Is the plane minus the integer lattice homeomorphic to the plane minus the integers?

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.

• You can probably construct a proof for "yes" by enumerating $\mathbb{Z}^2$ with a spiral, and using this to construct the homeomorphism on successive open squares. – Slade Oct 17 '14 at 22:39
• Or use classification of open surfaces: Both have genus zero and homeomorphic spaces of ends. – Moishe Kohan Oct 17 '14 at 22:42
• @Slade that was one approach that made me think 'yes, they are'. However, how do you strech the spiral onto the $x$-axis? In order to do so would require the spiral to be 'two-sided'; that is, two concentric spirals that meet at the origin. But how do you stretch that onto the $x$-axis without creating some tear between two lattice points that are close together? – Robert Wolfe Oct 17 '14 at 22:42
• @Bryan You don't need to stretch it onto the $x$-axis, just some neighborhood of the $x$-axis. It should probably correspond to the integers $0,1,-1,2,-2,\cdots$. This is messy, of course. – Slade Oct 17 '14 at 22:53
• What is the fundamental group? – Seth Oct 17 '14 at 23:08

1. Open connected surfaces are classified by a theorem of Brown and Messer "The classification of two-dimensional manifolds", Transactions of AMS, 1979. The first invariant is the genus (and genera of ends): In your case genus equals $0$ since both surfaces are obtained by removing compact subsets $E_1, E_2$ from $S^2$. The next invariant is orientability (and orientability of ends): Both surfaces are oriented. The last and the most interesting invariant is the "set of ends" of the surface, which is a certain compact Hausdorff topological space. In the case of surfaces $S^2\setminus E$, the set of ends is $E$ with the subspace topology, provided that $E$ is compact and totally disconnected. In the examples you have, both sets $E_1, E_2$ are 1-point compactifications of countably infinite discrete spaces, hence $E_1$ is homeomorphic to $E_2$. In particular, $S^2 \setminus E_1$ is homeomorphic to $S^2\setminus E_2$.

2. For the fundamental group: It is a corollary of one of Whitehead's theorems that each open connected surface is homotopy equivalent to a bouquet of circles, hence, has free fundamental group. To compute rank of the group, it suffices to look at the 1st homology group, which, in your case has countably infinite rank (say, by the Alexander duality). Hence, $\pi_1(R^2\setminus Z^2)\cong F_{\aleph_0}$, where $\aleph_0$ is the cardinality of the set of natural numbers.

It's easy enough to explicitly construct a homeomorphism for this. All we require is that there is some homeomorphism $f:\mathbb{R^2\rightarrow R^2}$ such that the set $$S=\{x:\exists y\in\mathbb{R}[f(x,y)\in\mathbb{Z^2}]\}$$ has no accumulation points and has that the associated $y$ to each $x$ is unique - that is, we imagine that, for a given $x$, the image of $f(x,y)$ where $y\in\mathbb{R}$ is a curve in $\mathbb{R}^2$. As we vary $x$, this curve sweeps through the space, and, at a series of values $x\in S$, it encounters, on this curve, a single integer point.

We can create such a function by choosing some $g(x)$ such that the set $\{\frac{1}{g(n)}\:n\in \mathbb{N}\}$ contains no pair of elements whose ratio is rational and has no accumulation point. Something like $g(n)=e^{-4n^2-n}$ would suffice, since the exponent is never equal for distinct integers and $e^{x}$ is never rational for rational, non-zero $x$. Then, we define $f(x,y)=xg(y)$ - representing sweeping the plane by curves which are simply scalings of the curve $(y,g(y))$. It is easy enough to see that the set $S$ of $x$ such that the curve $f(x,y)$ passes through a point on the integer lattice has no accumulation points.

Thus, what is remaining is to find a map $p:\mathbb{Z}\rightarrow S$ which preserves order. This clearly exists, since $S$ has no accumulation points, meaning each element has a "next" element, and enumerating elements like so would reach every element. We can extend this (for instance as a piecewise linear function) to a continuous bijection $p':\mathbb{R}\rightarrow \mathbb{R}$ such that $p'|_{\mathbb{Z}}=p$ - that is, it interpolates all the pairs of coordinates $(n,p(n))$. We can also create a continuous function $s$, by similar means, such that, for integer $n$, the function $s(n)$ is the solution for $y$ to $f(p(n),y)\in \mathbb{Z}^2$.

From here, we just note that the following function is a homeomorphism from $\mathbb{R^2 - Z\times \{0\}}$ to $\mathbb{R^2-Z^2}$: $$(x,y)\mapsto f(p'(x),y+s(x))$$ since the image $p'[\mathbb{R}-\mathbb{Z}]=\mathbb{R}-S$ and the image $\mathbb{R}+s(x)=\mathbb{R}$ as well, and the image $f[\mathbb{(R}-S)\mathbb{\times R}]=\mathbb{R}^2-\mathbb{Z}^2$ and all functions involved are continuous.

This generalizes to show that the plane minus any countably large set of points with no accumulation point is homeomorphic.

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 $\Bbb R^2-\Bbb Z\times\{0\}$.