a continuous bijection from $A$ to $B$.

To get a continuous bijection from $B$ to $A$, we do followings...maps infinity lines onto itself homeomorphically, and maps big circles homeomorphically onto the alternative shrinking circles, and maps the circles of Hawiian earrings homeomorphically onto those other alternative circles.

• This yould be closed if you did add the $y$ axis - but that would soil the $A\to B$ map. Do you think there is still hop to mend this (i.e., actually get a closed set example from this? – Hagen von Eitzen Jul 31 '16 at 20:27
• @HagenvonEitzen I actually thought about this, but still I couldnot manage to fix it. Do you have any idea? – Anubhav Mukherjee Aug 1 '16 at 9:48