Fun: Can you construct infinitely many non-homeomorphic connected subsets of the plane.

Question

Can you construct infinitely many non-homeomorphic connected subsets of the plane.

I think this is possible, would:

$I_N = \bigcup_{n=1}^{N} I_n$ for $N \in \mathbb{N}$

where

$$I_n = \{ (x,y) \in \mathbb{R}^2 : (x-\frac{1}{n})^2+y^2= \frac{1}{n}\}$$

work?

Answer 1

What you wrote is almost fine, except it should be $\frac{1}{n^2}$ (square of the radius of the circle) and you should also prove that the sets $I_N$ are pairwise nonhomeomorphic (this is not hard). More interestingly, there are
$$2^{2^{\aleph_0}}= 2^{{\mathfrak c}}$$ of pairwise nonhomeomorphic subsets of the real line. This is not easy, if you are interested, I can explain why.

Edit. Let me prove that there are $2^{{\mathfrak c}}$ of pairwise nonhomeomorphic subsets of the real line. First of all, clearly, there are $2^{{\mathfrak c}}$ subsets of the real line. Now, there is an interesting theorem of M.A.Lavrentiev (mostly famous for his contributions to applied mathematics):

Theorem. For any pair of completely metrizable spaces $X, Y$ (I will be using $X=Y={\mathbb R}$), for any two subsets $A\subset X, B\subset Y$ and any homeomorphism $h: A\to B$, there exist $G_\delta$-subsets $A', B'\subset {\mathbb R}$ such that $A\subset A', B\subset B'$ and $h$ extends to a homeomorphism $$ h': A'\to B'. $$ For a proof see for instance Theorem 4.3.21 in, R. Engelking, "General topology", Heldermann Verlag, Berlin, 1989.

Since there are exactly ${\mathfrak c}$ distinct $G_\delta$-subsets of ${\mathbb R}$ (every such subset is determined by a countable sequence of real numbers) and there is at most continuum of homeomorphisms between such subsets, it follows that each subset $A\subset {\mathbb R}$ is homeomorphic to at most ${\mathfrak c}$ distinct subsets of ${\mathbb R}$. Hence, it follows that ${\mathbb R}$ contains exactly $2^{{\mathfrak c}}$ pairwise nonhomeomorphic subsets.

Answer 2

One way: there is, for every countable ordinal $\alpha < \omega_!$ an embedding $i_\alpha: \alpha \to \mathbb{R} \times \{0\}$ and we can take the cone over $i_\alpha[\alpha]$ as one of the (path-)connected sets.

Answer 3

Another way: you can just connect the origin to $n$ distinct points on a circle around the it by a line segment.

As a side note, much stronger results are true: there are continuum-many pairwise non-homeomorphic planar continua (compact connected subsets of a plane), actually closures of piecewise linear graphs $[0, 1) → [0, 1]$ can be taken.