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?
 A: 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 $p$.
