Finding an example of nonhomeomorphic closed connected sets

Solution 1:

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 $A\not\approx B$.