Homeomorphism $\mathbb{R}^{2}\setminus \mathbb Z^2$ to $\mathbb{R}^{2}\setminus \{ (x,y) \ | \ (x-n)^2+(y-m)^2<\frac{1}{10}, n, m \in\mathbb Z \}$ Show that $\mathbb{R}^{2}\setminus \{(x,y)\, |\,  x \text{ and } y \text{ integers }\}$ is homeomorphic to the space $\mathbb{R}^{2}\setminus \big\{(x,y) \ | \text{ there are integers } $n$, $m$ \text{ such that } (x-n)^2+(y-m)^2\leq\dfrac{1}{10}\big\}$
 A: One approach: 
Let's rewrite the first set as 
$$X_1 = \mathbb{R}^{2}\setminus \left\{ (x,y) \ | \ (x-n)^2+(y-m)^2 = 0, n, m \in\mathbb Z \right\}$$
I assume the second set is
$$X_2 = \mathbb{R}^{2}\setminus \left\{ (x,y) \ | \ (x-n)^2+(y-m)^2 \color{red}{\leq}\frac{1}{10}, n, m \in\mathbb Z \right\}$$
The idea is to open up every 'point' hole at the integer lattice points into circular holes:
Carve up the plane into $1 \times 1$ squares parallel to the axes with each integer lattice point at the centers. For every point in the interior of each square let $(r,\theta)$ be its radial coordinates with respect to the integer lattice point $(n,m)$. Let $r_{max}$ be the sup of the radial distance along the direction $\theta$ within the square. E.g., for $\theta = 0, \pi/2$, $r_{max} = 1$; for $\theta = \pi/4, 3\pi/4, r_{max} = \sqrt 2$. 
Then map
$$(r,\theta) \mapsto \left( \frac{r_{max} - \frac{1}{\sqrt{10}}}{r_{max}} r + \frac{1}{\sqrt{10}}, \theta \right)$$
Leave any point on the boundary of the squares in place. 
This defines a homeomorphism (in the usual topology) between your two sets, $X_1 \to X_2$.
