# Find homeomorphic and non-homeomorphic pairs among the spaces $A, B$ and $C$.

Problem:

We define three topological spaces as follows: $A - \mathbb{R}^2/\tilde{}$ is the identification space by identifying the subset $(\mathbb{Z} \times \mathbb{R}) \cup (\mathbb{R} \times \mathbb{Z})$ as a single point and each other point of $\mathbb{R}^2$ as itself.

$B \subset \mathbb{R}^3$ is the union of spheres of radius $1/n$ tangential to the $xy$-plane at the origin (a spherical earring), and $C \subset \mathbb{R}^3$ is the union of spheres of radius $n$ tangential to the $xy$-plane at the origin.

Find homeomorphic and non-homeomorphic pairs among the spaces $A, B$ and $C$.

Solution:

$A$ and $C$ are homeomorphic, $A$ and $B$ are not homeomorphic and $B$ and $C$ are not homeomorphic.

Non homeomorphic pairs:

$B$ is a compact set but $A$ and $C$ are non-compact.

Homeomorphic pairs:

Let $\mathbb{R}^2 \rightarrow$ $A$ be the identification map.

The plane $\mathbb{R}^2$ is a union of unit squares, enumerated by $U_n$.

First we identify each square with a sphere $S_n$ with its boundary being glued together as a marked point $p_n$. Denote $q$ the distinguished point in $A$ obtained by identifying the integral lines.

We define a map $f$ from $A$ to $C$ by sending $p_n$ to the origin and sending the spheres $S_n$ to the sphere of radius $n$. $f$ is bijective and a homeomorphic mapping.

You find any flaws in this solution?

• @BrianM.Scott Yes, Brian. Fixed. May 7, 2016 at 12:21

The proof that $B$ is not homeomorphic to $A$ or $C$ is fine. However, $A$ has a strictly finer topology than $C$, so $A$ and $C$ cannot be homeomorphic. Specifically, $C$ is clearly metrizable and hence first countable, but $A$ does not have a countable local base at the ‘fancy’ point, which I’ll call $p$.

To see this, let $\mathscr{U}=\{U_n:n\in\Bbb N\}$ be a countable family of nbhds of $p$. Without loss of generality we may suppose that each $U_n$ has the form

$$U_n=\bigcup_{\langle k,\ell\rangle\in\Bbb Z\times\Bbb Z}q\left[B\big(\langle k,\ell\rangle,\epsilon(\langle k,\ell\rangle,n)\big)\right]\;,$$

where $q$ is the quotient map,

$$B(\langle k,\ell\rangle,\epsilon)=(k-\epsilon,k+\epsilon)\times(\ell-\epsilon,\ell+\epsilon)$$

for each $\langle k,\ell\rangle\in\Bbb Z\times\Bbb Z$ and $\epsilon>0$, and each $\epsilon(\langle k,\ell\rangle,n)<\frac12$.

Now enumerate $\Bbb Z\times\Bbb Z=\{p_n:n\in\Bbb N\}$, let

$$U=\bigcup_{n\in\Bbb N}q\left[B\left(p_n,\frac12\epsilon(p_n,n)\right)\right]\;,$$

and check that $U_n\nsubseteq U$ for each $n\in\Bbb N$. This shows that $\mathscr{U}$ is not a local base at $p$ and hence that $A$ is not first countable at $p$.