TITLE: Is the plane minus the integer lattice homeomorphic to the plane minus the integers?
QUESTION [16 upvotes]: The question, more rigorously posed, is:

Is $\Bbb R^2-\Bbb Z^2$ homeomorphic to $\Bbb R^2-\Bbb Z\times\{0\}$?

This question has been bugging me in the back of my head for a while now. Sometimes, I think it's clear that they are, and then I think it's clear that they aren't. Nothing from Point-set topology is helping me solve this, at least as far as I see. And the only help from algebraic topology that I'm familiar with is the fundamental group; however, these two spaces seem to have the same fundamental group.
Any help in alleviating my torment is appreciated.

REPLY [3 votes]: Alright, I came up with my own proof.
A couple of lemmas:

Lemma 1: Let $X$ and $Y$ be topological spaces. And let $\mathcal{B}$ be a locally-finite collection of closed sets such that $\bigcup\mathcal{B}=X$. Suppose, in addition, that there is a continuous function $f_B:B\rightarrow Y$ for each $B\in\mathcal{B}$ such that $f_B(x)=f_C(x)$ for all $B\in\mathcal{B}$, all $C\in\mathcal{B}$ such that $C\cap B\neq\varnothing$, and all $x\in B\cap C$. Then there is a unique continuous function $f:X\rightarrow Y$ such that $f|_B=f_B$ for all $B\in\mathcal{B}$. (This is an infinite pasting lemma)
Lemma 2: Let $v\in B_1(0)\subseteq\Bbb R^2$ (this is the open ball). Then there is a homeomorphism $h:\Bbb R^2\rightarrow \Bbb R^2$ such that $h(x)=x$ for all $x\in\Bbb R^2-B_1(0)$ and such that $h(0)=v$.
Lemma 3: Let $U\subseteq \Bbb R^2$ be a connected open set. For any two $u$ and $v\in U$ there is a finite number of points $\{x_1, x_2, \ldots, x_n\}\subseteq U$ and a finite number of strictly positive real numbers $\{\epsilon_1, \epsilon_2, \ldots, \epsilon_n\}$ such that $u=x_1$, and $v=x_n$, and $B_{\epsilon_{i}}(x_i)\cap B_{\epsilon_{i+1}}(x_{i+1})\neq\varnothing$ and $B_{\epsilon_i}(x_i)\subseteq U$ for all $i$.
Lemma 4: Let $U\subseteq \Bbb R^2$ be a connected, open subset. Then $U$ is strongly $n$-homogeneous for all $n\geq 1$.

I leave these nice lemmas for the interested reader. They are fun.
Now choose a series of radii $\{r_n\}_{n=0}^\infty$ such that $B_{r_n}(0)$ contains the square of inradius $n$ (for $n=0$ choose $r_0=1/2$) and such that $\overline{B_{r_n}}(0)$ is disjoint from the square of inradius $n+1$. Finally, consider the collection $\{\overline{B_{r_0}}(0), A_1, A_2, A_3, \ldots\}$ where we have $A_i=\overline{B_{r_i}}(0)-B_{r_{i-1}}(0)$. This collection consists of a closed ball containing the point $(0,0)$ and a sequence of concentric closed annuli which contain one of the squares in question in their interior.
Now we use the above lemmas to create a homeomorphism on each of the annuli (leave the first closed ball alone) to bring the lattice points they contain onto the $x$-axis while still fixing the boundaries of each annulus. We then stitch these together using the first lemma to create a global homeomorphism which pulls the integer lattice onto a discrete countable subset of the $x$-axis, which is clearly homeomorphic to the $\Bbb R^2-\Bbb Z\times\{0\}$.